Intermediate algebra [Eleventh edition, Pearson new international edition] 1292022736, 9781292022734, 9781292035932, 1292035935

Table of contents :
Cover......Page 1
Table of Contents......Page 4
1. Review of the Real Number System......Page 8
1. Basic Concepts......Page 9
2. Operations on Real Numbers......Page 21
3. Exponents, Roots, and Order of Operations......Page 31
4. Properties of Real Numbers......Page 39
2. Linear Equations, Inequalities, and Applications......Page 56
1. Linear Equations in One Variable......Page 57
2. Formulas and Percent......Page 65
3. Applications of Linear Equations......Page 76
4. Further Applications of Linear Equations......Page 90
5. Linear Inequalities in One Variable......Page 100
6. Set Operations and Compound Inequalities......Page 112
7. Absolute Value Equations and Inequalities......Page 121
3. Graphs, Linear Equations, and Functions......Page 146
1. The Rectangular Coordinate System......Page 147
2. The Slope of a Line......Page 159
3. Linear Equations in Two Variables......Page 172
4. Linear Inequalities in Two Variables......Page 186
5. Introduction to Relations and Functions......Page 192
6. Function Notation and Linear Functions......Page 201
4. Systems of Linear Equations......Page 222
1. Systems of Linear Equations in Two Variables......Page 223
2. Systems of Linear Equations in Three Variables......Page 239
3. Applications of Systems of Linear Equations......Page 246
4. Solving Systems of Linear Equations by Matrix Methods......Page 260
5. Exponents, Polynomials, and Polynomial Functions......Page 276
1. Integer Exponents and Scientific Notation......Page 277
2. Adding and Subtracting Polynomials......Page 291
3. Polynomial Functions, Graphs, and Composition......Page 297
4. Multiplying Polynomials......Page 306
5. Dividing Polynomials......Page 315
6. Factoring......Page 332
1. Greatest Common Factors and Factoring by Grouping......Page 333
2. Factoring Trinomials......Page 339
3. Special Factoring......Page 346
4. A General Approach to Factoring......Page 352
5. Solving Equations by Factoring......Page 356
7. Rational Expressions and Functions......Page 374
1. Rational Expressions and Functions; Multiplying and Dividing......Page 375
2. Adding and Subtracting Rational Expressions......Page 384
3. Complex Fractions......Page 393
4. Equations with Rational Expressions and Graphs......Page 399
5. Applications of Rational Expressions......Page 409
6. Variation......Page 420
8. Roots, Radicals, and Root Functions......Page 442
1. Radical Expressions and Graphs......Page 443
2. Rational Exponents......Page 450
3. Simplifying Radical Expressions......Page 458
4. Adding and Subtracting Radical Expressions......Page 468
5. Multiplying and Dividing Radical Expressions......Page 473
6. Solving Equations with Radicals......Page 483
7. Complex Numbers......Page 489
9. Quadratic Equations, Inequalities, and Functions......Page 510
1. The Square Root Property and Completing the Square......Page 511
2. The Quadratic Formula......Page 520
3. Equations Quadratic in Form......Page 527
4. Formulas and Further Applications......Page 538
5. Graphs of Quadratic Functions......Page 546
6. More About Parabolas and Their Applications......Page 556
7. Polynomial Rational Inequalities......Page 567
10. Inverse, Exponential, and Logarithmic Functions......Page 588
1. Inverse Functions......Page 589
2. Exponential Functions......Page 597
3. Logarithmic Functions......Page 605
4. Properties of Logarithms......Page 612
5. Common and Natural Logarithms......Page 621
6. Exponential and Logarithmic Equations; Further Applications......Page 630
11. Nonlinear Functions, Conic Sections, and Nonlinear Systems......Page 654
1. Additional Graphs and Functions......Page 655
2. The Circle and the Ellipse......Page 661
3. The Hyperbola and Functions Defined by Radicals......Page 669
4. Nonlinear Systems of Equations......Page 676
5. Second-Degree Inequalities and Systems of Inequalities......Page 683
Appendix: Determinants and Cramer's Rule......Page 698
Appendix: Synthetic Division......Page 708
Glossary......Page 714
Useful Mathematical References......Page 722
12. Student's Solutions Manual to accompany Review of the Real Number System......Page 726
13. Student's Solutions Manual to accompany Linear Equations, Inequalities, and Applications......Page 742
14. Student's Solutions Manual to accompany Graphs, Linear Equations, and Functions......Page 788
15. Student's Solutions Manual to accompany Systems of Linear Equations......Page 826
16. Student's Solutions Manual to accompany Exponents, Polynomials, and Polynomial Functions......Page 868
17. Student's Solutions Manual to accompany Factoring......Page 897
18. Student's Solutions Manual to accompany Rational Expressions and Functions......Page 920
19. Student's Solutions Manual to accompany Roots, Radicals, and Root Functions......Page 964
20. Student's Solutions Manual to accompany Quadratic Equations, Inequalities, and Functions......Page 1002
21. Student's Solutions Manual to accompany Inverse, Exponential, and Logarithmic Functions......Page 1064
22. Student's Solutions Manual to accompany Nonlinear Functions, Conic Sections, and Nonlinear Systems......Page 1094
C......Page 1130
E......Page 1131
G......Page 1132
L......Page 1133
N......Page 1134
P......Page 1135
R......Page 1136
S......Page 1137
Z......Page 1138

Citation preview

Intermediate Algebra Lial Hornsby McGinnis

9 781292 022734

Eleventh Edition

ISBN 978-1-29202-273-4

Intermediate Algebra M. Lial J. Hornsby T. McGinnis Eleventh Edition

Intermediate Algebra M. Lial J. Hornsby T. McGinnis Eleventh Edition

Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoned.co.uk © Pearson Education Limited 2014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS. All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affiliation with or endorsement of this book by such owners.

ISBN 10: 1-292-02273-6 ISBN 13: 978-1-292-02273-4

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Printed in the United States of America

P

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Table of Contents 1. Review of the Real Number System Margaret L. Lial/John Hornsby/Terry McGinnis 1. Basic Concepts

1 2

2. Operations on Real Numbers

14

3. Exponents, Roots, and Order of Operations

24

4. Properties of Real Numbers

32

2. Linear Equations, Inequalities, and Applications Margaret L. Lial/John Hornsby/Terry McGinnis 1. Linear Equations in One Variable

49 50

2. Formulas and Percent

58

3. Applications of Linear Equations

69

4. Further Applications of Linear Equations

83

5. Linear Inequalities in One Variable

93

6. Set Operations and Compound Inequalities

105

7. Absolute Value Equations and Inequalities

114

3. Graphs, Linear Equations, and Functions Margaret L. Lial/John Hornsby/Terry McGinnis 1. The Rectangular Coordinate System

139 140

2. The Slope of a Line

152

3. Linear Equations in Two Variables

165

4. Linear Inequalities in Two Variables

179

5. Introduction to Relations and Functions

185

6. Function Notation and Linear Functions

194

4. Systems of Linear Equations Margaret L. Lial/John Hornsby/Terry McGinnis

215

I

1. Systems of Linear Equations in Two Variables

216

2. Systems of Linear Equations in Three Variables

232

3. Applications of Systems of Linear Equations

239

4. Solving Systems of Linear Equations by Matrix Methods

253

5. Exponents, Polynomials, and Polynomial Functions Margaret L. Lial/John Hornsby/Terry McGinnis 1. Integer Exponents and Scientific Notation

269 270

2. Adding and Subtracting Polynomials

284

3. Polynomial Functions, Graphs, and Composition

290

4. Multiplying Polynomials

299

5. Dividing Polynomials

308

6. Factoring Margaret L. Lial/John Hornsby/Terry McGinnis 1. Greatest Common Factors and Factoring by Grouping

325 326

2. Factoring Trinomials

332

3. Special Factoring

339

4. A General Approach to Factoring

345

5. Solving Equations by Factoring

349

7. Rational Expressions and Functions Margaret L. Lial/John Hornsby/Terry McGinnis 1. Rational Expressions and Functions; Multiplying and Dividing

367 368

2. Adding and Subtracting Rational Expressions

377

3. Complex Fractions

386

4. Equations with Rational Expressions and Graphs

392

5. Applications of Rational Expressions

402

6. Variation

413

8. Roots, Radicals, and Root Functions Margaret L. Lial/John Hornsby/Terry McGinnis

II

1. Radical Expressions and Graphs

435 436

2. Rational Exponents

443

3. Simplifying Radical Expressions

451

4. Adding and Subtracting Radical Expressions

461

5. Multiplying and Dividing Radical Expressions

466

6. Solving Equations with Radicals

476

7. Complex Numbers

482

9. Quadratic Equations, Inequalities, and Functions Margaret L. Lial/John Hornsby/Terry McGinnis 1. The Square Root Property and Completing the Square

503 504

2. The Quadratic Formula

513

3. Equations Quadratic in Form

520

4. Formulas and Further Applications

531

5. Graphs of Quadratic Functions

539

6. More About Parabolas and Their Applications

549

7. Polynomial Rational Inequalities

560

10. Inverse, Exponential, and Logarithmic Functions Margaret L. Lial/John Hornsby/Terry McGinnis 1. Inverse Functions

581 582

2. Exponential Functions

590

3. Logarithmic Functions

598

4. Properties of Logarithms

605

5. Common and Natural Logarithms

614

6. Exponential and Logarithmic Equations; Further Applications

623

11. Nonlinear Functions, Conic Sections, and Nonlinear Systems Margaret L. Lial/John Hornsby/Terry McGinnis 1. Additional Graphs and Functions

647 648

2. The Circle and the Ellipse

654

3. The Hyperbola and Functions Defined by Radicals

662

4. Nonlinear Systems of Equations

669

5. Second-Degree Inequalities and Systems of Inequalities

676

Appendix: Determinants and Cramer's Rule Margaret L. Lial/John Hornsby/Terry McGinnis

691

Appendix: Synthetic Division Margaret L. Lial/John Hornsby/Terry McGinnis

701

Glossary Margaret L. Lial/John Hornsby/Terry McGinnis

707

Useful Mathematical References Margaret L. Lial/John Hornsby/Terry McGinnis

715

12. Student's Solutions Manual to accompany Review of the Real Number System Margaret L. Lial/John Hornsby/Terry McGinnis

719

III

13. Student's Solutions Manual to accompany Linear Equations, Inequalities, and Applications Margaret L. Lial/John Hornsby/Terry McGinnis

735

14. Student's Solutions Manual to accompany Graphs, Linear Equations, and Functions Margaret L. Lial/John Hornsby/Terry McGinnis

781

15. Student's Solutions Manual to accompany Systems of Linear Equations Margaret L. Lial/John Hornsby/Terry McGinnis

819

16. Student's Solutions Manual to accompany Exponents, Polynomials, and Polynomial Functions Margaret L. Lial/John Hornsby/Terry McGinnis

861

17. Student's Solutions Manual to accompany Factoring Margaret L. Lial/John Hornsby/Terry McGinnis

889

18. Student's Solutions Manual to accompany Rational Expressions and Functions Margaret L. Lial/John Hornsby/Terry McGinnis

913

19. Student's Solutions Manual to accompany Roots, Radicals, and Root Functions Margaret L. Lial/John Hornsby/Terry McGinnis

957

20. Student's Solutions Manual to accompany Quadratic Equations, Inequalities, and Functions Margaret L. Lial/John Hornsby/Terry McGinnis

995

21. Student's Solutions Manual to accompany Inverse, Exponential, and Logarithmic Functions Margaret L. Lial/John Hornsby/Terry McGinnis

1057

22. Student's Solutions Manual to accompany Nonlinear Functions, Conic Sections, and Nonlinear Systems

IV

Margaret L. Lial/John Hornsby/Terry McGinnis

1087

Index

1123

Review of the Real Number System Basic Concepts

2

Operations on Real Numbers

3

Exponents, Roots, and Order of Operations

4

Properties of Real Numbers

Kati Molin/Shutterstock

1

Americans love their pets. Over 71 million U.S. households owned pets in 2008. Combined, these households spent more than $44 billion pampering their animal friends. The fastest-growing segment of the pet industry is the high-end luxury area, which includes everything from gourmet pet foods, designer toys, and specialty furniture to groomers, dog walkers, boarding in posh pet hotels, and even pet therapists. (Source: American Pet Products Manufacturers Association.) In Exercise 101 of Section 3, we use an algebraic expression, one of the topics of this chapter, to determine how much Americans have spent annually on their pets in recent years. From Chapter 1 of Intermediate Algebra, Eleventh Edition, Margaret L. Lial, John Hornsby and Terry McGinnis. Copyright © 2012 by Pearson Education, Inc. All rights reserved.

1

Review of the Real Number System

1

Basic Concepts

OBJECTIVES 1 2 3 4 5 6 7

Write sets using set notation. Use number lines. Know the common sets of numbers. Find additive inverses. Use absolute value. Use inequality symbols. Graph sets of real numbers.

OBJECTIVE 1 Write sets using set notation. A set is a collection of objects called the elements or members of the set. In algebra, the elements of a set are usually numbers. Set braces, { }, are used to enclose the elements. For example, 2 is an element of the set 51, 2, 36. Since we can count the number of elements in the set 51, 2, 36, it is a finite set. In our study of algebra, we refer to certain sets of numbers by name. The set

N ⴝ 51, 2, 3, 4, 5, 6,

Á6

Natural (counting) numbers

is called the natural numbers, or the counting numbers. The three dots (ellipsis points) show that the list continues in the same pattern indefinitely. We cannot list all of the elements of the set of natural numbers, so it is an infinite set. Including 0 with the set of natural numbers gives the set of whole numbers. W ⴝ 50, 1, 2, 3, 4, 5, 6,

Á6

Whole numbers

The set containing no elements, such as the set of whole numbers less than 0, is called the empty set, or null set, usually written 0 or { }. CAUTION Do not write 506 for the empty set. 506 is a set with one element: 0.

Use the notation 0 or { } for the empty set.

To write the fact that 2 is an element of the set 51, 2, 36, we use the symbol 僆 (read “is an element of ”). 2 僆 51, 2, 36

The number 2 is also an element of the set of natural numbers N. 2僆N To show that 0 is not an element of set N, we draw a slash through the symbol 僆. 0僆N Two sets are equal if they contain exactly the same elements. For example, 51, 26 = 52, 16. (Order doesn’t matter.) However, 51, 26 Z 50, 1, 26 ( Z means “is not equal to”), since one set contains the element 0 while the other does not. In algebra, letters called variables are often used to represent numbers or to define sets of numbers. For example, 5x | x is a natural number between 3 and 156

(read “the set of all elements x such that x is a natural number between 3 and 15”) defines the set 54, 5, 6, 7, Á , 146.

The notation 5x | x is a natural number between 3 and 156 is an example of setbuilder notation. 5x | x has property P6 ⎧ ⎪ ⎨ ⎪ ⎩

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

⎧ ⎪ ⎨ ⎪ ⎩

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

2

the set of

all elements x

such that

x has a given property P

Review of the Real Number System

NOW TRY EXERCISE 1

List the elements in

5 p | p is a natural number less than 66.

EXAMPLE 1

Listing the Elements in Sets

List the elements in each set.

(a) 5x | x is a natural number less than 46 The natural numbers less than 4 are 1, 2, and 3. This set is 51, 2, 36.

(b) 5x | x is one of the first five even natural numbers6 is 52, 4, 6, 8, 106.

(c) 5x | x is a natural number greater than or equal to 76 The set of natural numbers greater than or equal to 7 is an infinite set, written with ellipsis points as 57, 8, 9, 10, Á 6.

NOW TRY EXERCISE 2

Use set-builder notation to describe the set. 59, 10, 11, 126

EXAMPLE 2

NOW TRY

Using Set-Builder Notation to Describe Sets

Use set-builder notation to describe each set.

(a) 51, 3, 5, 7, 96 There are often several ways to describe a set in set-builder notation. One way to describe the given set is 5x | x is one of the first five odd natural numbers6.

(b) 55, 10, 15, Á 6 This set can be described as 5x | x is a multiple of 5 greater than 06.

NOW TRY

Use number lines. A good way to get a picture of a set of numbers is to use a number line. See FIGURE 1 . OBJECTIVE 2

To draw a number line, choose any point on the line and label it 0. Then choose any point to the right of 0 and label it 1. Use the distance between 0 and 1 as the scale to locate, and then label, other points.

The number 0 is neither positive nor negative. Negative numbers

–5

–4

–3

–2

Positive numbers

–1

0

1

2

3

4

5

FIGURE 1

The set of numbers identified on the number line in FIGURE 1 , including positive and negative numbers and 0, is part of the set of integers. I ⴝ 5 Á , ⴚ3, ⴚ2, ⴚ1, 0, 1, 2, 3,

Á6

Integers

Each number on a number line is called the coordinate of the point that it labels, while the point is the graph of the number. FIGURE 2 shows a number line with several points graphed on it. Graph of –1

–1

3 4

2

–3 –2 –1

NOW TRY ANSWERS 1. 51, 2, 3, 4, 56

2. 5x | x is a natural number between 8 and 136

0

1

2

3

Coordinate FIGURE 2

3

Review of the Real Number System

The fractions - 12 and 43 , graphed on the number line in FIGURE 2 , are rational numbers. A rational number can be expressed as the quotient of two integers, with denominator not 0. The set of all rational numbers is written as follows. e

p ` p and q are integers, q ⴝ 0 f q

Rational numbers

The set of rational numbers includes the natural numbers, whole numbers, and integers, since these numbers can be written as fractions. For example, 14 =

14 , 1

-3 , 1

-3 =

0 =

and

0 . 1

A rational number written as a fraction, such as 18 or 32, can also be expressed as a decimal by dividing the numerator by the denominator. 0.666 Á 3冄2.000 Á 18 20 18 20 18 2 2 = 0.6 3

Terminating decimal 0.125 (rational number) 8冄1.000 8 20 16 40 40 Remainder is 0. 0 1 = 0.125 8

Repeating decimal (rational number)

Remainder is never 0. A bar is written over the repeating digit(s).

Thus, terminating decimals, such as 0.125 = 18, 0.8 = 45, and 2.75 = 11 4 , and repeating 2 3 decimals, such as 0.6 = 3 and 0.27 = 11, are rational numbers. Decimal numbers that neither terminate nor repeat, which include many square roots, are irrational numbers. d

=C d ␲ is approximately 3.141592653. . . . FIGURE 3

22 = 1.414213562 Á

and

- 27 = - 2.6457513 Á

NOTE Some square roots, such as 216 = 4 and

9

225

Irrational numbers

= 35 , are rational.

Another irrational number is p, the ratio of the circumference of a circle to its diameter. See FIGURE 3 . Some rational and irrational numbers are graphed on the number line in FIGURE 4 . The rational numbers together with the irrational numbers make up the set of real numbers. Every point on a number line corresponds to a real number, and every real number corresponds to a point on the number line.

Real numbers Irrational numbers –4 Rational numbers

√2

–√7 –3

–2

–1

0 0.27

3 5

FIGURE 4

4

1

␲ 2

3 2.75

4

√16

Review of the Real Number System

Know the common sets of numbers.

OBJECTIVE 3

Sets of Numbers

Natural numbers, or counting numbers Whole numbers

51, 2, 3, 4, 5, 6,

Á6

50, 1, 2, 3, 4, 5, 6,

Á6

5 Á , ⴚ3, ⴚ2, ⴚ1, 0, 1, 2, 3,

Integers Rational numbers

p Eq

Á6

円 p and q are integers, q ⴝ 0 F

Examples: 14 or 4, 1.3, - 92 or - 4 12 , 16 8 or 2, 兹 9 or 3, 0.6

Irrational numbers

5x円 x is a real number that is not rational6

Examples: 兹3, - 兹2, p

5x円 x is a rational number or an irrational number6*

Real numbers

FIGURE 5 shows the set of real numbers. Every real number is either rational or irrational. Notice that the integers are elements of the set of rational numbers and that the whole numbers and natural numbers are elements of the set of integers. Real numbers

Rational numbers

4 –1 4 9 –0.125 1.5

11 7 0.18

Irrational numbers

2 –3 5 4

–

8 15 23

Integers ..., –3, –2, –1

π π 4

Whole numbers 0 Natural numbers 1, 2, 3, ...

FIGURE 5

NOW TRY EXERCISE 3

List the numbers in the following set that are elements of each set.

E - 2.4, - 兹1, - 12 , 0, 0.3, 兹5, p, 5 F

(a) Whole numbers (b) Rational numbers NOW TRY ANSWERS 3. (a) 50, 56

(b) E - 2.4, - 兹 1, - 12 , 0, 0.3, 5 F

EXAMPLE 3

Identifying Examples of Number Sets

List the numbers in the following set that are elements of each set. e - 8, - 兹5, -

9 1 , 0, 0.5, , 1.12, 兹3, 2, p f 64 3

(a) Integers - 8, 0, and 2

(b) Rational numbers 9 , 0, 0.5, 13 , 1.12, and 2 - 8, - 64

(c) Irrational numbers - 兹 5, 兹 3, and p

(d) Real numbers All are real numbers.

NOW TRY

*An example of a number that is not real is 兹 - 1. This number is part of the complex number system.

5

Review of the Real Number System

NOW TRY EXERCISE 4

EXAMPLE 4

Decide whether each statement is true or false. If it is false, tell why. (a) All integers are irrational numbers. (b) Every whole number is an integer.

Determining Relationships Between Sets of Numbers

Decide whether each statement is true or false. (a) All irrational numbers are real numbers. This is true. As shown in FIGURE 5 , the set of real numbers includes all irrational numbers. (b) Every rational number is an integer. This statement is false. Although some rational numbers are integers, other NOW TRY rational numbers, such as 23 and - 14 , are not. Find additive inverses. Look at For each positive number, there is a negative number on the opposite side of 0 that lies the same distance from 0. These pairs of numbers are called additive inverses, opposites, or negatives of each other. For example, 3 and - 3 are additive inverses. OBJECTIVE 4

FIGURE 6 .

–3 –2 –1

0

1

2

3

Additive inverses (opposites) FIGURE 6

Additive Inverse

For any real number a, the number - a is the additive inverse of a. We change the sign of a number to find its additive inverse. As we shall see later, the sum of a number and its additive inverse is always 0. Uses of the Symbol ⴚ

The symbol “ - ” is used to indicate any of the following: 1. a negative number, such as - 9 or - 15; 2. the additive inverse of a number, as in “ - 4 is the additive inverse of 4”; 3. subtraction, as in 12 - 3. In the expression - 1 - 52, the symbol “ - ” is being used in two ways. The first indicates the additive inverse (or opposite) of - 5, and the second indicates a negative number, - 5. Since the additive inverse of - 5 is 5, it follows that - 1- 52 = 5.

Number

Additive Inverse

6

-6

-4

4

2 3

- 23

- 8.7

8.7

0

0

The number 0 is its own additive inverse.

NOW TRY ANSWERS 4. (a) false; All integers are rational numbers. (b) true

6

ⴚ1ⴚa2

For any real number a,

ⴚ1ⴚa2 ⴝ a.

Numbers written with positive or negative signs, such as +4, +8, - 9, and - 5, are called signed numbers. A positive number can be called a signed number even though the positive sign is usually left off. The table in the margin shows the additive inverses of several signed numbers. Use absolute value. Geometrically, the absolute value of a number a, written | a |, is the distance on the number line from 0 to a. For example, the absolute value of 5 is the same as the absolute value of - 5 because each number lies five units from 0. See FIGURE 7 on the next page. OBJECTIVE 5

Review of the Real Number System

Distance is 5, so ⏐–5⏐ = 5.

Distance is 5, so ⏐5⏐ = 5.

–5

5

0 FIGURE 7

CAUTION Because absolute value represents distance, and distance is never

negative, the absolute value of a number is always positive or 0. The formal definition of absolute value follows. Absolute Value

For any real number a,

円 a円 ⴝ e

a if a is positive or 0 ⴚa if a is negative.

The second part of this definition, | a | = - a if a is negative, requires careful thought. If a is a negative number, then - a, the additive inverse or opposite of a, is a positive number. Thus, | a | is positive. For example, if a = - 3, then

| a | = | - 3 | = - 1- 32 = 3. NOW TRY EXERCISE 5

Simplify by finding each absolute value. (a) | - 7 | (b) - | - 15 | (c) | 4 | - | - 4 |

EXAMPLE 5

| a | = - a if a is negative.

Finding Absolute Value

Simplify by finding each absolute value. (a) | 13 | = 13

(b) | - 2 | = - 1- 22 = 2

(c) | 0 | = 0

(d) | - 0.75 | = 0.75

(e) - | 8 | = - 182 = - 8

(f ) - | - 8 | = - 182 = - 8

Evaluate the absolute value. Then find the additive inverse. Work as in part (e); | - 8 | = 8.

(g) | - 2 | + | 5 | = 2 + 5 = 7

Evaluate each absolute value, and then add.

(h) - | 5 - 2 | = - | 3 | = - 3

Subtract inside the bars first.

EXAMPLE 6

NOW TRY

Comparing Rates of Change in Industries

The projected total rates of change in employment (in percent) in some of the fastestgrowing and in some of the most rapidly declining occupations from 2006 through 2016 are shown in the table.

Customer service representatives

NOW TRY ANSWERS 5. (a) 7 (b) - 15 (c) 0

Total Rate of Change (in percent) 24.8

Home health aides

48.7

Security guards

16.9

Word processors and typists

- 11.6

File clerks

- 41.3

Sewing machine operators

- 27.2

Source: Bureau of Labor Statistics.

Iofoto/Shutterstock

Occupation (2006–2016)

7

Review of the Real Number System

NOW TRY EXERCISE 6

Refer to the table in Example 6 on the preceding page. Of the security guards, file clerks, and customer service representatives, which occupation is expected to see the least change (without regard to sign)?

What occupation in the table on the preceding page is expected to see the greatest change? The least change? We want the greatest change, without regard to whether the change is an increase or a decrease. Look for the number in the table with the greatest absolute value. That number is for home health aides, since | 48.7 | = 48.7. Similarly, the least change is for word processors and typists: | - 11.6 | = 11.6. NOW TRY Use inequality symbols. The statement

OBJECTIVE 6

4 + 2 = 6 is an equation—a statement that two quantities are equal. The statement 4 Z 6

(read “4 is not equal to 6”)

is an inequality—a statement that two quantities are not equal. If two numbers are not equal, one must be less than the other. When reading from left to right, the symbol 6 means “is less than.” 8 6 9,

- 6 6 15,

- 6 6 - 1, and

0 6

4 3

All are true.

Reading from left to right, the symbol 7 means “is greater than.” 12 7 5,

9 7 - 2,

6 7 0 5

- 4 7 - 6, and

All are true.

In each case, the symbol “points” toward the lesser number. The number line in FIGURE 8 shows the graphs of the numbers 4 and 9. We know that 4 6 9. On the graph, 4 is to the left of 9. The lesser of two numbers is always to the left of the other on a number line. 4 is greater than » is greater than or equal to

infinity ⴚˆ negative infinity 1ⴚˆ, ˆ2 set of all real numbers 1a, ˆ2 the interval 5x | x 7 a6 1ⴚˆ, a2 the interval 5x | x 6 a6

ˆ

1a, b4 the interval 5x | a 6 x … b6 m factors of a am radical symbol 兹 兹a positive (or principal) square root of a

TEST YOUR WORD POWER See how well you have learned the vocabulary in this chapter. 1. The empty set is a set A. with 0 as its only element B. with an infinite number of elements C. with no elements D. of ideas. 2. A variable is A. a symbol used to represent an unknown number B. a value that makes an equation true C. a solution of an equation D. the answer in a division problem. 3. The absolute value of a number is A. the graph of the number B. the reciprocal of the number C. the opposite of the number D. the distance between 0 and the number on a number line.

4. The reciprocal of a nonzero number a is A. a B. 1a C. - a D. 1. 5. A factor is A. the answer in an addition problem B. the answer in a multiplication problem C. one of two or more numbers that are added to get another number D. any number that divides evenly into a given number. 6. An exponential expression is A. a number that is a repeated factor in a product B. a number or a variable written with an exponent C. a number that shows how many times a factor is repeated in a product

D. an expression that involves addition. 7. A term is A. a numerical factor B. a number or a product of a number and one or more variables raised to powers C. one of several variables with the same exponents D. a sum of numbers and variables raised to powers. 8. A numerical coefficient is A. the numerical factor in a term B. the number of terms in an expression C. a variable raised to a power D. the variable factor in a term.

39

Review of the Real Number System

ANSWERS (to Test Your Word Power) 1. C; Example: The set of whole numbers less than 0 is the empty set, written 0. 2. A; Examples: a, b, c 3. D; Examples: | 2 | = 2 and | - 2 | = 2 4. B; Examples: 3 is the reciprocal of 13 ; - 52 is the reciprocal of - 25 . 5. D; Example: 2 and 5 are factors of 10, since both divide evenly (without remainder) into 10. 6. B; Examples: 34 and x 10 7. B; Examples: 6, 2x , - 4ab 2 8. A; Examples: The term 8z has numerical coefficient 8, and - 10x 3y has numerical coefficient - 10.

QUICK REVIEW CONCEPTS

1

EXAMPLES

Basic Concepts

Sets of Numbers Natural Numbers

51, 2, 3, 4,

Á6 Whole Numbers 50, 1, 2, 3, 4, Á 6 Integers 5 Á , ⴚ2, ⴚ1, 0, 1, 2, Á 6 Rational Numbers

E q | p and q are integers, q ⴝ 0 F p

10, 25, 143

Natural Numbers

0, 8, 47

Whole Numbers

- 22, - 7, 0, 4, 9

Integers

2 15 - , - 0.14, 0, , 6, 0.33333 Á , 兹4 3 8

Rational Numbers

- 兹22, 兹3, p

Irrational Numbers

(all terminating or repeating decimals) Irrational Numbers

5x|x is a real number that is not rational6

(all nonterminating, nonrepeating decimals) Real Numbers

5x|x is a rational or an irrational number6

Absolute Value 円a円 ⴝ e

2

a ⴚa

2 - 3, - , 0.7, p, 兹11 7

Real Numbers

| 12 | = 12

if a is positive or 0 if a is negative

| - 12 | = 12

Operations on Real Numbers

Addition Same Sign: Add the absolute values. The sum has the same sign as the given numbers.

- 2 + 1- 72 = - 12 + 72 = - 9

Different Signs: Find the absolute values of the numbers, and subtract the lesser absolute value from the greater. The sum has the same sign as the number with the greater absolute value. Subtraction For all real numbers a and b,

a ⴚ b ⴝ a ⴙ 1ⴚb2.

Multiplication and Division Same Sign: The answer is positive when multiplying or dividing two numbers with the same sign. Different Signs: The answer is negative when multiplying or dividing two numbers with different signs.

-5 + 8 = 8 - 5 = 3 - 12 + 4 = - 112 - 42 = - 8

- 5 - 1- 32 = - 5 + 3 = - 2 - 15 = 3 -5

- 31 - 82 = 24 - 7152 = - 35

- 24 = -2 12

Division For all real numbers a and b (where b Z 0), aⴜbⴝ

a ⴝa b

#

1 . b

2 5 2 , = 3 6 3

#

6 4 = 5 5

Multiply by the reciprocal of the divisor.

Note (continued)

40

Review of the Real Number System

CONCEPTS

EXAMPLES

3

Exponents, Roots, and Order of Operations The product of an even number of negative factors is positive. The product of an odd number of negative factors is negative. Order of Operations 1. Work separately above and below any fraction bar. 2. If parentheses, brackets, or absolute value bars are present, start with the innermost set and work outward. 3. Evaluate all exponents, roots, and absolute values. 4. Multiply or divide in order from left to right. 5. Add or subtract in order from left to right.

1- 522 = 1- 521 - 52 = 25

1- 522 is positive:

1- 523 is negative: 1- 523 = 1- 521 - 521 - 52 = - 125 15 3 12 + 3 = = 5 # 2 10 2 1- 6232 2 - 13 + 424 + 3 = 1- 6232 2 - 74 + 3

= 1- 6234 - 74 + 3

= 1- 623- 34 + 3 = 18 + 3 = 21

4 Properties of Real Numbers For real numbers a, b, and c, the following are true. Distributive Property 1b ⴙ c2a ⴝ ba ⴙ ca

a1b ⴙ c2 ⴝ ab ⴙ ac and

1214 + 22 = 12

#

#

4 + 12

2

Identity Properties a ⴙ 0 ⴝ 0 ⴙ a ⴝ a and

a

# 1ⴝ1 # aⴝa

Inverse Properties

a ⴙ 1ⴚa2 ⴝ 0 and ⴚa ⴙ a ⴝ 0 1 1 a ⴝ 1 and aⴝ1 a a

#

#

Associative Properties

a ⴙ 1b ⴙ c2 ⴝ 1a ⴙ b2 ⴙ c and

a1bc2 ⴝ 1ab2c

#

1 = 17.5

- 12 + 12 = 0

1 = 1 5

1 - 1- 32 = 1 3

#

9 + 1- 32 = - 3 + 9

ab ⴝ ba

17.5

5 + 1- 52 = 0 5

Commutative Properties a ⴙ b ⴝ b ⴙ a and

- 32 + 0 = - 32

7 + 15 + 32 = 17 + 52 + 3

61 - 42 = 1- 426 - 416

#

32 = 1- 4

#

623

Multiplication Property of 0 a

# 0ⴝ0

and

0

# aⴝ0

4

#

0 = 0

01- 32 = 0

41

Review of the Real Number System

REVIEW EXERCISES 1

Graph the elements of each set on a number line.

9 1. e - 4, - 1, 2, , 4 f 4

2. e - 5, -

11 13 , - 0.5, 0, 3, f 4 3

Find the value of each expression. 3. | - 16 |

4. - | - 8 |

5. | - 8 | - | - 3 |

Let S = E - 9, - 43 , - 兹4, - 0.25, 0, 0.35, 53 , 兹7, 兹 - 9, 12 3 F . Simplify the elements of S as necessary, and then list those elements of S which belong to the specified set. 6. Whole numbers

7. Integers

8. Rational numbers

9. Real numbers

Write each set by listing its elements.

10. 5x | x is a natural number between 3 and 96 11. 5 y | y is a whole number less than 46 Write true or false for each inequality.

#

2 … | 12 - 4 |

13. 2 + | - 2 | 7 4

The graph shows the percent change in passenger car production at U.S. plants from 2006 to 2007 for various automakers. Use this graph to work Exercises 15–18. 15. Which automaker had the greatest change in sales? What was that change? 16. Which automaker had the least change in sales? What was that change? 17. True or false: The absolute value of the percent change for Chrysler was greater than the absolute value of the percent change for Hyundai. 18. True or false: The percent change for Subaru was more than twice the percent change for Chrysler.

14. 413 + 72 7 - | 40 |

Car Production, 2007 21.3%

Chrysler

Automakers

12. 4

Ford

–44.1%

GM

–18.8%

Honda

–2.8%

Hyundai –23.8% 45.9%

Subaru 2.14%

Toyota – 50

– 25

0

25

50

Percent Change from 2006 Source: World Almanac and Book of Facts.

Write each set in interval notation and graph the interval. 19. 5x | x 6 - 56

2

Add or subtract as indicated.

21. -

5 7 - a- b 8 3

23. - 5 + 1 - 112 + 20 - 7

25. - 15 + 1- 132 + 1- 112 27.

20. 5x | - 2 6 x … 36

3 1 9 - a b 4 2 10

22. -

4 3 - a- b 5 10

24. - 9.42 + 1.83 - 7.6 - 1.8 26. - 1 - 3 - 1- 102 + 1- 62

28. - | - 12 | - | - 9 | + 1- 42 - | 10 |

29. Telescope Peak, altitude 11,049 ft, is next to Death Valley, 282 ft below sea level. Find the difference between these altitudes. (Source: World Almanac and Book of Facts.)

42

Review of the Real Number System

Multiply or divide as indicated. 30. 21 - 521 - 321 - 32 31. 34. Concept Check

3

3 14 a- b 7 9

32.

75 -5

33. 5 7 - 7

Which one of the following is undefined:

- 2.3754 - 0.74

or

7 - 7 ? 5

Evaluate each expression. 3 3 36. a b 7

35. 10 4

37. 1- 523

38. - 53

Find each square root. If it is not a real number, say so. 39. 兹400

40.

64 B 121

41. - 兹0.81

42. 兹 - 49

Simplify each expression. 3 43. - 14 a b + 6 , 3 7

2 44. - 351- 22 + 8 - 434 3

45.

- 51322 + 9 A 兹4 B - 5 6 - 51- 22

Evaluate each expression for k = - 4, m = 2, and n = 16. 46. 4k - 7m

47. - 3兹n + m + 5k

48.

4m 3 - 3n 7k 2 - 10

49. The following expression for body mass index (BMI) can help determine ideal body weight. 704 * 1weight in pounds2 , 1height in inches22

A BMI of 19 to 25 corresponds to a healthy weight. (Source: The Washington Post.) John Hornsby

(a) Baseball player Grady Sizemore is 6 ft, 2 in., tall and weighs 200 lb. (Source: www.mlb.com) Find his BMI (to the nearest whole number). (b) Calculate your BMI.

4

Simplify each expression.

50. 2q + 18q

51. 13z - 17z

52. - m + 4m

53. 5p - p

54. - 21k + 32

55. 61r + 32

56. 912m + 3n2

57. - 1- p + 6q2 - 12p - 3q2

58. - 3y + 6 - 5 + 4y

59. 2a + 3 - a - 1 - a - 2

60. - 314m - 22 + 213m - 12 - 413m + 12 Complete each statement so that the indicated property is illustrated. Simplify each answer if possible. 61. 2x + 3x =

62. - 5

#

1 =

(distributive property)

(identity property) 64. - 3 + 13 =

63. 214x2 =

(commutative property)

(associative property) 65. - 3 + 3 =

66. 61x + z2 = (inverse property)

(distributive property) 68. 4

67. 0 + 7 = (identity property)

#

1 = 4

(inverse property)

43

Review of the Real Number System

MIXED REVIEW EXERCISES* The table gives U.S. exports and imports with Spain, in millions of U.S. dollars. Year

Exports

Imports

2007

9766

10,498

2008

12,190

11,094

2009

7294

6495

Source: U.S. Census Bureau.

Determine the absolute value of the difference between imports and exports for each year. Is the balance of trade (exports minus imports) in each year positive or negative? 69. 2007

70. 2008

71. 2009

Perform the indicated operations. 4 4 72. a - b 5

5 73. - 1- 402 8

4 74. - 25 a- b + 33 - 32 , 兹4 5

75. - 8 + | - 14 | + | - 3 |

6 # 兹4 - 3 # 兹16 - 2 # 5 + 71- 32 - 10

10 5 , a- b 21 14

77. - 兹25

78. -

79. 0.8 - 4.9 - 3.2 + 1.14

80. - 32

81.

82. - 21k - 12 + 3k - k

83. - 兹 - 100

84. - 13k - 6h2

76.

- 38 - 19

85. - 4.612.482

2 86. - 1- 152 + 12 4 - 8 , 42 3

87. - 2x + 5 - 4x - 1

88. -

2 1 5 - a - b 3 6 9

89. Evaluate - m13k 2 + 5m2 for (a) k = - 4 and m = 2 and (b) k =

1 2

and m = - 34 .

90. Concept Check To evaluate 13 + 522, should you work within the parentheses first, or should you square 3 and square 5 and then add? *The order of exercises in this final group does not correspond to the order in which topics occur in the chapter. This random ordering should help you prepare for the chapter test in yet another way.

TEST

CHAPTER

VIDEOS

Step-by-step test solutions are found on the Chapter Test Prep Videos available in or on (search “LialIntermediateAlg”).

5 1. Graph e - 3, 0.75, , 5, 6.3 f on a number line. 3 Let A = E - 兹6, - 1, - 0.5, 0, 3, 兹25, 7.5, 24 2 , 兹 - 4 F . Simplify the elements of A as necessary, and then list those elements of A which belong to the specified set.

44

2. Whole numbers

3. Integers

4. Rational numbers

5. Real numbers

Review of the Real Number System

Write each set in interval notation and graph the interval. 6. 5x | x 6 - 36

7. 5x | - 4 6 x … 26

Perform the indicated operations. 8. - 6 + 14 + 1- 112 - 1 - 32

9. 10 - 4

10. 7 - 42 + 2162 + 1- 422 12.

11.

- 233 - 1- 1 - 22 + 24

13.

兹91 - 32 - 1- 22

The table shows the heights in feet of some selected mountains and the depths in feet (as negative numbers) of some selected ocean trenches. 14. What is the difference between the height of Mt. Foraker and the depth of the Philippine Trench?

#

3 + 61 - 42

10 - 24 + 1- 62 兹161- 52

8

#

4 - 32 -3

#

#

5 - 21- 12

23

+ 1

Mountain

Height

Trench

Depth

Foraker

17,400

Philippine

- 32,995

Wilson

14,246

Cayman

- 24,721

Pikes Peak

14,110

Java

- 23,376

Source: World Almanac and Book of Facts.

15. What is the difference between the height of Pikes Peak and the depth of the Java Trench? 16. How much deeper is the Cayman Trench than the Java Trench? Find each square root. If the number is not real, say so. 17. 兹196

18. - 兹225

19. 兹 - 16

20. Concept Check For the expression 1a, under what conditions will its value be each of the following? (b) not real

(a) positive 21. Evaluate

2m 2

8k + r - 2

(c) 0

for k = - 3, m = - 3, and r = 25.

22. Simplify - 312k - 42 + 413k - 52 - 2 + 4k. 23. How does the subtraction sign affect the terms - 4r and 6 when 13r + 82 - 1- 4r + 62 is simplified? What is the simplified form? Match each statement in Column I with the appropriate property in Column II. Answers may be used more than once. 24. 6 + 1 - 62 = 0

I

25. - 2 + 13 + 62 = 1- 2 + 32 + 6 26. 5x + 15x = 15 + 152x 27. 13

#

0 = 0

28. - 9 + 0 = - 9 29. 4

#

1 = 4

II A. Distributive property B. Inverse property C. Identity property D. Associative property E. Commutative property F. Multiplication property of 0

30. 1a + b2 + c = 1b + a2 + c

45

Review of the Real Number System

Answers to Selected Exercises In this section, we provide the answers that we think most students will obtain when they work the exercises using the methods explained in the text. If your answer does not look exactly like the one given here, it is not necessarily wrong. In many cases, there are equivalent forms of the answer that are correct. For example, if the answer section shows 34

and your answer

5. greater; 15 + 1 - 22 = 13 7. the number with lesser absolute value is subtracted from the one with greater absolute value;

- 15 - 1 - 32 = - 12 9. negative; - 51152 = - 75 11. - 19 15. - 19 12

13. 9

25. - 10.18

is 0.75, you have obtained the right answer, but written it in a different (yet

39. - 4

equivalent) form. Unless the directions specify otherwise, 0.75 is just as

53.

valid an answer as

3 4.

13 2,

17. - 1.85 67 30

27.

41. 8

or 6 12

29. 14

19. - 11

21. 21

31. - 5

33. - 6

45. - 74

43. 3.018

47. - 78

23. - 13 35. - 11

49. 1

37. 16

51. 6

55. It is true for multiplication (and division). It is false

for addition and subtraction when the number to be subtracted has the

In general, if your answer does not agree with the one given in the

lesser absolute value. A more precise statement is, “The product or quo-

text, see whether it can be transformed into the other form. If it can, then it

tient of two negative numbers is positive.” 57. - 35 59. 40

is the correct answer. If you still have doubts, talk with your instructor.

63. - 12

65.

77. - 4

79. 0

Section 1 1. 51, 2, 3, 4, 56

3. 55, 6, 7, 8, Á 6

7. 510, 12, 14, 16, Á 6

89. 10,000

5. 5 Á , - 1, 0, 1, 2, 3, 46

103. - 35 27

11. 5- 4, 46

9. 0

6 5

67. 1 17 18 - 49

105.

121. (a) $466.02

15. 5x | x is a multiple of 4 greater than 06

(b) $262 thousand

–6 –1 5

17.

–1

21. (a) 8, 13, 75 5 (or 15)

13 4

0

1

2

(b) 0, 8, 13, 75 5

75 (d) - 9, - 0.7, 0, 67, 4.6, 8, 21 2 , 13, 5

87. - 2.1

2 99. - 15

101.

3 5

107. - 12.351 109. - 15.876 111. - 4.14 117. 112°F

(b) $190.68

119. $30.13

123. (a) - $475 thousand

(c) - $83 thousand

125. (a) 2000: $129 billion; (b) The

11

cost of Social Security will exceed revenue in 2030 by $501 billion.

5.2 2

3

4

5

Section 3

(c) - 9, 0, 8, 13, 75 5

(e) - 兹6, 兹7 (f ) All are real

1. false; - 76 = - 1762 3. true 5. true 7. true 9. false; The base

numbers. 23. yes 25. false; Some are whole numbers, but negative

is 8.

integers are not. 27. false; No irrational number is an integer.

17. 1- 923

29. true 31. true 33. true 35. (a) A

(b) A

37. (a) - 6 (b) 6

41. (a) - 65

3 2

39. (a) 12

(b) 12

(c) B (d) B

47. - 5 49. - 2 51. - 4.5 53. 5 55. 6

(b)

6 5

43. 8

57. 0

The population decreased by 0.6%.

61. Pacific Ocean, Indian Ocean,

Caribbean Sea, South China Sea, Gulf of California 63. true 65. true 67. false 69. true 71. true 73. 2 6 6 75. 4 7 - 9

97. - 3 Ú - 3; true 99. - 8 7 - 6; false

105. 10, 3.52 109. 1 - 4, 34

103. 1- q , 64

0

0

–4

107. 32, 74

3.5

0

3

0

111. 10, 34

10 11

43.

23. 0.021952 25.

33. - 729

45. - 0.7

63. - 48

73. - 2

75. undefined 77. - 7

85.

87. 8

65. - 2 89.

99. 0.035

5 - 16

1 125

27.

5

256 625

37. 9 39. 13

47. not a real number 49. (a) B

61. - 8 15 - 238

35. - 4096

15. A 34 B

13. 10 4

(d) - 64

51. negative 53. 24 67. - 79

55. 15

57. 55

69. - 10

79. - 1

71. 2

81. 17

91. - 2.75 93.

3 - 16

101. (a) $19.4 billion

59. - 91

83. - 96 95. $1572

(b) $30.8 billion

(d) The amount spent on pets more than doubled

Section 4 0

6

2

7

1. B 3. A 21. 8a

0

3

5. product; 0 7. grouping 9. like

13. - 12x + 12y 15. 8k

115. x 6 y

Section 2 1. additive inverses; 4 + 1- 42 = 0 3. negative; - 7 + 1 - 212 = - 28

11. 2m + 2p

17. - 2r 19. cannot be simplified

23. - 2d + ƒ 25. x + y 27. - 6y + 3 29. p + 11

31. - 2k + 15 33. - 3m + 2 35. - 1 37. 2p + 7 39. - 6z - 39

113. Iowa (IA), Ohio (OH), Pennsylvania (PA)

21. 16

31. 256

41. - 20

(c) 64

from 1996 to 2008.

91. - 3 … 3x 6 4 93. - 6 6 10; true 95. 10 Ú 10; true

–1

29. - 125

(c) $42.3 billion

85. 3t - 4 … 10 87. 5x + 3 Z 0 89. - 3 6 t 6 5

(b) - 64

19. z 7

97. $3296

77. - 10 6 - 5 79. x 7 0 81. 7 7 y 83. 5 Ú 5

101. 1 - 1, q 2

11. (a) 64

(b) C (c) A

59. (a) Las Vegas; The population increased by 35.6%. (b) Detroit;

46

97.

- 22 45

61. 2 75. 6

19. –6 –4 –2 0 2 4 6

45.

95.

9 85. - 13

25 102

- 19 24

73. - 7

2010: $206 billion; 2020: $74 billion; 2030: - $501 billion

5 6

4

93.

17 36

115. 51.495

In Exercises 13 and 15, we give one possible answer.

13. 5x | x is an even natural number less than or equal to 86

71. - 10.676

81. undefined 83.

91.

113. 4800

69. 5.88

41. 15 + 82x = 13x

45. 9y + 5x 47. 7

49. 0

43. 15

#

92r = 45r

51. 81 - 42 + 8x = - 32 + 8x 53. 0

55. Answers will vary. One example of commutativity is washing your face and brushing your teeth. An example of non-commutativity is

Review of the Real Number System putting on your socks and putting on your shoes. 61. 431

57. 1900

59. 75

63. associative property 64. associative property

65. commutative property 66. associative property 67. distributive property 68. arithmetic facts 69. No. One example is

7 + 15

#

–4 –2

0

– 11 4

4

13 3

–0.5 –1 0 1

–5 –3

- 兹 4 (or - 2), - 0.25, 0, 0.35, 53 , 12 3 (or 4) 10. 54, 5, 6, 7, 86

3

24. - 16.99 30. - 90 36.

27 343

0

31.

25. - 39 2 3

3

33.

38. - 125

42. not a real number 43. - 4 47. - 30

48.

8 - 51

51. - 4z 52. 3m

49. (a) 26 53. 4p

8. - 9, - 43,

73. 25

74. 31

16. Toyota; 2.14%

41 24

27.

23 20

0

22. - 12

23. - 3

28. - 35

3.21 34. 7 -5 7 8 39. 20 40. 11 44. 44

(b) - 94

75. 9 76. 0

82. 2

83. not a real

87. - 6x + 4

90. Work within the

Test 5

–2

45. - 2

7.5,

24 2

6. 1 - q, - 32

18. - 15

(b) Answers will vary. 50. 20q

2

4

6

4. - 1, - 0.5, 0, 3, 兹 25 (or 5),

5. All are real numbers except 兹 - 4. –3

9. - 26 10. 19

35. 10,000

54. - 2k - 6 55. 6r + 18

56. 18m + 27n 57. - p - 3q 58. y + 1 59. 0

(or 12)

14. 50,395 ft

46. - 30

0

2. 0, 3, 兹 25 (or 5), 24 2 (or 12)

3. - 1, 0, 3, 兹 25 (or 5), 24 2 (or 12)

29. 11,331 ft 41. - 0.9

6.3

0.75 3

1.

8. 0

26. 0

32. - 15

37. - 125

256 625

79. - 6.16 80. - 9 81. 2

89. (a) - 116

12. true

–5

21. –2

72.

4. - 8

5

11. 50, 1, 2, 36

17. false 18. true 19. 1- q, - 52

68. 1

parentheses first. 3. 16

9. All are real numbers

13. false 14. true 15. Subaru; 45.9%

20. 1 - 2, 34

4 3

78.

5 88. - 18

12 5. 5 6. 0, 12 3 (or 4) 7. - 9, - 兹 4 (or - 2), 0, 3 (or 4)

except 兹 - 9.

42x = 8x

number 84. - 3k + 6h 85. - 11.408 86. 24

2.

2

#

66. 6x + 6z 67. 7

69. 732 million; negative 70. 1096 million; positive 77. - 5

Review Exercises 1.

64. 13 + 1 - 32 = 10 65. 0 71. 799 million; positive

32 = 17 + 5217 + 32, which is false.

9 4

61. 12 + 32x = 5x 62. - 5 63. 12

7. 1- 4, 24

0

11. 1

15. 37,486 ft

12.

16 7

16. 1345 ft

13.

–4

0

2

11 23

17. 14

19. not a real number 20. (a) a must be positive.

6 (b) a must be negative. (c) a must be 0. 21. - 23

22. 10k - 10 23. It changes the sign of each term. The simplified form is 7r + 2. 24. B 29. C

25. D 26. A 27. F 28. C

30. E

60. - 18m

47

48

Linear Equations, Inequalities, and Applications 1

Linear Equations in One Variable

2

Formulas and Percent

3

Applications of Linear Equations

4

Further Applications of Linear Equations

5

Linear Inequalities in One Variable

6

Set Operations and Compound Inequalities

7

Absolute Value Equations and Inequalities

Summary Exercises on Solving Linear and Absolute Value Equations and Inequalities

Kelsey McNeal/ABC/Courtesy/Everett Collection

Summary Exercises on Solving Applied Problems

Despite increasing competition from the Internet and video games, television remains a popular form of entertainment. In 2009, 114.9 million American households owned at least one TV set, and average viewing time for all viewers was almost 34 hours per week. During the 2008–2009 season, favorite prime-time television programs were American Idol and Dancing with the Stars. (Source: Nielsen Media Research.) In Section 2 we discuss the concept of percent—one of the most common everyday applications of mathematics—and use it in Exercises 43–46 to determine additional information about television ownership and viewing in U.S. households.

From Chapter 2 of Intermediate Algebra, Eleventh Edition, Margaret L. Lial, John Hornsby and Terry McGinnis, Copyright © 2012 by Pearson Education, Inc. All rights reserved.

49

Linear Equations, Inequalities, and Applications

Linear Equations in One Variable

1

2

3

4

5

6

Distinguish between expressions and equations. Identify linear equations, and decide whether a number is a solution of a linear equation. Solve linear equations by using the addition and multiplication properties of equality. Solve linear equations by using the distributive property. Solve linear equations with fractions or decimals. Identify conditional equations, contradictions, and identities.

Distinguish between expressions and equations. Equations and inequalities compare algebraic expressions, just as a balance scale compares the weights of two quantities. An equation is a statement that two algebraic expressions are equal. An equation always contains an equals symbol, while an expression does not. OBJECTIVE 1

EXAMPLE 1

Distinguishing between Expressions and Equations

Decide whether each of the following is an expression or an equation. (a) 3x - 7 = 2 (b) 3x - 7 In part (a) we have an equation, because there is an equals symbol. In part (b), there is no equals symbol, so it is an expression. See the diagram below. 3x - 7 = 2 Left side

3x - 7

{

OBJECTIVES

  

1

Right side

Equation (to solve)

Expression (to simplify or evaluate) NOW TRY

OBJECTIVE 2 Identify linear equations, and decide whether a number is a solution of a linear equation. A linear equation in one variable involves only real numbers and one variable raised to the first power.

x + 1 = - 2, x - 3 = 5, and

2k + 5 = 10

Examples of linear equations

Linear Equation in One Variable NOW TRY EXERCISE 1

Decide whether each of the following is an expression or an equation. (a) 2x + 17 - 3x (b) 2x + 17 = 3x

A linear equation in one variable can be written in the form Ax ⴙ B ⴝ C, where A , B, and C are real numbers, with A Z 0.

A linear equation is a first-degree equation, since the greatest power on the variable is 1. Some equations that are not linear (that is, nonlinear) follow. x 2 + 3y = 5,

NOW TRY ANSWERS 1. (a) expression (b) equation

50

8 = - 22, x

and

2x = 6

Examples of nonlinear equations

If the variable in an equation can be replaced by a real number that makes the statement true, then that number is a solution of the equation. For example, 8 is a solution of the equation x - 3 = 5, since replacing x with 8 gives a true statement. An equation is solved by finding its solution set, the set of all solutions. The solution set of the equation x - 3 = 5 is 586.

Linear Equations, Inequalities, and Applications

Equivalent equations are related equations that have the same solution set. To solve an equation, we usually start with the given equation and replace it with a series of simpler equivalent equations. For example, 5x + 2 = 17,

5x = 15, and

x = 3

are all equivalent, since each has the solution set 536.

Equivalent equations

OBJECTIVE 3 Solve linear equations by using the addition and multiplication properties of equality. We use two important properties of equality to produce equivalent equations. Addition and Multiplication Properties of Equality

Addition Property of Equality For all real numbers A, B, and C, the equations AⴝB

AⴙCⴝBⴙC

and

are equivalent.

That is, the same number may be added to each side of an equation without changing the solution set. Multiplication Property of Equality For all real numbers A and B, and for C Z 0, the equations AⴝB

and

AC ⴝ BC

are equivalent.

That is, each side of an equation may be multiplied by the same nonzero number without changing the solution set.

Because subtraction and division are defined in terms of addition and multiplication, respectively, the preceding properties can be extended. The same number may be subtracted from each side of an equation, and each side of an equation may be divided by the same nonzero number, without changing the solution set. EXAMPLE 2

Using the Properties of Equality to Solve a Linear Equation

Solve 4x - 2x - 5 = 4 + 6x + 3. The goal is to isolate x on one side of the equation. 4x - 2x - 5 = 4 + 6x + 3 2x - 5 = 7 + 6x

Combine like terms.

2x - 5 + 5 = 7 + 6x + 5

Add 5 to each side.

2x = 12 + 6x 2x - 6x = 12 + 6x - 6x

Combine like terms. Subtract 6x from each side.

- 4x = 12

Combine like terms.

12 - 4x = -4 -4

Divide each side by - 4.

x = -3 Check by substituting - 3 for x in the original equation.

51

Linear Equations, Inequalities, and Applications

NOW TRY EXERCISE 2

CHECK

Solve. 5x + 11 = 2x - 13 - 3x

4x - 2x - 5 = 4 + 6x + 3 41- 32 - 21- 32 - 5 ⱨ 4 + 61- 32 + 3 - 12 + 6 - 5 ⱨ 4 - 18 + 3

Use parentheses around substituted values to avoid errors.

- 11 = - 11 ✓

Original equation Let x = - 3. Multiply.

This is not the solution.

True

The true statement indicates that 5 - 36 is the solution set.

NOW TRY

CAUTION In Example 2, the equality symbols are aligned in a column. Use only one equality symbol in a horizontal line of work when solving an equation.

Solving a Linear Equation in One Variable

Step 1

Clear fractions or decimals. Eliminate fractions by multiplying each side by the least common denominator. Eliminate decimals by multiplying by a power of 10.

Step 2

Simplify each side separately. Use the distributive property to clear parentheses and combine like terms as needed.

Step 3

Isolate the variable terms on one side. Use the addition property to get all terms with variables on one side of the equation and all numbers on the other.

Step 4

Isolate the variable. Use the multiplication property to get an equation with just the variable (with coefficient 1) on one side.

Step 5

Check. Substitute the proposed solution into the original equation.

Solve linear equations by using the distributive property. In Example 2, we did not use Step 1 or the distributive property in Step 2 as given in the box. Many equations, however, will require one or both of these steps. OBJECTIVE 4

EXAMPLE 3

Using the Distributive Property to Solve a Linear Equation

Solve 21x - 52 + 3x = x + 6. Step 1 Since there are no fractions in this equation, Step 1 does not apply. Step 2 Use the distributive property to simplify and combine like terms on the left. Be sure to distribute over all terms within the parentheses.

21x - 52 + 3x = x + 6 2x + 21- 52 + 3x = x + 6 2x - 10 + 3x = x + 6 5x - 10 = x + 6

Distributive property Multiply. Combine like terms.

Step 3 Next, use the addition property of equality. 5x - 10 + 10 = x + 6 + 10 5x = x + 16 NOW TRY ANSWER 2. 5 - 46

52

5x - x = x + 16 - x 4x = 16

Add 10. Combine like terms. Subtract x. Combine like terms.

Linear Equations, Inequalities, and Applications

NOW TRY EXERCISE 3

Step 4 Use the multiplication property of equality to isolate x on the left. 16 4x = 4 4

Solve. 51x - 42 - 9 = 3 - 21x + 162

Divide by 4.

x = 4 Step 5 Check by substituting 4 for x in the original equation. CHECK Always check your work.

21x - 52 + 3x = x + 6 214 - 52 + 3142 ⱨ 4 + 6 21- 12 + 12 ⱨ 10 10 = 10 ✓

The solution checks, so 546 is the solution set.

Original equation Let x = 4. Simplify. True NOW TRY

OBJECTIVE 5 Solve linear equations with fractions or decimals. When fractions or decimals appear as coefficients in equations, our work can be made easier if we multiply each side of the equation by the least common denominator (LCD) of all the fractions. This is an application of the multiplication property of equality. NOW TRY EXERCISE 4

Solve.

EXAMPLE 4

Solve

x + 7 6

2x + 4 x - 4 + = 5 4 8

+

Solving a Linear Equation with Fractions 2x - 8 2

6a

Step 1 Step 2 6 a

= - 4.

2x - 8 x + 7 + b = 61- 42 6 2

x + 7 2x - 8 b + 6a b = 61- 42 6 2 x + 7 + 312x - 82 = - 24

x + 7 + 312x2 + 31- 82 = - 24 x + 7 + 6x - 24 = - 24 7x - 17 = - 24 7x - 17 + 17 = - 24 + 17

Step 3

Step 4

Eliminate the fractions. Multiply each side by the LCD, 6. Distributive property Multiply. Distributive property Multiply. Combine like terms. Add 17.

7x = - 7

Combine like terms.

-7 7x = 7 7

Divide by 7.

x = -1 x + 7 2x - 8 + = -4 6 2 21- 12 - 8 -1 + 7 ⱨ -4 + 6 2

Step 5 CHECK

6 - 10 ⱨ + -4 6 2 1 - 5 ⱨ -4 NOW TRY ANSWERS 3. 506

4. 5116

-4 = -4 ✓

The solution checks, so the solution set is 5 - 16.

Let x = - 1. Add and subtract in the numerators. Simplify each fraction. True NOW TRY

53

Linear Equations, Inequalities, and Applications

Some equations have decimal coefficients. We can clear these decimals by multiplying by a power of 10, such as 10 1 = 10, 10 2 = 100,

and so on.

This allows us to obtain integer coefficients.

NOW TRY EXERCISE 5

Solve. 0.08x - 0.121x - 42 = 0.031x - 52

EXAMPLE 5

Solving a Linear Equation with Decimals

Solve 0.06x + 0.09115 - x2 = 0.071152. Because each decimal number is given in hundredths, multiply each side of the equation by 100. A number can be multiplied by 100 by moving the decimal point two places to the right. 0.06x + 0.09115 - x2 = 0.071152 0.06x + 0.09115 - x2 = 0.071152 Move decimal points 2 places to the right.

Multiply each term by 100.

6x + 9115 - x2 = 71152 6x + 91152 - 9x = 71152 - 3x + 135 = 105 - 3x + 135 - 135 = 105 - 135

Distributive property Combine like terms and multiply. Subtract 135.

- 3x = - 30

Combine like terms.

- 30 - 3x = -3 -3

Divide by - 3.

x = 10 CHECK

0.06x + 0.09115 - x2 = 0.071152 0.061102 + 0.09115 - 102 ⱨ 0.071152 0.6 + 0.09152 ⱨ 1.05 0.6 + 0.45 ⱨ 1.05 1.05 = 1.05 3

The solution set is 5106.

Let x = 10. Multiply and subtract. Multiply. True NOW TRY

NOTE Because of space limitations, we will not always show the check when solving an equation. To be sure that your solution is correct, you should always check your work.

NOW TRY ANSWER 5. 596

54

OBJECTIVE 6 Identify conditional equations, contradictions, and identities. In Examples 2–5, all of the equations had solution sets containing one element, such as 5106 in Example 5. Some equations, however, have no solutions, while others have an infinite number of solutions. The table on the next page gives the names of these types of equations.

Linear Equations, Inequalities, and Applications

Type of Linear Equation

Number of Solutions

Indication when Solving

Conditional

One

Final line is x = a number. (See Example 6(a).)

Identity

Infinite; solution set

Final line is true, such as 0 = 0. (See Example 6(b).)

{all real numbers} Contradiction

NOW TRY EXERCISE 6

Solve each equation. Decide whether it is a conditional equation, an identity, or a contradiction. (a) 9x - 31x + 42 = 61x - 22 (b) - 312x - 12 - 2x = 3 + x (c) 10x - 21 = 21x - 52 + 8x

EXAMPLE 6

None; solution set 0

Final line is false, such as - 15 = - 20. (See Example 6(c).)

Recognizing Conditional Equations, Identities, and Contradictions

Solve each equation. Decide whether it is a conditional equation, an identity, or a contradiction. (a)

512x + 62 - 2 = 71x + 42 10x + 30 - 2 = 7x + 28

Distributive property

10x + 28 = 7x + 28

Combine like terms.

10x + 28 - 7x - 28 = 7x + 28 - 7x - 28

Subtract 7x. Subtract 28.

3x = 0

Combine like terms.

0 3x = 3 3

Divide by 3.

x = 0 The solution set, 506, has only one element, so 512x + 62 - 2 = 71x + 42 is a conditional equation. (b)

5x - 15 = 51x - 32 5x - 15 = 5x - 15

Distributive property

5x - 15 - 5x + 15 = 5x - 15 - 5x + 15

Subtract 5x. Add 15.

0 = 0

True

The final line, 0 = 0, indicates that the solution set is 5all real numbers6, and the equation 5x - 15 = 51x - 32 is an identity. (The first step yielded 5x - 15 = 5x - 15, which is true for all values of x. We could have identified the equation as an identity at that point.) (c)

5x - 15 = 51x - 42 5x - 15 = 5x - 20 5x - 15 - 5x = 5x - 20 - 5x - 15 = - 20

Distributive property Subtract 5x. False

Since the result, - 15 = - 20, is false, the equation has no solution. The solution set is 0, so the equation 5x - 15 = 51x - 42 is a contradiction. NOW TRY

NOW TRY ANSWERS

6. (a) 5all real numbers6; identity (b) 506; conditional equation (c) 0; contradiction

CAUTION A common error in solving an equation like that in Example 6(a) is to think that the equation has no solution and write the solution set as 0. This equation has one solution, the number 0, so it is a conditional equation with solution set 506.

55

Linear Equations, Inequalities, and Applications

1 EXERCISES 1. Concept Check

Which equations are linear equations in x?

A. 3x + x - 1 = 0

B. 8 = x 2

C. 6x + 2 = 9

D.

1 1 x - = 0 x 2

2. Which of the equations in Exercise 1 are nonlinear equations in x? Explain why. 3. Decide whether 6 is a solution of 31x + 42 = 5x by substituting 6 for x. If it is not a solution, explain why. 4. Use substitution to decide whether - 2 is a solution of 51x + 42 - 31x + 62 = 91x + 12. If it is not a solution, explain why. Decide whether each of the following is an expression or an equation. See Example 1. 5. - 3x + 2 - 4 = x

6. - 3x + 2 - 4 - x = 4

7. 41x + 32 - 21x + 12 - 10

8. 41x + 32 - 21x + 12 + 10

9. - 10x + 12 - 4x = - 3

10. - 10x + 12 - 4x + 3 = 0

Solve each equation, and check your solution. If applicable, tell whether the equation is an identity or a contradiction. See Examples 2, 3, and 6. 11. 7x + 8 = 1

12. 5x - 4 = 21

13. 5x + 2 = 3x - 6

14. 9x + 1 = 7x - 9

15. 7x - 5x + 15 = x + 8

16. 2x + 4 - x = 4x - 5

17. 12w + 15w - 9 + 5 = - 3w + 5 - 9

18. - 4x + 5x - 8 + 4 = 6x - 4

19. 312t - 42 = 20 - 2t

20. 213 - 2x2 = x - 4

21. - 51x + 12 + 3x + 2 = 6x + 4

22. 51x + 32 + 4x - 5 = 4 - 2x

23. - 2x + 5x - 9 = 31x - 42 - 5

24. - 6x + 2x - 11 = - 212x - 32 + 4

25. 21x + 32 = - 41x + 12

26. 41x - 92 = 81x + 32

27. 312x + 12 - 21x - 22 = 5

28. 41x - 22 + 21x + 32 = 6

29. 2x + 31x - 42 = 21x - 32

30. 6x - 315x + 22 = 411 - x2

31. 6x - 413 - 2x2 = 51x - 42 - 10

32. - 2x - 314 - 2x2 = 21x - 32 + 2

33. - 21x + 32 - x - 4 = - 31x + 42 + 2

34. 412x + 72 = 2x + 25 + 312x + 12

37. - 32x - 15x + 224 = 2 + 12x + 72

38. - 36x - 14x + 824 = 9 + 16x + 32

35. 23x - 12x + 42 + 34 = 21x + 12

36. 432x - 13 - x2 + 54 = - 12 + 7x2

39. - 3x + 6 - 51x - 12 = - 5x - 12x - 42 + 5 40. 41x + 22 - 8x - 5 = - 3x + 9 - 21x + 62 41. 732 - 13 + 4x24 - 2x = - 9 + 211 - 15x2

42. 436 - 11 + 2x24 + 10x = 2110 - 3x2 + 8x

43. - 33x - 12x + 524 = - 4 - 3312x - 42 - 3x4 44. 23- 1x - 12 + 44 = 5 + 3- 16x - 72 + 9x4 45. Concept Check

To solve the linear equation 8x 5x = - 13, 3 4

we multiply each side by the least common denominator of all the fractions in the equation. What is this least common denominator?

56

Linear Equations, Inequalities, and Applications

46. Suppose that in solving the equation 1 1 1 x + x = x, 3 2 6 we begin by multiplying each side by 12, rather than the least common denominator, 6. Would we get the correct solution? Explain. 47. Concept Check To solve a linear equation with decimals, we usually begin by multiplying by a power of 10 so that all coefficients are integers. What is the least power of 10 that will accomplish this goal in each equation? (a) 0.05x + 0.121x + 50002 = 940 (Exercise 63) (b) 0.0061x + 22 = 0.007x + 0.009 (Exercise 69) 48. Concept Check following?

The expression 0.06110 - x211002 is equivalent to which of the

A. 0.06 - 0.06x

B. 60 - 6x

C. 6 - 6x

D. 6 - 0.06x

Solve each equation, and check your solution. See Examples 4 and 5. 5 49. - x = 2 9

50.

3 x = -5 11

51.

6 x = -1 5

7 52. - x = 6 8

53.

x x + = 5 2 3

54.

x x - = 1 5 4

55.

3x 5x + = 13 4 2

56.

8x x - = - 13 3 2

57.

x - 10 2 x + = 5 5 3

58.

2x - 3 3 x + = 7 7 3

59.

3x - 1 x + 3 + = 3 4 6

60.

3x + 2 x + 4 = 2 7 5

61.

x + 5 x - 3 4x + 1 = + 3 6 6

63. 0.05x + 0.121x + 50002 = 940

62.

2x + 5 3x + 1 -x + 7 = + 5 2 2

64. 0.09x + 0.131x + 3002 = 61

65. 0.021502 + 0.08x = 0.04150 + x2 66. 0.20114,0002 + 0.14x = 0.18114,000 + x2 67. 0.05x + 0.101200 - x2 = 0.45x 68. 0.08x + 0.121260 - x2 = 0.48x 69. 0.0061x + 22 = 0.007x + 0.009 70. 0.004x + 0.006150 - x2 = 0.0041682

PREVIEW EXERCISES Use the given value(s) to evaluate each expression. 71. 2L + 2W; 73.

1 Bh; 3

75.

5 1F - 322; 9

L = 10,

W = 8

B = 27, h = 8 F = 122

72. rt; 74. prt; 76.

r = 0.15, t = 3 p = 8000, r = 0.06, t = 2

9 C + 32; 5

C = 60

57

Linear Equations, Inequalities, and Applications

STUDY

SKILLS

Tackling Your Homework You are ready to do your homework AFTER you have read the corresponding material and worked through the examples and Now Try exercises. Graphs, Linear Equat ions, and Functions

Homework Tips

6 EXE RCI SES

N Work problems neatly. Use pencil and write legibly, so

Complete solution availa ble on the Video Resources on DVD

1. Concept Check Choos e the correct response: The notation ƒ132 means A. the variable ƒ times 3, or 3ƒ. B. the value of the depen dent variable when the indep endent variable is 3. C. the value of the indep endent variable when the dependent variable is 3. D. ƒ equals 3. 2. Concept Check Give an example of a function from everyday life. (Hint: blanks: depends on Fill in the , so is a function of .) Let ƒ1x2 = - 3x + 4 and g1x2 = - x 2 + 4x + 1. Find the following. See Examples 1–3. 3. ƒ102 4. ƒ1- 32 5. g1- 22 6. g1102 1 7 7. ƒ a b 8. ƒ a b 3 9. g10.52 3 10. g11.52 11. ƒ1 p2 12. g1k2 13. ƒ1- x2 14. g1- x2 15. ƒ1x + 22 16. ƒ1x - 22 17. g1p2 18. g1e2 19. ƒ1x + h2 20. ƒ1x + h2 - ƒ1x2 21. ƒ142 - g142 22. ƒ1102 - g1102 For each function, find (a) ƒ122 and (b) ƒ1- 12. See Examples 4 and 5. 23. ƒ = 51- 2, 22, 1- 1, - 12, 12, - 126 24. ƒ = 51- 1, - 52, 10, 52, 12, - 526 25. ƒ = 51- 1, 32, 14, 72, 10, 62, 12, 226 26. ƒ = 512, 52, 13, 92, 1- 1, 112, 15, 326 27. f 28. f

others can read your work. Skip lines between steps. Clearly separate problems from each other.

N Show all your work. It is tempting to take shortcuts. Include ALL steps.

N Check your work frequently to make sure you are on the right track. It is hard to unlearn a mistake. For all oddnumbered problems, answers are given at the end of the chapter.

N If you have trouble with a problem, refer to the corresponding worked example in the section. The exercise directions will often reference specific examples to review. Pay attention to every line of the worked example to see how to get from step to step.

N If you are having trouble with an even-numbered problem, work the corresponding odd-numbered problem. Check your answer at the end of the chapter, and apply the same steps to work the even-numbered problem.

N Mark any problems you don’t understand. Ask your

–1 2 3 5

29.

10 15 19 27

x

y = ƒ1x2

2 1 0 -1 -2

4 1 0 1 4

31.

30.

y

2

3 4

58

y = ƒ1x2

8 5 2 -1 -4

6 3 0 -3 -6 y

–2

0

2

x

2

0

y = f(x)

33.

y

0

x

2 y = f(x)

34.

y

y = f(x) 2

y = f(x)

x

–2

0

2

x

–2

Formulas and Percent

OBJECTIVES 1

x

32.

Select several homework tips to try this week.

2

1 7 20

2

2

instructor about them.

2 5 –1 3

Solve a formula for a specified variable. Solve applied problems by using formulas. Solve percent problems. Solve problems involving percent increase or decrease.

A mathematical model is an equation or inequality that describes a real situation. Models for many applied problems, called formulas, already exist. A formula is an equation in which variables are used to describe a relationship. For example, the formula for finding the area a of a triangle is a = 12 bh. Here, b is the length of the base and h is the height. See FIGURE 1 .

h b FIGURE 1

Linear Equations, Inequalities, and Applications

OBJECTIVE 1 Solve a formula for a specified variable. The formula I = prt says that interest on a loan or investment equals principal (amount borrowed or invested) times rate (percent) times time at interest (in years). To determine how long it will take for an investment at a stated interest rate to earn a predetermined amount of interest, it would help to first solve the formula for t. This process is called solving for a specified variable or solving a literal equation. When solving for a specified variable, the key is to treat that variable as if it were the only one. Treat all other variables like numbers (constants). The steps used in the following examples are very similar to those used in solving linear equations from Section 1. NOW TRY EXERCISE 1

Solve the formula I = prt for p.

EXAMPLE 1

Solving for a Specified Variable

Solve the formula I = prt for t. We solve this formula for t by treating I, p, and r as constants (having fixed values) and treating t as the only variable. prt = I

1 pr2t = I

1 pr2t I = pr pr t =

Our goal is to isolate t.

Associative property Divide by pr.

I pr

The result is a formula for t, time in years.

NOW TRY

Solving for a Specified Variable

Step 1

If the equation contains fractions, multiply both sides by the LCD to clear the fractions.

Step 2

Transform so that all terms containing the specified variable are on one side of the equation and all terms without that variable are on the other side.

Step 3

Divide each side by the factor that is the coefficient of the specified variable.

EXAMPLE 2

Solving for a Specified Variable

Solve the formula P = 2L + 2W for W. This formula gives the relationship between perimeter of a rectangle, P, length of the rectangle, L, and width of the rectangle, W. See FIGURE 2 . L

W

W

Perimeter, P, distance around a rectangle, is given by P = 2L + 2W.

L FIGURE 2

NOW TRY ANSWER 1. p =

I rt

We solve the formula for W by isolating W on one side of the equals symbol.

59

Linear Equations, Inequalities, and Applications

P = 2L + 2W

NOW TRY EXERCISE 2

Solve the formula for b. P = a + 2b + c

Solve for W.

Step 1 is not needed here, since there are no fractions in the formula. Step 2

Step 3

P - 2L = 2L + 2W - 2L

Subtract 2L.

P - 2L = 2W

Combine like terms.

P - 2L 2W = 2 2

Divide by 2.

P - 2L = W, or 2

W =

P - 2L 2

NOW TRY

CAUTION In Step 3 of Example 2, we cannot simplify the fraction by dividing 2 into the term 2L. The fraction bar serves as a grouping symbol. Thus, the subtraction in the numerator must be done before the division.

P - 2L Z P - L 2

NOW TRY EXERCISE 3

EXAMPLE 3

Solve P = 21L + W 2 for L.

Solving a Formula Involving Parentheses

The formula for the perimeter of a rectangle is sometimes written in the equivalent form P = 21L + W 2. Solve this form for W. One way to begin is to use the distributive property on the right side of the equation to get P = 2L + 2W, which we would then solve as in Example 2. Another way to begin is to divide by the coefficient 2. P = 21L + W 2

P = L + W 2 P - L = W, or 2

Divide by 2.

W =

P - L 2

Subtract L.

We can show that this result is equivalent to our result in Example 2 by rewriting L as 22 L. P - L = W 2 2 P - 1L2 = W 2 2

2 2

= 1, so L =

2 2 1L2.

2L P = W 2 2 P - 2L = W 2

Subtract fractions.

The final line agrees with the result in Example 2. NOW TRY ANSWERS 2. b = 3. L =

60

P - a - c 2 P W, 2

or L =

P - 2W 2

NOW TRY

In Examples 1–3, we solved formulas for specified variables. In Example 4, we solve an equation with two variables for one of these variables.

Linear Equations, Inequalities, and Applications

NOW TRY EXERCISE 4

EXAMPLE 4

Solve the equation for y. 5x - 6y = 12

Solving an Equation for One of the Variables

Solve the equation 3x - 4y = 12 for y. Our goal is to isolate y on one side of the equation. 3x - 4y = 12 3x - 4y - 3x = 12 - 3x

Subtract 3x.

- 4y = 12 - 3x

Combine like terms.

- 4y 12 - 3x = -4 -4

Divide by - 4.

y =

12 - 3x -4

There are other equivalent forms of the final answer that are also correct. For example, since -ab = -ba, we rewrite the fraction by moving the negative sign from the denominator to the numerator, taking care to distribute to both terms. y =

12 - 3x -4

- 112 - 3x2

can be written as y =

4

, or

y =

Multiply both terms of the numerator by - 1.

3x - 12 . 4 NOW TRY

OBJECTIVE 2 Solve applied problems by using formulas. The distance formula, d = rt, relates d, the distance traveled, r, the rate or speed, and t, the travel time.

NOW TRY EXERCISE 5

EXAMPLE 5

It takes 12 hr for Dorothy Easley to drive 21 mi to work each day. What is her average rate?

Finding Average Rate

Phyllis Koenig found that on average it took her 34 hr each day to drive a distance of 15 mi to work. What was her average rate (or speed)? Find the formula for rate r by solving d = rt for r. d = rt rt d = t t

Divide by t.

d = r, t

or

r =

d t

Notice that only Step 3 was needed to solve for r in this example. Now find the rate by substituting the given values of d and t into this formula. r =

15

Let d = 15, t =

3 4

r = 15

#

4 3

3 4.

Multiply by the reciprocal of 34 .

r = 20 NOW TRY ANSWERS 4. y =

12 - 5x -6 ,

5. 42 mph

or y =

5x - 12 6

Her average rate was 20 mph. (That is, at times she may have traveled a little faster or NOW TRY slower than 20 mph, but overall her rate was 20 mph.)

61

Linear Equations, Inequalities, and Applications

OBJECTIVE 3 Solve percent problems. An important everyday use of mathematics involves the concept of percent. Percent is written with the symbol %. The word percent means “per one hundred.” One percent means “one per one hundred” or “one one-hundredth.”

1% ⴝ 0.01 or 1% ⴝ

1 100

Solving a Percent Problem

Let a represent a partial amount of b, the base, or whole amount. Then the following equation can be used to solve a percent problem. partial amount a ⴝ percent (represented as a decimal) base b

For example, if a class consists of 50 students and 32 are males, then the percent of males in the class is found as follows. partial amount a 32 = base b 50 =

Let a = 32, b = 50.

64 100

32 50

= 0.64, or 64%

NOW TRY EXERCISE 6

Solve each problem. (a) A 5-L mixture of water and antifreeze contains 2 L of antifreeze. What is the percent of antifreeze in the mixture? (b) If a savings account earns 2.5% interest on a balance of $7500 for one year, how much interest is earned?

EXAMPLE 6

#

2 2

=

64 100

Write as a decimal and then a percent.

Solving Percent Problems

(a) A 50-L mixture of acid and water contains 10 L of acid. What is the percent of acid in the mixture? The given amount of the mixture is 50 L, and the part that is acid is 10 L. Let x represent the percent of acid in the mixture. x =

10 50

partial amount whole amount (base)

x = 0.20, or 20% The mixture is 20% acid. (b) If a savings account balance of $4780 earns 5% interest in one year, how much interest is earned? Let x represent the amount of interest earned (that is, the part of the whole amount invested). Since 5% = 0.05, the equation is written as follows. x = 0.05 4780 x = 0.05147802

partial amount a base b

= percent

Multiply by 4780.

x = 239 NOW TRY ANSWERS 6. (a) 40% (b) $187.50

62

The interest earned is $239.

NOW TRY

Linear Equations, Inequalities, and Applications

NOW TRY EXERCISE 7

Refer to FIGURE 3 . How much was spent on vet care? Round your answer to the nearest tenth of a billion dollars.

EXAMPLE 7

Interpreting Percents from a Graph

In 2007, Americans spent about $41.2 billion on their pets. Use the graph in FIGURE 3 to determine how much of this amount was spent on pet food. Spending on Kitty and Rover Grooming/boarding 7.3%

Vet care 24.5%

Live animal purchases 5.1%

John Hornsby

Supplies/ medicine 23.8%

Food 39.3% Pythagoras

Source: American Pet Products Manufacturers Association Inc. FIGURE 3

Since 39.3% was spent on food, let x = this amount in billions of dollars. x = 0.393 41.2

39.3% = 0.393

x = 41.210.393)

Multiply by 41.2.

x L 16.2

Nearest tenth

Therefore, about $16.2 billion was spent on pet food.

NOW TRY

Solve problems involving percent increase or decrease. Percent is often used to express a change in some quantity. Buying an item that has been marked up and getting a raise at a job are applications of percent increase. Buying an item on sale and finding population decline are applications of percent decrease. To solve problems of this type, we use the following form of the percent equation. OBJECTIVE 4

percent change ⴝ EXAMPLE 8

amount of change base

Subtract to find this.

Solving Problems about Percent Increase or Decrease

(a) An electronics store marked up a laptop computer from their cost of $1200 to a selling price of $1464. What was the percent markup? “Markup” is a name for an increase. Let x = the percent increase (as a decimal). percent increase = Subtract to find the amount of increase.

x = x =

amount of increase base 1464 - 1200 1200 264 1200

Substitute the given values. Use the original cost.

x = 0.22, or 22%

Use a calculator.

NOW TRY ANSWER 7. $10.1 billion

The computer was marked up 22%.

63

Linear Equations, Inequalities, and Applications

NOW TRY EXERCISE 8

(a) Jane Brand bought a jacket on sale for $56. The regular price of the jacket was $80. What was the percent markdown? (b) When it was time for Horatio Loschak to renew the lease on his apartment, the landlord raised his rent from $650 to $689 a month. What was the percent increase?

(b) The enrollment at a community college declined from 12,750 during one school year to 11,350 the following year. Find the percent decrease to the nearest tenth. Let x = the percent decrease (as a decimal). percent decrease = Subtract to find the amount of decrease.

x =

amount of decrease base 12,750 - 11,350 12,750

Substitute the given values. Use the original number.

1400 x = 12,750 x L 0.11, or 11%

Use a calculator.

The college enrollment decreased by about 11%.

NOW TRY

CAUTION When calculating a percent increase or decrease, be sure that you use NOW TRY ANSWERS 8. (a) 30% (b) 6%

the original number (before the increase or decrease) as the base. A common error is to use the final number (after the increase or decrease) in the denominator of the fraction.

2 EXERCISES Solve each formula for the specified variable. See Examples 1–3. 1. I = prt for r (simple interest)

2. d = rt for t (distance)

3. P = 2L + 2W for L (perimeter of a rectangle)

4. a = bh for b (area of a parallelogram)*

L h W b

5. V = LWH (volume of a rectangular solid) (a) for W (b) for H

6. P = a + b + c (perimeter of a triangle) (a) for b

H

(b) for c b

a W

c

L

7. C = 2pr for r (circumference of a circle)

r

1 bh for h 2 (area of a triangle)

8. a =

h b

*We use a to denote area.

64

Linear Equations, Inequalities, and Applications

9. a =

1 h1b + B2 (area of a trapezoid) 2

(a) for h

10. S = 2prh + 2pr 2 for h (surface area of a right circular cylinder)

(b) for B b

h

h

r

B

5 1F - 322 for F 9 (Fahrenheit to Celsius)

9 C + 32 for C 5 (Celsius to Fahrenheit)

11. F =

12. C =

13. Concept Check When a formula is solved for a particular variable, several different equivalent forms may be possible. If we solve a = 12 bh for h, one possible correct answer is h =

2a . b

Which one of the following is not equivalent to this? a A. h = 2a b b

1 B. h = 2aa b b

a C. h =

D. h =

1 2b

1 2a

b

14. Concept Check The answer to Exercise 11 is given as C = 59 1F - 322. Which one of the following is not equivalent to this? A. C =

160 5 F 9 9

B. C =

5F 160 9 9

C. C =

5F - 160 9

D. C =

5 F - 32 9

Solve each equation for y. See Example 4. 15. 4x + 9y = 11

16. 7x + 8y = 11

17. - 3x + 2y = 5

18. - 5x + 3y = 12

19. 6x - 5y = 7

20. 8x - 3y = 4

Solve each problem. See Example 5.

Paul Sancya/AP Images

Kevin C. Cox/Allsport Photography/ Getty Images

21. Ryan Newman won the Daytona 500 (mile) 22. In 2007, rain shortened the Indianapolis race with a rate of 152.672 mph in 2008. 500 race to 415 mi. It was won by Dario Find his time to the nearest thousandth. Franchitti, who averaged 151.774 mph. (Source: www.daytona500.com) What was his time to the nearest thousandth? (Source: www.indy500.com)

23. Nora Demosthenes traveled from Kansas City to Louisville, a distance of 520 mi, in 10 hr. Find her rate in miles per hour. 24. The distance from Melbourne to London is 10,500 mi. If a jet averages 500 mph between the two cities, what is its travel time in hours? 25. As of 2009, the highest temperature ever recorded in Tennessee was 45°C. Find the corresponding Fahrenheit temperature. (Source: National Climatic Data Center.) 65

Linear Equations, Inequalities, and Applications

26. As of 2009, the lowest temperature ever recorded in South Dakota was - 58°F. Find the corresponding Celsius temperature. (Source: National Climate Data Center.) 27. The base of the Great Pyramid of Cheops is a square whose perimeter is 920 m. What is the length of each side of this square? (Source: Atlas of Ancient Archaeology.) x Perimeter = 920 m

29. The circumference of a circle is 480p in. What is the radius? What is the diameter?

Steve Broer/Shutterstock

28. Marina City in Chicago is a complex of two residential towers that resemble corncobs. Each tower has a concrete cylindrical core with a 35-ft diameter and is 588 ft tall. Find the volume of the core of one of the towers to the nearest whole number. (Hint: Use the p key on your calculator.) (Source: www.architechgallery.com; www.aviewoncities.com)

30. The radius of a circle is 2.5 in. What is the diameter? What is the circumference?

r r = 2.5 in. d

31. A sheet of standard-size copy paper measures 8.5 in. by 11 in. If a ream (500 sheets) of this paper has a volume of 187 in.3, how thick is the ream? 11 in.

32. Copy paper (Exercise 31) also comes in legal size, which has the same width, but is longer than standard size. If a ream of legalsize paper has the same thickness as standard-size paper and a volume of 238 in.3, what is the length of a sheet of legal paper? 8.5 in.

Solve each problem. See Example 6. 33. A mixture of alcohol and water contains a total of 36 oz of liquid. There are 9 oz of pure alcohol in the mixture. What percent of the mixture is water? What percent is alcohol? 34. A mixture of acid and water is 35% acid. If the mixture contains a total of 40 L, how many liters of pure acid are in the mixture? How many liters of pure water are in the mixture? 35. A real-estate agent earned $6300 commission on a property sale of $210,000. What is her rate of commission? 36. A certificate of deposit for 1 yr pays $221 simple interest on a principal of $3400. What is the interest rate being paid on this deposit? When a consumer loan is paid off ahead of schedule, the finance charge is less than if the loan were paid off over its scheduled life. By one method, called the rule of 78, the amount of unearned interest (the finance charge that need not be paid) is given by k1k ⴙ 12 uⴝƒ . n1n ⴙ 12

#

66

Linear Equations, Inequalities, and Applications

In the formula, u is the amount of unearned interest (money saved) when a loan scheduled to run for n payments is paid off k payments ahead of schedule. The total scheduled finance charge is ƒ. Use the formula for the rule of 78 to work Exercises 37–40. 37. Sondra Braeseker bought a new car and agreed to pay it off in 36 monthly payments. The total finance charge was $700. Find the unearned interest if she paid the loan off 4 payments ahead of schedule. 38. Donnell Boles bought a truck and agreed to pay it off in 36 monthly payments. The total finance charge on the loan was $600. With 12 payments remaining, he decided to pay the loan in full. Find the amount of unearned interest. 39. The finance charge on a loan taken out by Kha Le is $380.50. If 24 equal monthly installments were needed to repay the loan, and the loan is paid in full with 8 months remaining, find the amount of unearned interest. 40. Maky Manchola is scheduled to repay a loan in 24 equal monthly installments. The total finance charge on the loan is $450. With 9 payments remaining, he decides to repay the loan in full. Find the amount of unearned interest. In baseball, winning percentage (Pct.) is commonly expressed as a decimal rounded to the nearest thousandth. To find the winning percentage of a team, divide the number of wins 1W2 by the total number of games played 1W + L2. 41. The final 2009 standings of the Eastern Division of the American League are shown in the table. Find the winning percentage of each team. (a) Boston

(b) Tampa Bay

(c) Toronto

(d) Baltimore

(a) Philadelphia

(b) Atlanta

(c) New York Mets

(d) Washington W

L

Philadelphia

93

69

Florida

87

75

67

Atlanta

86

76

84

78

New York Mets

70

92

75

87

Washington

59

103

64

98

W

L

Pct.

103

59

.636

Boston

95

Tampa Bay Toronto Baltimore

New York Yankees

42. Repeat Exercise 41 for the following standings for the Eastern Division of the National League.

Pct. .537

Source: World Almanac and Book of Facts.

As mentioned in the chapter introduction, 114.9 million U.S. households owned at least one TV set in 2009. (Source: Nielsen Media Research.) Use this information to work Exercises 43–46. Round answers to the nearest percent in Exercises 43–44, and to the nearest tenth million in Exercises 45–46. See Example 6. 43. About 62.0 million U.S. households owned 3 or more TV sets in 2009. What percent of those owning at least one TV set was this? 44. About 102.2 million households that owned at least one TV set in 2009 had a DVD player. What percent of those owning at least one TV set had a DVD player?

45. Of the households owning at least one TV set in 2009, 88% received basic cable. How many households received basic cable? 46. Of the households owning at least one TV set in 2009, 35% received premium cable. How many households received premium cable?

67

Kelsey McNeal/ABC/Everett Collection

Source: World Almanac and Book of Facts.

Linear Equations, Inequalities, and Applications

An average middle-income family will spend $221,190 to raise a child born in 2008 from birth through age 17. The graph shows the percents spent for various categories. Use the graph to answer Exercises 47–50. See Example 7.

The Cost of Parenthood Housing 32%

Child care/ education 16%

47. To the nearest dollar, how much will be spent to provide housing for the child? 48. To the nearest dollar, how much will be spent for health care? 49. Use your answer from Exercise 48 to find how much will be spent for child care and education.

Miscellaneous 8%

Health care 8% Food 16%

Clothing 6% Transportation 14%

Source: U.S. Department of Agriculture.

50. About $35,000 will be spent for food. To the nearest percent, what percent of the cost of raising a child from birth through age 17 is this? Does your answer agree with the percent shown in the graph? Solve each problem about percent increase or percent decrease. See Example 8. 52. Clayton bought a ticket to a rock concert at a discount. The regular price of the ticket was $70.00, but he only paid $59.50. What was the percent discount?

53. Between 2000 and 2007, the estimated population of Pittsfield, Massachusetts, declined from 134,953 to 129,798. What was the percent decrease to the nearest tenth? (Source: U.S. Census Bureau.)

54. Between 2000 and 2007, the estimated population of Anchorage, Alaska, grew from 320,391 to 362,340. What was the percent increase to the nearest tenth? (Source: U.S. Census Bureau.)

55. In April 2008, the audio CD of the Original Broadway Cast Recording of the musical Wicked was available for $9.97. The list price (full price) of this CD was $18.98. To the nearest tenth, what was the percent discount? (Source: www.amazon.com)

56. In April 2008, the DVD of the movie Alvin and the Chipmunks was released. This DVD had a list price of $29.99 and was on sale for $15.99. To the nearest tenth, what was the percent discount? (Source: www.amazon.com)

ZanyZeus/Shutterstock

TM & 20th Century Fox/Everett Collection

51. After 1 yr on the job, Grady got a raise from $10.50 per hour to $11.34 per hour. What was the percent increase in his hourly wage?

PREVIEW EXERCISES Solve each equation. See Section 1.

68

57. 4x + 41x + 72 = 124

58. x + 0.20x = 66

59. 2.4 + 0.4x = 0.2516 + x2

60. 0.07x + 0.0519000 - x2 = 510

Linear Equations, Inequalities, and Applications

Evaluate. 61. The product of - 3 and 5, divided by 1 less than 6 62. Half of - 18, added to the reciprocal of

1 5

63. The sum of 6 and - 9, multiplied by the additive inverse of 2 64. The product of - 2 and 4, added to the product of - 9 and - 3

3

Applications of Linear Equations

OBJECTIVES 1

2

3

4

5 6 7

Translate from words to mathematical expressions. Write equations from given information. Distinguish between simplifying expressions and solving equations. Use the six steps in solving an applied problem. Solve percent problems. Solve investment problems. Solve mixture problems.

OBJECTIVE 1

Translate from words to mathematical expressions.

PROBLEM-SOLVING HINT

There are usually key words and phrases in a verbal problem that translate into mathematical expressions involving addition, subtraction, multiplication, and division. Translations of some commonly used expressions follow.

Translating from Words to Mathematical Expressions Verbal Expression

Mathematical Expression (where x and y are numbers)

Addition The sum of a number and 7

x + 7

6 more than a number

x + 6

3 plus a number

3 + x

24 added to a number

x + 24

A number increased by 5

x + 5

The sum of two numbers

x + y

Subtraction 2 less than a number

x - 2

2 less a number

2 - x

12 minus a number

12 - x

A number decreased by 12

x - 12

A number subtracted from 10

10 - x

10 subtracted from a number

x - 10

The difference between two numbers

x - y

Multiplication 16 times a number

16x

A number multiplied by 6

6x

2 3

2 3x

of a number (used with fractions and percent) as much as a number

3 4x

Twice (2 times) a number

2x

The product of two numbers

xy

3 4

Division The quotient of 8 and a number

8 x

1x Z 02

x y

1y Z 02

x 13

A number divided by 13 The ratio of two numbers or the quotient of two numbers

69

Linear Equations, Inequalities, and Applications

CAUTION Because subtraction and division are not commutative operations, it is important to correctly translate expressions involving them. For example,

“2 less than a number” is translated as

x - 2, not

“A number subtracted from 10” is expressed as

2 - x.

10 - x, not

x - 10.

For division, the number by which we are dividing is the denominator, and the number into which we are dividing is the numerator. “A number divided by 13” and “13 divided into x” both translate as

x 13 .

“The quotient of x and y” is translated as xy . OBJECTIVE 2 Write equations from given information. The symbol for equality, = , is often indicated by the word is. NOW TRY EXERCISE 1

Translate each verbal sentence into an equation, using x as the variable. (a) The quotient of a number and 10 is twice the number. (b) The product of a number and 5, decreased by 7, is zero.

EXAMPLE 1

Translating Words into Equations

Translate each verbal sentence into an equation. Verbal Sentence

Equation 2x - 3 = 42

Twice a number, decreased by 3, is 42. The product of a number and 12,

12x - 7 = 105

decreased by 7, is 105. The quotient of a number and the number plus 4 is 28. The quotient of a number and 4, plus the number, is 10.

x = 28 x + 4

Any words that indicate the idea of “sameness” translate as =.

x + x = 10 4

NOW TRY

OBJECTIVE 3 Distinguish between simplifying expressions and solving equations. An expression translates as a phrase. An equation includes the = symbol, with expressions on both sides, and translates as a sentence. EXAMPLE 2

Distinguishing between Simplifying Expressions and Solving Equations

Decide whether each is an expression or an equation. Simplify any expressions, and solve any equations. (a) 213 + x2 - 4x + 7 There is no equals symbol, so this is an expression. 213 + x2 - 4x + 7 = 6 + 2x - 4x + 7

Distributive property

= - 2x + 13

Simplified expression

(b) 213 + x2 - 4x + 7 = - 1 Because there is an equals symbol with expressions on both sides, this is an equation. NOW TRY ANSWERS 1. (a)

70

x 10

= 2x (b) 5x - 7 = 0

213 + x2 - 4x + 7 = - 1 6 + 2x - 4x + 7 = - 1

Distributive property

Linear Equations, Inequalities, and Applications

- 2x + 13 = - 1

NOW TRY EXERCISE 2

Decide whether each is an expression or an equation. Simplify any expressions, and solve any equations. (a) 31x - 52 + 2x - 1 (b) 31x - 52 + 2x = 1

Combine like terms.

- 2x = - 14

Subtract 13.

x = 7

The solution set is 576.

Divide by - 2. NOW TRY

OBJECTIVE 4 Use the six steps in solving an applied problem. While there is no one method that allows us to solve all types of applied problems, the following six steps are helpful. Solving an Applied Problem

Step 1

Read the problem, several times if necessary. What information is given? What is to be found?

Step 2

Assign a variable to represent the unknown value. Use a sketch, diagram, or table, as needed. Write down what the variable represents. If necessary, express any other unknown values in terms of the variable.

Step 3

Write an equation using the variable expression(s).

Step 4

Solve the equation.

Step 5

State the answer. Label it appropriately. Does it seem reasonable?

Step 6

Check the answer in the words of the original problem.

EXAMPLE 3

Solving a Perimeter Problem

The length of a rectangle is 1 cm more than twice the width. The perimeter of the rectangle is 110 cm. Find the length and the width of the rectangle. Step 1 Read the problem. What must be found? The length and width of the rectangle. What is given? The length is 1 cm more than twice the width and the perimeter is 110 cm. Step 2 Assign a variable. Let W = the width. Then 2W + 1 = the length. Make a sketch, as in FIGURE 4 .

W

2W + 1 FIGURE 4

Step 3 Write an equation. Use the formula for the perimeter of a rectangle. P = 2L + 2W 110 = 212W + 12 + 2W

Perimeter of a rectangle Let L = 2W + 1 and P = 110.

Step 4 Solve the equation obtained in Step 3. 110 = 4W + 2 + 2W

Distributive property

110 = 6W + 2

Combine like terms.

110 - 2 = 6W + 2 - 2 108 = 6W NOW TRY ANSWERS 2. (a) expression; 5x - 16 (b) equation; E 16 5 F

6W 108 = 6 6 18 = W

Subtract 2. Combine like terms.

We also need to find the length.

Divide by 6.

71

Linear Equations, Inequalities, and Applications

NOW TRY EXERCISE 3

The length of a rectangle is 2 ft more than twice the width. The perimeter is 34 ft. Find the length and width of the rectangle.

NOW TRY EXERCISE 4

During the 2008 regular NFL football season, Drew Brees of the New Orleans Saints threw 4 more touchdown passes than Kurt Warner of the Arizona Cardinals. Together, these two quarterbacks completed a total of 64 touchdown passes. How many touchdown passes did each player complete? (Source: www.nfl.com)

Step 5 State the answer. The width of the rectangle is 18 cm and the length is 21182 + 1 = 37 cm. Step 6 Check. The length, 37 cm, is 1 cm more than 21182 cm (twice the width). The perimeter is 21372 + 21182 = 74 + 36 = 110 cm,

EXAMPLE 4

as required.

NOW TRY

Finding Unknown Numerical Quantities

During the 2009 regular season, Justin Verlander of the Detroit Tigers and Tim Lincecum of the San Francisco Giants were the top major league pitchers in strikeouts. The two pitchers had a total of 530 strikeouts. Verlander had 8 more strikeouts than Lincecum. How many strikeouts did each pitcher have? (Source: www.mlb.com) Step 1 Read the problem. We are asked to find the number of strikeouts each pitcher had. Step 2 Assign a variable to represent the number of strikeouts for one of the men. Let s = the number of strikeouts for Tim Lincecum. We must also find the number of strikeouts for Justin Verlander. Since he had 8 more strikeouts than Lincecum, s + 8 = the number of strikeouts for Verlander. Step 3 Write an equation. The sum of the numbers of strikeouts is 530. Lincecum’s strikeouts

+

Verlander’s strikeouts

=

Total

s

+

1s + 82

=

530

Step 4 Solve the equation. Brad Mangin/Getty Images

s + 1s + 82 = 530 2s + 8 = 530 2s + 8 - 8 = 530 - 8

Tim Lincecum

Don’t stop here.

Combine like terms. Subtract 8.

2s = 522

Combine like terms.

522 2s = 2 2

Divide by 2.

s = 261

Step 5 State the answer. We let s represent the number of strikeouts for Lincecum, so Lincecum had 261. Then Verlander had s + 8 = 261 + 8 = 269 strikeouts. Step 6 Check. 269 is 8 more than 261, and 261 + 269 = 530. The conditions of NOW TRY the problem are satisfied, and our answer checks.

NOW TRY ANSWERS 3. width: 5 ft; length: 12 ft 4. Drew Brees: 34; Kurt Warner: 30

72

CAUTION Be sure to answer all the questions asked in the problem. In Example 4, we were asked for the number of strikeouts for each player, so there was extra work in Step 5 in order to find Verlander’s number.

Linear Equations, Inequalities, and Applications

OBJECTIVE 5 Solve percent problems. Recall from Section 2 that percent means “per one hundred,” so 5% means 0.05, 14% means 0.14, and so on.

NOW TRY EXERCISE 5

In the fall of 2009, there were 96 Introductory Statistics students at a certain community college, an increase of 700% over the number of Introductory Statistics students in the fall of 1992. How many Introductory Statistics students were there in the fall of 1992?

EXAMPLE 5

Solving a Percent Problem

In 2006, total annual health expenditures in the United States were about $2000 billion (or $2 trillion). This was an increase of 180% over the total for 1990. What were the approximate total health expenditures in billions of dollars in the United States in 1990? (Source: U.S. Centers for Medicare & Medicaid Services.) Step 1 Read the problem. We are given that the total health expenditures increased by 180% from 1990 to 2006, and $2000 million was spent in 2006. We must find the expenditures in 1990. Step 2 Assign a variable. Let x represent the total health expenditures for 1990. 180% = 18010.012 = 1.8, so 1.8x represents the additional expenditures since 1990. Step 3 Write an equation from the given information. the expenditures in 1990 + the increase = 2000

x

+

1.8x

= 2000 Note the x in 1.8x.

Step 4 Solve the equation. 1x + 1.8x = 2000 2.8x = 2000 x L 714

Identity property Combine like terms. Divide by 2.8.

Step 5 State the answer. Total health expenditures in the United States for 1990 were about $714 billion. Step 6 Check that the increase, 2000 - 714 = 1286, is about 180% of 714. NOW TRY

CAUTION Avoid two common errors that occur in solving problems like the one in Example 5.

1. Do not try to find 180% of 2000 and subtract that amount from 2000. The 180% should be applied to the amount in 1990, not the amount in 2006. 2. Do not write the equation as x + 1.8 = 2000.

Incorrect

The percent must be multiplied by some number. In this case, the number is the amount spent in 1990, giving 1.8x.

Solve investment problems. The investment problems in this chapter deal with simple interest. In most real-world applications, compound interest is used. OBJECTIVE 6

NOW TRY ANSWER 5. 12

73

Linear Equations, Inequalities, and Applications

NOW TRY EXERCISE 6

Gary Jones received a $20,000 inheritance from his grandfather. He invested some of the money in an account earning 3% annual interest and the remaining amount in an account earning 2.5% annual interest. If the total annual interest earned is $575, how much is invested at each rate?

EXAMPLE 6

Solving an Investment Problem

Thomas Flanagan has $40,000 to invest. He will put part of the money in an account paying 4% interest and the remainder into stocks paying 6% interest. The total annual income from these investments should be $2040. How much should he invest at each rate? Step 1 Read the problem again. We must find the two amounts. Step 2 Assign a variable. Let

x = the amount to invest at 4% ;

40,000 - x = the amount to invest at 6%. The formula for interest is I = prt. Here the time t is 1 yr. Use a table to organize the given information.

Rate (as a decimal)

Principal

Interest

x

0.04

0.04 x

40,000 - x

0.06

0.06140,000 - x2

40,000

Multiply principal, rate, and time (here, 1 yr) to get interest. Total

2040

Step 3 Write an equation. The last column of the table gives the equation. interest at 4%

0.04x

+

interest at 6%

=

+ 0.06140,000 - x2 =

total interest

2040

Step 4 Solve the equation. 0.04x + 0.06140,0002 - 0.06x = 2040

Distributive property.

0.04x + 2400 - 0.06x = 2040

Multiply.

- 0.02x + 2400 = 2040

Combine like terms.

- 0.02x = - 360

Subtract 2400.

x = 18,000

Divide by - 0.02.

Step 5 State the answer. Thomas should invest $18,000 of the money at 4%. At 6%, he should invest $40,000 - $18,000 = $22,000. Step 6 Check. Find the annual interest at each rate. The sum of these two amounts should total $2040. 0.041$18,0002 = $720

and

0.061$22,0002 = $1320

$720 + $1320 = $2040,

as required.

NOW TRY

PROBLEM-SOLVING HINT

NOW TRY ANSWER 6. $15,000 at 3%; $5000 at 2.5%

74

In Example 6, we chose to let the variable represent the amount invested at 4%. Students often ask, “Can I let the variable represent the other unknown?” The answer is yes. The equation will be different, but in the end the answers will be the same.

Linear Equations, Inequalities, and Applications

Solve mixture problems.

OBJECTIVE 7

NOW TRY EXERCISE 7

EXAMPLE 7

Solving a Mixture Problem

A chemist must mix 8 L of a 40% acid solution with some 70% solution to get a 50% solution. How much of the 70% solution should be used?

How many liters of a 20% acid solution must be mixed with 5 L of a 30% acid solution to get a 24% acid solution?

Step 1 Read the problem. The problem asks for the amount of 70% solution to be used. Step 2 Assign a variable. Let x = the number of liters of 70% solution to be used. The information in the problem is illustrated in FIGURE 5 and organized in the table. After mixing Number of Liters

+ 40% 8L

70%

=

Unknown number of liters, x

50%

From 70% From 40%

(8 + x) L

Percent (as a decimal)

Liters of Pure Acid

8

0.40

0.40182 = 3.2

x

0.70

0.70x

8 + x

0.50

0.5018 + x2

Sum must equal

FIGURE 5

The numbers in the last column of the table were found by multiplying the strengths by the numbers of liters. The number of liters of pure acid in the 40% solution plus the number of liters in the 70% solution must equal the number of liters in the 50% solution. Step 3 Write an equation. 3.2 + 0.70x = 0.5018 + x2 Step 4 Solve. 3.2 + 0.70x = 4 + 0.50x 0.20x = 0.8 x = 4

Distributive property Subtract 3.2 and 0.50x. Divide by 0.20.

Step 5 State the answer. The chemist should use 4 L of the 70% solution. Step 6 Check. 8 L of 40% solution plus 4 L of 70% solution is 810.402 + 410.702 = 6 L of acid. Similarly, 8 + 4 or 12 L of 50% solution has 1210.502 = 6 L of acid. The total amount of pure acid is 6 L both before and after mixing, so the answer checks. NOW TRY

PROBLEM-SOLVING HINT NOW TRY ANSWER

Remember that when pure water is added to a solution, water is 0% of the chemical (acid, alcohol, etc.). Similarly, pure chemical is 100% chemical.

7. 7 12 L

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Linear Equations, Inequalities, and Applications

NOW TRY EXERCISE 8

How much pure antifreeze must be mixed with 3 gal of a 30% antifreeze solution to get a 40% antifreeze solution?

EXAMPLE 8

Solving a Mixture Problem When One Ingredient Is Pure

The octane rating of gasoline is a measure of its antiknock qualities. For a standard fuel, the octane rating is the percent of isooctane. How many liters of pure isooctane should be mixed with 200 L of 94% isooctane, referred to as 94 octane, to get a mixture that is 98% isooctane? Step 1 Read the problem. The problem asks for the amount of pure isooctane.

Step 2 Assign a variable. Let x = the number of liters of pure 1100%2 isooctane. Complete a table. Recall that 100% = 10010.012 = 1. Number of Liters

Percent (as a decimal)

Liters of Pure Isooctane

x

1

200

0.94

x 0.9412002

x + 200

0.98

0.981x + 2002

Step 3 Write an equation. The equation comes from the last column of the table. x + 0.9412002 = 0.981x + 2002 Step 4 Solve. x + 0.9412002 = 0.98x + 0.9812002 x + 188 = 0.98x + 196 0.02x = 8

8.

1 2

gal

Multiply. Subtract 0.98x and 188.

x = 400 NOW TRY ANSWER

Distributive property

Divide by 0.02.

Step 5 State the answer. 400 L of isooctane is needed. Step 6 Check by showing that 400 + 0.9412002 = 0.981400 + 2002 is true. NOW TRY

3 EXERCISES Concept Check In each of the following, (a) translate as an expression and (b) translate as an equation or inequality. Use x to represent the number. 1. (a) 15 more than a number

2. (a) 5 greater than a number

(b) 15 is more than a number.

(b) 5 is greater than a number.

3. (a) 8 less than a number

4. (a) 6 less than a number

(b) 8 is less than a number.

(b) 6 is less than a number.

5. Concept Check Which one of the following is not a valid translation of “40% of a number,” where x represents the number? A. 0.40x

B. 0.4x

C.

2x 5

D. 40x

6. Explain why 13 - x is not a correct translation of “13 less than a number.” Translate each verbal phrase into a mathematical expression. Use x to represent the unknown number. See Example 1. 7. Twice a number, decreased by 13 9. 12 increased by four times a number

76

8. The product of 6 and a number, decreased by 14 10. 15 more than one-half of a number

Linear Equations, Inequalities, and Applications

11. The product of 8 and 16 less than a number

12. The product of 8 more than a number and 5 less than the number

13. The quotient of three times a number and 10

14. The quotient of 9 and five times a nonzero number

Use the variable x for the unknown, and write an equation representing the verbal sentence. Then solve the problem. See Example 1. 15. The sum of a number and 6 is - 31. Find the number. 16. The sum of a number and - 4 is 18. Find the number. 17. If the product of a number and - 4 is subtracted from the number, the result is 9 more than the number. Find the number. 18. If the quotient of a number and 6 is added to twice the number, the result is 8 less than the number. Find the number. 19. When 23 of a number is subtracted from 14, the result is 10. Find the number. 20. When 75% of a number is added to 6, the result is 3 more than the number. Find the number. Decide whether each is an expression or an equation. Simplify any expressions, and solve any equations. See Example 2. 21. 51x + 32 - 812x - 62

22. - 71x + 42 + 131x - 62

23. 51x + 32 - 812x - 62 = 12

24. - 71x + 42 + 131x - 62 = 18

25.

1 1 3 x - x + - 8 2 6 2

26.

1 1 1 x + x - + 7 3 5 2

Concept Check Complete the six suggested problem-solving steps to solve each problem. 27. In 2008, the corporations securing the most U.S. patents were IBM and Samsung. Together, the two corporations secured a total of 7671 patents, with Samsung receiving 667 fewer patents than IBM. How many patents did each corporation secure? (Source: U.S. Patent and Trademark Office.) Step 1

Read the problem carefully. We are asked to find

.

Step 2 Assign a variable. Let x = the number of patents that IBM secured. Then . x - 667 = the number of Step 3

Write an equation.

Step 4

Solve the equation. x =

+

Step 5 State the answer. IBM secured patents.

= 7671 patents, and Samsung secured

Step 6 Check. The number of Samsung patents was fewer than the number of , and the total number of patents was 4169 + . = 28. In 2008, 7.8 million more U.S. residents traveled to Mexico than to Canada. There was a total of 32.8 million U.S. residents traveling to these two countries. How many traveled to each country? (Source: U.S. Department of Commerce.) Step 1

Read the problem carefully. We are asked to find

.

Step 2 Assign a variable. Let x = the number of travelers to Mexico (in millions). Then x - 7.8 = the number of . Step 3

Write an equation.

Step 4

Solve the equation. x =

Step 5 State the answer. There were to Canada.

+

= 32.8 travelers to Mexico and

travelers

Step 6 Check. The number of was more than the number of and the total number of these travelers was 20.3 + . =

,

77

Linear Equations, Inequalities, and Applications

Solve each problem. See Examples 3 and 4. 29. The John Hancock Center in Chicago has a rectangular base. The length of the base measures 65 ft less than twice the width. The perimeter of the base is 860 ft. What are the dimensions of the base? 30. The John Hancock Center (Exercise 29) tapers as it rises. The top floor is rectangular and has perimeter 520 ft. The width of the top floor measures 20 ft more than one-half its length. What are the dimensions of the top floor?

The perimeter of the L top floor is 520 ft.

1 L 2

+ 20

W 2W – 65 The perimeter of the base is 860 ft.

American Gothic (1930), Grant Wood. Oil on board (74.3  62.4 cm). The Art Institute of Chicago, Friends of American Art Collection, 1930.934. Art © Figge Art Museum, successors to the Estate of Nan Wood Graham/Licensed by VAGA, New York, NY/Superstock

31. Grant Wood painted his most famous work, American Gothic, in 1930 on composition board with perimeter 108.44 in. If the painting is 5.54 in. taller than it is wide, find the dimensions of the painting. (Source: The Gazette.)

32. The perimeter of a certain rectangle is 16 times the width. The length is 12 cm more than the width. Find the length and width of the rectangle. W

W + 12

American Gothic by Grant Wood. © Figge Art Museum/Estate of Nan Wood Graham/VAGA, NY

33. The Bermuda Triangle supposedly causes trouble for aircraft pilots. It has a perimeter of 3075 mi. The shortest side measures 75 mi less than the middle side, and the longest side measures 375 mi more than the middle side. Find the lengths of the three sides. 34. The Vietnam Veterans Memorial in Washington, DC, is in the shape of two sides of an isosceles triangle. If the two walls of equal length were joined by a straight line of 438 ft, the perimeter of the resulting triangle would be 931.5 ft. Find the lengths of the two walls. (Source: Pamphlet obtained at Vietnam Veterans Memorial.)

x

x 438 ft

36. Two of the longest-running Broadway shows were Cats, which played from 1982 through 2000, and Les Misérables, which played from 1987 through 2003. Together, there were 14,165 performances of these two shows during their Broadway runs. There were 805 fewer performances of Les Misérables than of Cats. How many performances were there of each show? (Source: The Broadway League.)

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Joseph Barrak/AFP/Getty Images

35. The two companies with top revenues in the Fortune 500 list for 2009 were Exxon Mobil and Wal-Mart. Their revenues together totaled $848.5 billion. Wal-Mart revenues were $37.3 billion less than Exxon Mobil revenues. What were the revenues of each corporation? (Source: www.money.cnn.com)

Linear Equations, Inequalities, and Applications

Luciano Mortula/Shutterstock

37. Galileo Galilei conducted experiments involving Italy’s famous Leaning Tower of Pisa to investigate the relationship between an object’s speed of fall and its weight. The Leaning Tower is 880 ft shorter than the Eiffel Tower in Paris, France. The two towers have a total height of 1246 ft. How tall is each tower? (Source: www.leaned.org, www.tour-eiffel.fr.)

39. In the 2008 presidential election, Barack Obama and John McCain together received 538 electoral votes. Obama received 192 more votes than McCain. How many votes did each candidate receive? (Source: World Almanac and Book of Facts.) 40. Ted Williams and Rogers Hornsby were two great hitters in Major League Baseball. Together, they got 5584 hits in their careers. Hornsby got 276 more hits than Williams. How many base hits did each get? (Source: Neft, D. S., and R. M. Cohen, The Sports Encyclopedia: Baseball, St. Martins Griffin; New York, 2007.)

Solve each percent problem. See Example 5.

Bruce Bennett Studios/Getty Images

38. In 2009, the New York Yankees and the New York Mets had the highest payrolls in Major League Baseball. The Mets’ payroll was $65.6 million less than the Yankees’ payroll, and the two payrolls totaled $337.2 million. What was the payroll for each team? (Source: Associated Press.)

41. In 2009, the number of graduating seniors taking the ACT exam was 1,480,469. In 2000, a total of 1,065,138 graduating seniors took the exam. By what percent did the number increase over this period of time, to the nearest tenth of a percent? (Source: ACT.) 42. Composite scores on the ACT exam rose from 20.8 in 2002 to 21.1 in 2009. What percent increase was this, to the nearest tenth of a percent? (Source: ACT.)

44. In 1995, the average cost of tuition and fees at private four-year universities in the United States was $12,216 for full-time students. By 2009, it had risen approximately 115.1%. To the nearest dollar, what was the approximate cost in 2009? (Source: The College Board.) 45. In 2009, the average cost of a traditional Thanksgiving dinner for 10, featuring turkey, stuffing, cranberries, pumpkin pie, and trimmings, was $42.91, a decrease of 3.8% over the cost in 2008. What was the cost, to the nearest cent, in 2008? (Source: American Farm Bureau.)

46. Refer to Exercise 45. The cost of a traditional Thanksgiving dinner in 2009 was $42.91, an increase of 60.4% over the cost in 1987 when data was first collected. What was the cost, to the nearest cent, in 1987? (Source: American Farm Bureau.) 47. At the end of a day, Lawrence Hawkins found that the total cash register receipts at the motel where he works amounted to $2725. This included the 9% sales tax charged. Find the amount of the tax.

79

Ryan McVay/Photodisc/Getty Images

43. In 1995, the average cost of tuition and fees at public four-year universities in the United States was $2811 for full-time students. By 2009, it had risen approximately 150%. To the nearest dollar, what was the approximate cost in 2009? (Source: The College Board.)

Linear Equations, Inequalities, and Applications

48. David Ruppel sold his house for $159,000. He got this amount knowing that he would have to pay a 6% commission to his agent. What amount did he have after the agent was paid? Solve each investment problem. See Example 6. 49. Mario Toussaint earned $12,000 last year by giving tennis lessons. He invested part of the money at 3% simple interest and the rest at 4%. In one year, he earned a total of $440 in interest. How much did he invest at each rate?

Principal

Rate (as a decimal)

x

0.03

Interest

50. Sheryl Zavertnik won $60,000 on a slot machine in Las Vegas. She invested part of the money at 2% simple interest and the rest at 3%. In one year, she earned a total of $1600 in interest. How much was invested at each rate?

Principal

Rate (as a decimal)

x

0.02

Interest

0.04

51. Jennifer Siegel invested some money at 4.5% simple interest and $1000 less than twice this amount at 3%. Her total annual income from the interest was $1020. How much was invested at each rate? 52. Piotr Galkowski invested some money at 3.5% simple interest, and $5000 more than three times this amount at 4%. He earned $1440 in annual interest. How much did he invest at each rate? 53. Dan Abbey has invested $12,000 in bonds paying 6%. How much additional money should he invest in a certificate of deposit paying 3% simple interest so that the total return on the two investments will be 4%? 54. Mona Galland received a year-end bonus of $17,000 from her company and invested the money in an account paying 6.5%. How much additional money should she deposit in an account paying 5% so that the return on the two investments will be 6%? Solve each problem involving rates of concentration and mixtures. See Examples 7 and 8. 55. Ten liters of a 4% acid solution must be mixed with a 10% solution to get a 6% solution. How many liters of the 10% solution are needed? Liters of Solution

Percent (as a decimal)

10

0.04

x

0.10

Liters of Pure Acid

56. How many liters of a 14% alcohol solution must be mixed with 20 L of a 50% solution to get a 30% solution?

Liters of Solution

Percent (as a decimal)

x

0.14

Liters of Pure Alcohol

0.50

0.06

58. How many liters of a 10% alcohol solution must be mixed with 40 L of a 50% solution to get a 40% solution? 59. How much pure dye must be added to 4 gal of a 25% dye solution to increase the solution to 40%? (Hint: Pure dye is 100% dye.) 60. How much water must be added to 6 gal of a 4% insecticide solution to reduce the concentration to 3%? (Hint: Water is 0% insecticide.)

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Brand X Pictures/Thinkstock

57. In a chemistry class, 12 L of a 12% alcohol solution must be mixed with a 20% solution to get a 14% solution. How many liters of the 20% solution are needed?

Linear Equations, Inequalities, and Applications

61. Randall Albritton wants to mix 50 lb of nuts worth $2 per lb with some nuts worth $6 per lb to make a mixture worth $5 per lb. How many pounds of $6 nuts must he use? Pounds of Nuts

Cost per Pound

62. Lee Ann Spahr wants to mix tea worth 2¢ per oz with 100 oz of tea worth 5¢ per oz to make a mixture worth 3¢ per oz. How much 2¢ tea should be used?

Total Cost

Ounces of Tea

Cost per Ounce

Total Cost

63. Why is it impossible to mix candy worth $4 per lb and candy worth $5 per lb to obtain a final mixture worth $6 per lb? 64. Write an equation based on the following problem, solve the equation, and explain why the problem has no solution: How much 30% acid should be mixed with 15 L of 50% acid to obtain a mixture that is 60% acid?

RELATING CONCEPTS

EXERCISES 65–68

FOR INDIVIDUAL OR GROUP WORK

Consider each problem. Problem A Jack has $800 invested in two accounts. One pays 5% interest per year and the other pays 10% interest per year. The amount of yearly interest is the same as he would get if the entire $800 was invested at 8.75%. How much does he have invested at each rate? Problem B Jill has 800 L of acid solution. She obtained it by mixing some 5% acid with some 10% acid. Her final mixture of 800 L is 8.75% acid. How much of each of the 5% and 10% solutions did she use to get her final mixture? In Problem A, let x represent the amount invested at 5% interest, and in Problem B, let y represent the amount of 5% acid used. Work Exercises 65–68 in order. 65. (a) Write an expression in x that represents the amount of money Jack invested at 10% in Problem A. (b) Write an expression in y that represents the amount of 10% acid solution Jill used in Problem B. 66. (a) Write expressions that represent the amount of interest Jack earns per year at 5% and at 10%. (b) Write expressions that represent the amount of pure acid in Jill’s 5% and 10% acid solutions. 67. (a) The sum of the two expressions in part (a) of Exercise 66 must equal the total amount of interest earned in one year. Write an equation representing this fact. (b) The sum of the two expressions in part (b) of Exercise 66 must equal the amount of pure acid in the final mixture. Write an equation representing this fact. 68. (a) Solve Problem A.

(b) Solve Problem B.

(c) Explain the similarities between the processes used in solving Problems A and B.

81

Linear Equations, Inequalities, and Applications

PREVIEW EXERCISES Solve each problem. See Section 2. 69. Use d = rt to find d if r = 50 and t = 4. 70. Use P = 2L + 2W to find P if L = 10 and W = 6. 71. Use P = a + b + c to find a if b = 13, c = 14, and P = 46. 72. Use a = 12 h1b + B2 to find h if a = 156, b = 12, and B = 14.

STUDY

SKILLS

Taking Lecture Notes Study the set of sample math notes given here.

N Use a new page for each day’s lecture. N Include the date and title of the day’s lecture topic. N Skip lines and write neatly to make reading easier. N Include cautions and warnings to emphasize common errors to avoid.

N Mark important concepts with stars, underlining, circling, boxes, etc.

N Use two columns, which allows an example and its explanation to be close together.

N Use brackets and arrows to clearly show steps, related material, etc. With a partner or in a small group, compare lecture notes. 1. What are you doing to show main points in your notes (such as boxing, using stars or capital letters, etc.)? 2. In what ways do you set off explanations from worked problems and subpoints (such as indenting, using arrows, circling, etc.)? 3. What new ideas did you learn by examining your classmates’ notes? 4. What new techniques will you try in your note taking?

82

Translating Words to Expression s Sept. 1 and Equations Problem solving: key words or phr ases translate to algebraic expressions. Caution Sub traction is not com mutative; the order does matter. Examples: 10 less than a number a number sub tracted from 10 10 minus a number A phrase (part of a sentence) algebraic expression Note difference

Correct x –10 10 – x 10 – x

Wrong 10 – x x –10 x –10

A sentence equation with = sign

No equal sign in an expression. Equation has an equal sign. 3x + 2 3x + 2 = 14

Pay close attention to exact wor ding of the sentence; watch for commas. The quotient of a number and the number plus 4 is 28. x x+4 = 28 The quotient of a number and 4, plus the number, is 28. x +x 4 = 28 Commas separate this from division par t

Linear Equations, Inequalities, and Applications

4

Further Applications of Linear Equations

OBJECTIVES 1

2

3

Solve problems about different denominations of money. Solve problems about uniform motion. Solve problems about angles.

NOW TRY EXERCISE 1

Steven Danielson has a collection of 52 coins worth $3.70. His collection contains only dimes and nickels. How many of each type of coin does he have?

OBJECTIVE 1

Solve problems about different denominations of money.

PROBLEM-SOLVING HINT

In problems involving money, use the following basic fact. number of monetary total monetary : denomination ⴝ units of the same kind value 30 dimes have a monetary value of 301$0.102 = $3.00. Fifteen 5-dollar bills have a value of 151$52 = $75.

EXAMPLE 1

Solving a Money Denomination Problem

For a bill totaling $5.65, a cashier received 25 coins consisting of nickels and quarters. How many of each denomination of coin did the cashier receive? Step 1 Read the problem. The problem asks that we find the number of nickels and the number of quarters the cashier received. Step 2 Assign a variable. Then organize the information in a table. Let

x = the number of nickels.

Then 25 - x = the number of quarters. Number of Coins

Denomination

Value

Nickels

x

0.05

0.05x

Quarters

25 - x

0.25

0.25125 - x2 5.65

Total

Step 3 Write an equation from the last column of the table. 0.05x + 0.25125 - x2 = 5.65 Step 4 Solve. 0.05x + 0.25125 - x2 = 5.65 5x + 25125 - x2 = 565 Move decimal points 2 places to the right.

5x + 625 - 25x = 565 - 20x = - 60 x = 3

Multiply by 100. Distributive property Subtract 625. Combine like terms. Divide by - 20.

Step 5 State the answer. The cashier has 3 nickels and 25 - 3 = 22 quarters. Step 6 Check. The cashier has 3 + 22 = 25 coins, and the value of the coins is $0.05132 + $0.251222 = $5.65,

NOW TRY ANSWER 1. 22 dimes; 30 nickels

as required.

NOW TRY

CAUTION Be sure that your answer is reasonable when you are working problems like Example 1. Because you are dealing with a number of coins, the correct answer can be neither negative nor a fraction.

83

Linear Equations, Inequalities, and Applications

OBJECTIVE 2

Solve problems about uniform motion.

PROBLEM-SOLVING HINT

Uniform motion problems use the distance formula d = rt. When rate (or speed) is given in miles per hour, time must be given in hours. Draw a sketch to illustrate what is happening. Make a table to summarize given information.

NOW TRY EXERCISE 2

Two trains leave a city traveling in opposite directions. One travels at a rate of 80 km per hr and the other at a rate of 75 km per hr. How long will it take before they are 387.5 km apart?

EXAMPLE 2

Solving a Motion Problem (Motion in Opposite Directions)

Two cars leave the same place at the same time, one going east and the other west. The eastbound car averages 40 mph, while the westbound car averages 50 mph. In how many hours will they be 300 mi apart? Step 1 Read the problem. We are looking for the time it takes for the two cars to be 300 mi apart. Step 2 Assign a variable. A sketch shows what is happening in the problem. The cars are going in opposite directions. See FIGURE 6 . 50 mph

40 mph Starting point

W

E

Total distance  300 mi FIGURE 6

Let x represent the time traveled by each car, and summarize the information of the problem in a table. Rate

Time

Distance

Eastbound Car

40

x

40x

Westbound Car

50

x

50x 300

Fill in each distance by multiplying rate by time, using the formula d = rt. The sum of the two distances is 300.

Step 3 Write an equation. The sum of the two distances is 300. 40x + 50x = 300 Step 4 Solve.

90x = 300 x =

10 300 = 90 3

Combine like terms. Divide by 90; lowest terms

1 Step 5 State the answer. The cars travel 10 3 = 3 3 hr, or 3 hr, 20 min.

Step 6 Check. The eastbound car traveled 40 A 10 3 B = traveled 50 A 10 3 B =

500 3

mi, for a total distance of

400 3 mi. The westbound 400 500 900 3 + 3 = 3 = 300

as required.

car mi,

NOW TRY

CAUTION It is a common error to write 300 as the distance traveled by each car in Example 2. Three hundred miles is the total distance traveled.

NOW TRY ANSWER 2. 2 12 hr

84

As in Example 2, in general, the equation for a problem involving motion in opposite directions is of the following form. partial distance ⴙ partial distance ⴝ total distance

Linear Equations, Inequalities, and Applications

NOW TRY EXERCISE 3

Michael Good can drive to work in 12 hr. When he rides his bicycle, it takes 1 12 hours. If his average rate while driving to work is 30 mph faster than his rate while bicycling to work, determine the distance that he lives from work.

EXAMPLE 3

Solving a Motion Problem (Motion in the Same Direction)

Jeff can bike to work in 34 hr. When he takes the bus, the trip takes 14 hr. If the bus travels 20 mph faster than Jeff rides his bike, how far is it to his workplace? Step 1 Read the problem. We must find the distance between Jeff’s home and his workplace. Step 2 Assign a variable. Although the problem asks for a distance, it is easier here to let x be Jeff’s rate when he rides his bike to work. Then the rate of the bus is x + 20.

#

3 3 = x. 4 4

For the trip by bike,

d = rt = x

For the trip by bus,

d = rt = 1x + 202

#

1 1 = 1x + 202. 4 4

Summarize this information in a table. Rate

Time

Bike

x

3 4

Bus

x + 20

1 4

Distance 3 x 4 1 1x + 202 4

Same

Step 3 Write an equation. The key to setting up the correct equation is to understand that the distance in each case is the same. See FIGURE 7 . Workplace

Home

FIGURE 7

1 3 x = 1x + 202 4 4 Step 4 Solve.

3 1 4 a x b = 4a b 1x + 202 4 4

The distance is the same in each case. Multiply by 4.

3x = x + 20

Multiply; 1x = x

2x = 20

Subtract x.

x = 10

Divide by 2.

Step 5 State the answer. The required distance is d =

3 3 30 x = 1102 = = 7.5 mi. 4 4 4

Step 6 Check by finding the distance using d =

The same result

1 1 30 1x + 202 = 110 + 202 = = 7.5 mi. 4 4 4 NOW TRY

NOW TRY ANSWER 3. 22.5 mi

As in Example 3, the equation for a problem involving motion in the same direction is usually of the following form. one distance ⴝ other distance 85

Linear Equations, Inequalities, and Applications

PROBLEM-SOLVING HINT

In Example 3, it was easier to let the variable represent a quantity other than the one that we were asked to find. It takes practice to learn when this approach works best.

OBJECTIVE 3 Solve problems about angles. An important result of Euclidean geometry (the geometry of the Greek mathematician Euclid) is that the sum of the angle measures of any triangle is 180°. This property is used in the next example. NOW TRY EXERCISE 4

EXAMPLE 4

Find the value of x, and determine the measure of each angle. (3x – 36)°

Finding Angle Measures

Find the value of x, and determine the measure of each angle in FIGURE 8 . Step 1 Read the problem. We are asked to find the measure of each angle. Step 2 Assign a variable. Let x = the measure of one angle.

x° (x + 11)°

(x + 20)°

Step 3 Write an equation. The sum of the three measures shown in the figure must be 180°.

Step 4 Solve.

- x + 230 = 180 - x = - 50 x = 50

(210 – 3x)°

x°

x + 1x + 202 + 1210 - 3x2 = 180

FIGURE 8

Combine like terms. Subtract 230. Multiply by - 1.

Step 5 State the answer. One angle measures 50°. The other two angles measure x + 20 = 50 + 20 = 70° and NOW TRY ANSWER

210 - 3x = 210 - 31502 = 60°.

Step 6 Check. Since 50° + 70° + 60° = 180°, the answers are correct. NOW TRY

4. 41°, 52°, 87°

4 EXERCISES Concept Check

Solve each problem.

1. What amount of money is found in a coin hoard containing 14 dimes and 16 quarters? 2. The distance between Cape Town, South Africa, and Miami is 7700 mi. If a jet averages 550 mph between the two cities, what is its travel time in hours? 3. Tri Phong traveled from Chicago to Des Moines, a distance of 300 mi, in 10 hr. What was his rate in miles per hour? 4. A square has perimeter 80 in. What would be the perimeter of an equilateral triangle whose sides each measure the same length as the side of the square? Concept Check

Answer the questions in Exercises 5–8.

5. Read over Example 3 in this section. The solution of the equation is 10. Why is 10 mph not the answer to the problem?

86

Linear Equations, Inequalities, and Applications

6. Suppose that you know that two angles of a triangle have equal measures and the third angle measures 36°. How would you find the measures of the equal angles without actually writing an equation? 7. In a problem about the number of coins of different denominations, would an answer that is a fraction be reasonable? Would a negative answer be reasonable? 8. In a motion problem the rate is given as x mph and the time is given as 10 min. What variable expression represents the distance in miles? Solve each problem. See Example 1. 9. Otis Taylor has a box of coins that he uses when he plays poker with his friends. The box currently contains 44 coins, consisting of pennies, dimes, and quarters. The number of pennies is equal to the number of dimes, and the total value is $4.37. How many of each denomination of coin does he have in the box? 10. Nana Nantambu found some coins while looking under her sofa pillows. There were equal numbers of nickels and quarters and twice as many half-dollars as quarters. If she found $2.60 in all, how many of each denomination of coin did she find?

Number of Coins

Denomination

Value

x

0.01

0.01x

x 0.25 4.37

Number of Coins

Denomination

Value

x

0.05

0.05x

Total

x 2x

0.50 2.60

Total

Beth Anderson

11. In Canada, $1 and $2 bills have been replaced by coins. The $1 coins are called “loonies” because they have a picture of a loon (a well-known Canadian bird) on the reverse, and the $2 coins are called “toonies.” When Marissa returned home to San Francisco from a trip to Vancouver, she found that she had acquired 37 of these coins, with a total value of 51 Canadian dollars. How many coins of each denomination did she have?

Photodisc/Getty Images

12. Dan Ulmer works at an ice cream shop. At the end of his shift, he counted the bills in his cash drawer and found 119 bills with a total value of $347. If all of the bills are $5 bills and $1 bills, how many of each denomination were in his cash drawer? 13. Dave Bowers collects U.S. gold coins. He has a collection of 41 coins. Some are $10 coins, and the rest are $20 coins. If the face value of the coins is $540, how many of each denomination does he have?

14. In the 19th century, the United States minted two-cent and three-cent pieces. Frances Steib has three times as many three-cent pieces as two-cent pieces, and the face value of these coins is $2.42. How many of each denomination does she have? 15. In 2010, general admission to the Art Institute of Chicago cost $18 for adults and $12 for children and seniors. If $22,752 was collected from the sale of 1460 general admission tickets, how many adult tickets were sold? (Source: www.artic.edu) 16. For a high school production of Annie Get Your Gun, student tickets cost $5 each while nonstudent tickets cost $8. If 480 tickets were sold for the Saturday night show and a total of $2895 was collected, how many tickets of each type were sold?

87

Jeff Gross/Getty Images Sport

Linear Equations, Inequalities, and Applications

In Exercises 17–20, find the rate on the basis of the information provided. Use a calculator and round your answers to the nearest hundredth. All events were at the 2008 Summer Olympics in Beijing, China. (Source: World Almanac and Book of Facts.) Event

17. 18. 19. 20.

Participant

Distance

Time 12.54 sec

100-m hurdles, women

Dawn Harper, USA

100 m

400-m hurdles, women

Melanie Walker, Jamaica

400 m

52.64 sec

400-m hurdles, men

Angelo Taylor, USA

400 m

47.25 sec

400-m run, men

LaShawn Merritt, USA

400 m

43.75 sec

Solve each problem. See Examples 2 and 3. 21. Two steamers leave a port on a river at the same time, traveling in opposite directions. Each is traveling 22 mph. How long will it take for them to be 110 mi apart? Rate

Time

First Steamer

22. A train leaves Kansas City, Kansas, and travels north at 85 km per hr. Another train leaves at the same time and travels south at 95 km per hr. How long will it take before they are 315 km apart?

Distance

t

Second Steamer

First Train

22

Rate

Time

85

t

Distance

Second Train 110

23. Mulder and Scully are driving to Georgia to investigate “Big Blue,” a giant reptile reported in one of the local lakes. Mulder leaves the office at 8:30 A.M. averaging 65 mph. Scully leaves at 9:00 A.M., following the same path and averaging 68 mph. At what time will Scully catch up with Mulder? Rate

Time

315

24. Lois and Clark, two elderly reporters, are covering separate stories and have to travel in opposite directions. Lois leaves the Daily Planet building at 8:00 A.M. and travels at 35 mph. Clark leaves at 8:15 A.M. and travels at 40 mph. At what time will they be 140 mi apart?

Distance

Rate

Mulder

Lois

Scully

Clark

25. It took Charmaine 3.6 hr to drive to her mother’s house on Saturday morning for a weekend visit. On her return trip on Sunday night, traffic was heavier, so the trip took her 4 hr. Her average rate on Sunday was 5 mph slower than on Saturday. What was her average rate on Sunday? 26. Sharon Kobrin commutes to her office by train. When she walks to the train station, it takes her 40 min. When she rides her bike, it takes her 12 min. Her average walking rate is 7 mph less than her average biking rate. Find the distance from her house to the train station.

Time

Distance

Rate

Time

Distance

Rate

Time

Distance

Saturday Sunday

Walking Biking

27. Johnny leaves Memphis to visit his cousin, Anne Hoffman, who lives in the town of Hornsby, Tennessee, 80 mi away. He travels at an average rate of 50 mph. One-half hour later, Anne leaves to visit Johnny, traveling at an average rate of 60 mph. How long after Anne leaves will it be before they meet? 28. On an automobile trip, Laura Iossi maintained a steady rate for the first two hours. Rushhour traffic slowed her rate by 25 mph for the last part of the trip. The entire trip, a distance of 125 mi, took 2 21 hr. What was her rate during the first part of the trip?

88

Linear Equations, Inequalities, and Applications

Find the measure of each angle in the triangles shown. See Example 4. 29.

30.

(x + 15)°

(2x – 120)°

(x + 5)°

( 12 x + 15)°

(10x – 20)°

(x – 30)°

31.

32.

(x + 61)°

(9x – 4)° x° (2x + 7)°

(3x + 7)°

(4x + 1)°

RELATING CONCEPTS

EXERCISES 33–36

FOR INDIVIDUAL OR GROUP WORK

Consider the following two figures. Work Exercises 33–36 in order.

2x°

x°

60°

60° FIGURE A

y°

FIGURE B

33. Solve for the measures of the unknown angles in FIGURE A . 34. Solve for the measure of the unknown angle marked y° in FIGURE B . 35. Add the measures of the two angles you found in Exercise 33. How does the sum compare to the measure of the angle you found in Exercise 34? 36. Based on the answers to Exercises 33–35, make a conjecture (an educated guess) about the relationship among the angles marked 1 , 2 , and 3 in the figure shown below. 2 1

3

In Exercises 37 and 38, the angles marked with variable expressions are called vertical angles. It is shown in geometry that vertical angles have equal measures. Find the measure of each angle. 37.

38. (7x + 17)° (9 – 5x)°

(25 – 3x)°

(8x + 2)°

89

Linear Equations, Inequalities, and Applications

39. Two angles whose sum is 90° are called complementary angles. Find the measures of the complementary angles shown in the figure.

40. Two angles whose sum is 180° are called supplementary angles. Find the measures of the supplementary angles shown in the figure.

(5x – 1)°

(3x + 5)°

(5x + 15)°

(2x)°

Consecutive Integer Problems

Consecutive integers are integers that follow each other in counting order, such as 8, 9, and 10. Suppose we wish to solve the following problem: Find three consecutive integers such that the sum of the first and third, increased by 3, is 50 more than the second. Let x = the first of the unknown integers, x + 1 = the second, and x + 2 = the third. We solve the following equation. Sum of the first and third

increased by 3

is

x + 1x + 22

+ 3

=

50 more than the second.

1x + 12 + 50

2x + 5 = x + 51 x = 46 The solution of this equation is 46, so the first integer is x = 46, the second is x + 1 = 47, and the third is x + 2 = 48. The three integers are 46, 47, and 48. Check by substituting these numbers back into the words of the original problem.

Solve each problem involving consecutive integers. 41. Find three consecutive integers such that the sum of the first and twice the second is 17 more than twice the third. 42. Find four consecutive integers such that the sum of the first three is 54 more than the fourth. 43. If I add my current age to the age I will be next year on this date, the sum is 103 yr. How old will I be 10 yr from today? 44. Two pages facing each other in this book have 193 as the sum of their page numbers. What are the two page numbers? 45. Find three consecutive even integers such that the sum of the least integer and the middle integer is 26 more than the greatest integer. 46. Find three consecutive even integers such that the sum of the least integer and the greatest integer is 12 more than the middle integer. 47. Find three consecutive odd integers such that the sum of the least integer and the middle integer is 19 more than the greatest integer. 48. Find three consecutive odd integers such that the sum of the least integer and the greatest integer is 13 more than the middle integer.

90

Linear Equations, Inequalities, and Applications

PREVIEW EXERCISES Graph each interval. 49. 14, q2

50. 1- q, - 24

51. 1- 2, 62

52. 3- 1, 64

SUMMARY EXERCISES on Solving Applied Problems Solve each problem. 1. The length of a rectangle is 3 in. more than its width. If the length were decreased by 2 in. and the width were increased by 1 in., the perimeter of the resulting rectangle would be 24 in. Find the dimensions of the original rectangle.

2. A farmer wishes to enclose a rectangular region with 210 m of fencing in such a way that the length is twice the width and the region is divided into two equal parts, as shown in the figure. What length and width should be used?

x+3

Width

x Length

3. After a discount of 46%, the sale price for a Harry Potter Paperback Boxed Set (Books 1–7) by J. K. Rowling was $46.97. What was the regular price of the set of books to the nearest cent? (Source: www.amazon.com) 4. An electronics store offered a Blu-ray player for $255, the sale price after the regular price was discounted 40%. What was the regular price? 5. An amount of money is invested at 4% annual simple interest, and twice that amount is invested at 5%. The total annual interest is $112. How much is invested at each rate? 6. An amount of money is invested at 3% annual simple interest, and $2000 more than that amount is invested at 4%. The total annual interest is $920. How much is invested at each rate? 8. Before being overtaken by Avatar, the two all-time top-grossing American movies were Titanic and The Dark Knight. Titanic grossed $67.5 million more than The Dark Knight. Together, the two films brought in $1134.1 million. How much did each movie gross? (Source: www.imdb.com)

Warner Brothers/Everett Collection

Kolvenbach/Alamy Images

7. LeBron James of the Cleveland Cavaliers was the leading scorer in the NBA for the 2007–2008 season, and Dwyane Wade was the leading scorer for the 2008–2009 season. Together, they scored 4636 points, with James scoring 136 points fewer than Wade. How many points did each of them score?

(continued)

91

Linear Equations, Inequalities, and Applications

9. Atlanta and Cincinnati are 440 mi apart. John leaves Cincinnati, driving toward Atlanta at an average rate of 60 mph. Pat leaves Atlanta at the same time, driving toward Cincinnati in her antique auto, averaging 28 mph. How long will it take them to meet? Pat ! %&

Atlanta

John Cincinnati 440 mi

10. Deriba Merga from Ethiopia won the 2009 men’s Boston Marathon with a winning time of 2 hr, 8 min, 42 sec, or 2.145 hr. The women’s race was won by Salina Kosgei from Kenya, whose winning time was 2 hr, 32 min, 16 sec, or 2.538 hr. Kosgei’s average rate was 1.9 mph slower than Merga’s. Find the average rate for each runner, to the nearest hundredth. (Source: World Almanac and Book of Facts.) 11. A pharmacist has 20 L of a 10% drug solution. How many liters of 5% solution must be added to get a mixture that is 8%? 12. A certain metal is 20% tin. How many kilograms of this metal must be mixed with 80 kg of a metal that is 70% tin to get a metal that is 50% tin? 13. A cashier has a total of 126 bills in fives and tens. The total value of the money is $840. How many of each denomination of bill does he have? 14. The top-grossing domestic movie in 2008 was The Dark Knight. On the opening weekend, one theater showing this movie took in $20,520 by selling a total of 2460 tickets, some at $9 and the rest at $7. How many tickets were sold at each price? (Source: Variety.) 15. Find the measure of each angle.

16. Find the measure of each marked angle.

(6x – 50)° x°

(10x + 7)° (7x + 3)°

(x – 10)°

17. The sum of the least and greatest of three consecutive integers is 32 more than the middle integer. What are the three integers? 18. If the lesser of two consecutive odd integers is doubled, the result is 7 more than the greater of the two integers. Find the two integers. 19. The perimeter of a triangle is 34 in. The middle side is twice as long as the shortest side. The longest side is 2 in. less than three times the shortest side. Find the lengths of the three sides.

x inches

20. The perimeter of a rectangle is 43 in. more than the length. The width is 10 in. Find the length of the rectangle.

92

Linear Equations, Inequalities, and Applications

5

Linear Inequalities in One Variable

OBJECTIVES

We use interval notation to write solution sets of inequalities.

1

●

2

3

4

Solve linear inequalities by using the addition property. Solve linear inequalities by using the multiplication property. Solve linear inequalities with three parts. Solve applied problems by using linear inequalities.

●

A parenthesis indicates that an endpoint is not included. A square bracket indicates that an endpoint is included.

We summarize the various types of intervals here. Type of Interval

Set-Builder Notation

Open

Interval Notation

5x | a 6 x 6 b6

1a, b2

5x | a … x … b6

3a, b4

5x | a … x 6 b6

3a, b2

5x | a 6 x … b6

1a, b4

5x | x 6 a or x 7 b6

1 - q, a2 ´ 1b, q2

5x | x 7 a6

1a, q2

interval Closed interval Half-open (or half-closed) interval Disjoint interval*

Graph

a

b

a

b

a

b

a

b

a

b a

Infinite

5x | x Ú a6

3a, q2

5x | x 6 a6

1 - q, a2

5x | x … a6

1 - q, a4

5x | x is a real number6

1 - q, q2

a

interval

a a 0

NOTE A parenthesis is always used next to an infinity symbol, - q or q.

An inequality says that two expressions are not equal. Solving inequalities is similar to solving equations. Linear Inequality in One Variable

A linear inequality in one variable can be written in the form Ax ⴙ BC, or Ax ⴙ B » C, where A , B, and C are real numbers, with A Z 0. x + 5 6 2,

x - 3 Ú 5,

and 2k + 5 … 10

Examples of linear inequalities

*We will work with disjoint intervals in Section 6 when we study set operations and compound inequalities.

93

Linear Equations, Inequalities, and Applications

OBJECTIVE 1 Solve linear inequalities by using the addition property. We solve an inequality by finding all numbers that make the inequality true. Usually, an inequality has an infinite number of solutions. These solutions, like solutions of equations, are found by producing a series of simpler related equivalent inequalities. Equivalent inequalities are inequalities with the same solution set. We use two important properties to produce equivalent inequalities. The first is the addition property of inequality.

Addition Property of Inequality

For all real numbers A, B, and C, the inequalities A 6

x

(b) $2558 million y

0

x

2x + y ≤ 1 and x ≥ 2y

213

Graphs, Linear Equations, and Functions 40.

0

42. domain: 5- 4, 16; range:

41. D

y

52, - 2, 5, - 56; not a function

x 2

x

43. domain:

59, 11, 4, 17, 256; range: 532, 47, 69, 146; function 44. domain: 3- 4, 44; range: 30, 24; function

or y

5.

2

45. domain: 1- q , 04; range: 1- q , q 2; not a function

46. function;

domain: 1- q , q 2; linear function 47. not a function; domain:

1- q, q2

48. function; domain: 1- q , q 2

49. function; domain:

C - 74, q B 50. not a function; domain: 30, q 2 51. function;

domain: 1- q , 62 ´ 16, q 2 55. - 2k 2 + 3k - 6

52. - 6

53. - 8.52

56. ƒ1x2 = 2x 2; 18

57. C

6. It is a vertical line.

by about 929 each year from 1980 to 2008. 10. (a) y = - 5x + 19 (b) 5x + y = 19 11. (a) y = 14 (b) y = 14 12. (a) y = - 12 x + 2 (b) x + 2y = 4

60. - 32

slope is negative. 64. ƒ1x2 =

- 32 x

+

7 2

65.

67. 0

7 3

59. Because it falls from left to right, the 61. - 32 ; 32

- 17 2

62. A 73, 0 B

66. x =

68. E 73 F

69.

23 3 A 73, q B

63. A 0, 72 B

70. A - q , 73 B

19.

y 3x – 2y > 6

0

0 2

x

–3

20. D

y

21. D

3

x

y < 2x – 1 and x–y  % is less than or equal to "!" can be written as $>  % Ÿ "!.

87.

"&B  $ is not equal to !" can be written as &B  $ Á !.

89.

"> is between $ and &" can be written as $  >  &.

91.

"$B is between $ and %, including $ and excluding %" can be written as $ Ÿ $B  %.

93.

'  (  $ '  "!

105. eB ± !  B  $Þ&f includes all numbers between ! and $Þ&, written !ß $Þ& . Place parentheses at ! and $Þ& since these numbers are not elements of the set. The graph goes from ! to $Þ&.

107. eB ± # Ÿ B Ÿ (f includes all numbers from # to (, written Ò#ß (Ó. Place brackets at # and ( since these numbers are elements of the set. The graph goes from # to (.

109. eB ± %  B Ÿ $f includes all numbers between % and $, excluding %, but including $, written Ð%ß $Ó. Place a parenthesis at % and a bracket at $ to show that % is not included, but $ is. The graph goes from % to $.

111. eB ± !  B Ÿ $f includes all numbers between ! and $, excluding !, but including $, written Ð!ß $Ó. Place a parenthesis at ! and a bracket at $ to show that ! is not included, but $ is. The graph goes from ! to $.

?

True

113. 2006: IA œ "$,)"", OH œ (&!(, PA œ '')(

The last statement is true since ' is to the left of "! on a number line. 95.

97. 99.

In 2006, Iowa (IA), Ohio (OH), and Pennsylvania (PA) had production greater than '!!! million eggs.

?

#•&   %  ' "!   "!

True

k$k   $ $   $

True

115. 2006: TX œ B œ &!$*, OH œ C œ (&!( Since &!$*  (&!(, B  C is true.

?

)  k'k )  ' ?

2

Operations on Real Numbers

2 Now Try Exercises False

101. eB ± B  "f includes all numbers greater than ", written "ß ∞ . Place a parenthesis at " since " is not an element of the set. The graph extends from " to the right.

N1. (a) %  Ð*Ñ œ Ð%  *Ñ œ "$ (b) (Þ#&  Ð$Þ&(Ñ œ Ð(Þ#&  $Þ&(Ñ œ "!Þ)# $ (c)  #&  Ð "! Ñ œ Ð &# 

œ 103. eB ± B Ÿ 'f includes all numbers less than or equal to ', written Ð∞ß 'Ó. Place a bracket at ' since ' is an element of the set. The graph extends from ' to the left.

œ

$ "! Ñ % $ Ð "!  "! Ñ (  "!

N2. (a) "&  ( Subtract the lesser absolute value from the greater absolute value Ð"&  (Ñ, and take the sign of the larger. "&  ( œ Ð"&  (Ñ œ )

721

Review of the Real Number System (b) %Þ'  Ð#Þ)Ñ œ %Þ'  #Þ) œ "Þ)

7.

& # &•( #•* $& ") (c)   œ   œ  * ( *•( (•* '$ '$ $& ") "( œ Œ   œ  '$ '$ '$

The difference between two negative numbers is negative if the number with lesser absolute value is subtracted from the one with greater absolute value . For example, "&  Ð$Ñ œ "#.

9.

The product of two numbers with different signs is negative . For example, &Ð"&Ñ œ (&.

11.

'  Ð"$Ñ œ Ð'  "$Ñ œ "*

13.

"$  Ð%Ñ œ "$  % œ *

15.

 ($ 

17.

The difference between #Þ$ and !Þ%& is "Þ)&. The number with the greater absolute value, #Þ$, is negative, so the answer is negative. Thus, #Þ$  !Þ%& œ "Þ)&Þ

19.

'  & œ '  Ð&Ñ œ Ð'  &Ñ œ ""

21.

)  Ð"$Ñ œ )  "$ œ #"

23.

"'  Ð$Ñ œ "'  $ œ "$

25.

"#Þ$"  Ð#Þ"$Ñ œ "#Þ$"  #Þ"$ œ Ð"#Þ$"  #Þ"$Ñ œ "!Þ")

27.

* "!

N3. (a) %  "" œ %  Ð""Ñ œ "& (b) &Þ'(  Ð#Þ$%Ñ œ &Þ'(  #Þ$% œ Ð&Þ'(  #Þ$%Ñ œ $Þ$$ (c)

% $ % $ %•& $•*  œ  œ  * & * & *•& &•* #! #( ( œ  œ %& %& %&

N4. %  Ð#  (Ñ  "# œ %  Ð*Ñ  "# œ %  *  "# œ &  "# œ ( N5. k"#  Ð(Ñk œ k"#  (k œ k"*k œ "* N6. (a) $Ð"!Ñ œ $! The numbers have the same sign, so the product is positive. (b) !Þ(Ð"Þ#Ñ œ !Þ)% The numbers have different signs, so the product is negative. (c)  N7. (a) (b)

) ) • $ • "" Ð$$Ñ œ  œ #% "" ""

"! &

 "! $ $ )

œ "!Ð "& Ñ œ # "! $ "! ) œ ƒ œ • $ ) $ $ "! • ) )! œ œ $•$ *

( #•$•( ' $ ' (c)  ƒ Œ  œ  • Œ  œ $ &•$ & ( & #•( "% œ œ & &

29. 31.

3. 5.

722

The sum of a positive number and a negative number is ! if the numbers are additive inverses . For example, %  Ð%Ñ œ !.

 Ð %$ Ñ œ

* "!

* "#



% $

"* œ  "#

œ

#( $!



%! $!

œ

k)  'k œ k"%k œ Ð"%Ñ œ "%

'( $!

k%  *k œ k&k œ &

#  k%k œ #  % œ #  Ð%Ñ œ '

35.

(  &  * œ Ð(  &Ñ  * œ #  * œ ""

37.

'  Ð#Ñ  ) œ '  #  ) œ )  ) œ "'

39.

*  %  Ð$Ñ  ' œ Ð*  %Ñ  $  ' œ "$  $  ' œ "!  ' œ %

41.

)  Ð"#Ñ  Ð#  'Ñ œ )  "#  Ð%Ñ œ )  "#  % œ%% œ)

43.

!Þ$)#  %  !Þ' œ $Þ'")  !Þ' œ $Þ!")

45.

Ð &%  #$ Ñ 

" '

œ Ð "& "#  œ  #$ "# 

The sum of two negative numbers is a negative number. For example, (  Ð#"Ñ œ #). The sum of a positive number and a negative number is positive if the positive number has the greater absolute value. For example, "&  Ð#Ñ œ "$.

#) œ  "# 

33.

2 Section Exercises 1.

$ %

) "# Ñ # "#

#  Ð "# Ñ

( œ  #" "# , or  %

47.

 $%  Ð "#  $) Ñ œ  ')  Ð %)  $) Ñ œ  ')  œ  ()

" )

Review of the Real Number System 49.

51.

k""k  k&k  k(k  k#k œ ""  &  (  # œ'(# œ "  # œ "

75.

#% %

E œ % and F œ #. The distance between E and F is the absolute value of their difference. k%  #k œ k'k œ '

53.

H œ ' and J œ "# . The distance between H and J is the absolute value of their difference. "¸ ¸'  "# ¸ œ ¸ "# #  # "$ œ ¸ # ¸

œ

55.

57.

"$ # ,

It is true for multiplication (and division). It is false for addition and subtraction when the number to be subtracted has the lesser absolute value. A more precise statement is, "The product or quotient of two negative numbers is positive." The product of two numbers with different signs is negative, so

The product of two numbers with the same sign is positive, so )Ð&Ñ œ %!.

61.

The product of two numbers with the same sign is positive, so "!Ð "& Ñ œ # • & • Ð "& Ñ œ #.

63.

The product of two numbers with different signs is negative, so $ % Ð"'Ñ

65.

77.

œ  $% • % • % œ "#Þ

The product of two numbers with the same sign is positive, so

The product of two numbers with the same sign is positive, so $ #% $•$•)  Œ  œ œ ". ) * )•*

" " œ "!! • #& œ % • #& • #& œ %Þ

79.

! )

81.

Division by ! is undefined, so

83.

The quotient of two nonzero real numbers with the same sign is positive, so 

œ ! • Ð ") Ñ œ !

87.

#(Þ(# "$Þ#

œ

œ #Þ"

"!! "!! " œ " œ "!! ƒ œ "!! • "!! !Þ!" "!! "!! œ "!,!!!

91.

" '

 Ð *( Ñ œ

93.

 "* 

95.

 $) 

97.

(  $! 

( *

œ

$ ")



( "#

% œ  $' 

#" $'

œ

"( $'

& "#

* œ  #% 

"! #%

œ  "* #%

# %&



" '

$ "!



#" œ  *! 

% *!



œ

"( ")

#( *!

%% ## œ  *! œ  %&

) & #•%•& # # œ œ Œ  œ  #& "# &•&•$•% &•$ "&

101.

& * % &•$•$•#•# $ œ Œ Œ  œ ' "! & #•$•#•&•& &

103.

105.

"! (•#•& ( * ( ƒ Œ  œ • Œ  œ  * #•$•* ' "! ' (•& $& ) œ œ  , or " $•* #( #(  )* ) # ) " #•% % œ œ ƒ œ • œ *•# * # * " * #

107. )Þ'  $Þ(&" œ Ð)Þ'  $Þ(&"Ñ œ "#Þ$&" 109. Ð%Þ#ÑÐ"Þ%ÑÐ#Þ(Ñ œ Ð&Þ))ÑÐ#Þ(Ñ œ "&Þ)('

#Þ%Ð#Þ%&Ñ œ &Þ))

111. #%Þ)% ƒ ' œ %Þ"%

71.

$Þ%Ð$Þ"%Ñ œ "!Þ'('

113. #%*' ƒ Ð!Þ&#Ñ œ %)!!

73.

The quotient of two nonzero real numbers with different signs is negative, so œ "% • "# œ # • ( • "# œ (Þ

"% ")

99.

69.

"% #

is undefined.

"# % "# $ ƒ Œ  œ Œ  "$ $ "$ % $•%•$ * œ œ "$ • % "$

"# "$  %$

89.

& !

"! "# "! & #•&•& #& œ œ . ƒ Œ  œ • "( & "( "# "( • # • ' "!#

85.

& "# &•#•' '  Œ  œ œ . # #& #•&•& & 67.

œ #% • "% œ % • ' • "% œ 'Þ

The quotient of two nonzero real numbers with different signs is negative, so "!! #&

or ' "#

&Ð(Ñ œ $&Þ 59.

The quotient of two nonzero real numbers with the same sign is positive, so

115. "%Þ#$  *Þ)"  (%Þ'$  ")Þ("& œ %Þ%#  (%Þ'$  ")Þ("& œ (!Þ#"  ")Þ("& œ &"Þ%*&

723

Review of the Real Number System 117. To find the difference between these two temperatures, subtract the lowest temperature from the highest temperature. *!°  Ð##°Ñ œ *!°  ##° œ ""#° The difference is ""#°F. 119. %)Þ$&  $&Þ**  #!Þ!!  #)Þ&!  ''Þ#( œ "#Þ$'  #!Þ!!  #)Þ&!  ''Þ#( œ (Þ'%  #)Þ&!  ''Þ#( œ $'Þ"%  ''Þ#( œ $!Þ"$ His balance is $$!Þ"$. 121. $)#Þ%&  #&Þ"!  $%Þ&!  %&Þ!!  *)Þ"( œ %''Þ!# His balance is $%''Þ!#.

N2. (a) (# œ ( • ( œ %* (b) Ð(Ñ# œ Ð(ÑÐ(Ñ œ %* (c) (# œ Ð( • (Ñ œ %*

N3. (a) È"%% œ "# since the negative sign is outside the radical symbol. (b) É "!! * œ # Ð "! $ Ñ

"! $

N5. &#  "! ƒ &  k$  (k

œ &#  "! ƒ &  k%k œ &#  "! ƒ &  4

(b) %''Þ!#  $!!  #%Þ'' œ "*!Þ')

œ #&  "! ƒ &  4

His balance is $"*!Þ').

œ #&  #  4 œ #$  4 œ #(

The total loss for 2007–2010 was $%(& thousand. (b) ((  Ð")&Ñ œ ((  ")& œ #'#

is positive and

N4. "&  $ • %  # œ "&  "#  # œ$# œ&

123. (a) "%#  ##&  ")&  (( œ %(& N6.

È$'  % • $#

Multiply. Subtract. Add.

Work inside the absolute value bars. Subtract inside the absolute value bars. Take absolute value. Evaluate the power. Divide. Add. SubtractÞ

##  ) • $  "$

The difference between the profit or loss from 2009 to 2010 was $#'# thousand.

œ

'  %•* %  ) • $  "$

Evaluate powers and roots.

(c) ##&  Ð"%#Ñ œ ##&  "%# œ )$

œ

'  $' %  #%  "$

Multiply.

The difference between the profit or loss from 2007 to 2008 was $)$ thousand.

œ

$! "&

Add and subtract.

125. (a)

Year 2000 2010 2020 2030

Difference (in billions) $&$)  $%!* œ $"#* $*"'  $("! œ $#!' $"%(*  $"%!& œ $(% $#!%"  $#&%# œ $&!"

(b) The cost of Social Security will exceed revenue in 2030 by $&!" billion.

Exponents, Roots, and Order of Operations

3 Now Try Exercises N1. (a) Ð$ÑÐ$ÑÐ$Ñ œ Ð$Ñ$ (b) > • > • > • > • > œ >&

724

since

(c) È"%% is not a real number.

(a) To pay off the balance, his payment should be $%''Þ!#.

3

œ

"! $

"!! * .

Reduce.

œ#

N7. Let B œ %, C œ (, and D œ $'. B#  È D Ð%Ñ#  È$' œ $BC $Ð%ÑÐ(Ñ œ

"'  ' $Ð%ÑÐ(Ñ

œ

"'  ' "#Ð(Ñ

"'  ' )% "! & œ œ )% %# œ

Review of the Real Number System 35.

3 Section Exercises 1.

3.

(' œ Ð( • ( • ( • ( • ( • (Ñ œ ""(,'%*, whereas Ð(Ñ' œ Ð(ÑÐ(ÑÐ(ÑÐ(ÑÐ(ÑÐ(Ñ œ ""(,'%*.

37.

Thus, the original statement, (' œ Ð(Ñ' , is false. The following statement is true: (' œ Ð(' Ñ

43.

The statement "È#& is a positive number" is true. The symbol È" always gives a positive square root provided the radicand is positive.

5.

The statement "Ð'Ñ( is a negative number" is true. Ð'Ñ( gives an odd number of negative factors, so the product is negativeÞ

7.

The statement "The product of "! positive factors and "! negative factors is positive" is true. The product of an even number of negative factors is positive.

9.

The statement "In the exponential )& , ) is the base" is false. The base is ), not ). If the problem were written Ð)Ñ& , then ) would be the base.

11.

39. 41.

45. 47. 49.

51.

13.

"! • "! • "! • "! œ "!%

15.

$ $ $ $ $ % • % • % • % • %

17.

Ð*ÑÐ*ÑÐ*Ñ œ Ð*Ñ$

19.

D • D • D • D • D • D • D œ D(

21.

%# œ % • % œ "'

23.

!Þ#)$ œ Ð!Þ#)ÑÐ!Þ#)ÑÐ!Þ#)Ñ œ !Þ!#"*&#

25.

Ð "& Ñ$ œ

27.

Ð %& Ñ% œ Ð %& ÑÐ %& ÑÐ %& ÑÐ %& Ñ %•%•%•% #&' œ œ &•&•&•& '#&

29.

Ð&Ñ$ œ Ð&ÑÐ&ÑÐ&Ñ œ "#&

œ

" "#&

)

31.

Ð#Ñ œ Ð#ÑÐ#ÑÐ#ÑÐ#ÑÐ#ÑÐ#ÑÐ#ÑÐ#Ñ œ #&'

33.

$' œ Ð$ • $ • $ • $ • $ • $Ñ œ (#*

È%!! œ ÐÈ%!!Ñ œ Ð#!Ñ œ #! É "!! "#" œ

"! ""

since

"! ""

# is positive and Ð "! "" Ñ œ

"!! "#" Þ

È!Þ%* œ ÈÐ!Þ(Ñ# œ Ð!Þ(Ñ œ !Þ(

There is no real number whose square is negative, so È$' is not a real number. (a) È"%% œ "#; choice B

(b) È"%% is not a real number; choice C If B is a positive number, then ÈB is also a positive number, and ÈB is a negative number. Multiply. Add.

"#  $ • % œ "#  "# œ #%

55.

' • $  "# ƒ % œ ")  "# ƒ % œ ")  $ œ "&

Multiply. Divide. Subtract.

57.

"!  $! ƒ # • $ œ "!  "& • $ œ "!  %& œ &&

Divide. Multiply. Add.

59.

$Ð&Ñ#  Ð#ÑÐ)Ñ œ $Ð#&Ñ  Ð#ÑÐ)Ñ œ (&  "' œ *"

œ Ð $% Ñ&

" " " & • & • &

È"'* œ "$ since "$ is positive and "$# œ "'*Þ

53.

(c) Ð)Ñ# œ Ð)ÑÐ)Ñ œ '% (d) Ð)Ñ# œ Ð'%Ñ œ '%

È)" œ * since * is positive and *# œ )"Þ

(c) È"%% œ "#; choice A

(a) )# œ '% (b) )# œ Ð) • )Ñ œ '%

)% œ Ð) • ) • ) • )Ñ œ %!*'

61.

63.

&  ( • $  Ð#Ñ$ œ &  ( • $  Ð)Ñ œ &  #"  ) œ "'  ) œ )

Evaluate power. Multiply; change sign Subtract. Add.

(ŠÈ$'‹  Ð#ÑÐ$Ñ œ (Ð'Ñ  Ð#ÑÐ$Ñ œ %#  ' œ %)

65.

'k%  &k  #% ƒ $ œ 'k"k  #% ƒ $ œ 'Ð"Ñ  #% ƒ $ œ') œ #

Evaluate power. Multiply. Subtract.

Evaluate root. Multiply. Subtract.

Simplify within absolute value bars. Take absolute value. Multiply and divide. Subtract.

725

Review of the Real Number System 67.

k'  &kÐ)Ñ  $#

In Exercises 79–86, + œ $, , œ '%, and - œ '.

œ k""kÐ)Ñ  $# œ ""Ð)Ñ  $# œ ""Ð)Ñ  * œ ))  * œ (*

69.

71.

'  #$ Ð*Ñ 

& ) • "'

œ '  Ð'Ñ  "! œ !  "! œ "!

"%Ð #( Ñ ƒ Ð# • '  "!Ñ œ % ƒ Ð"#  "!Ñ œ%ƒ# œ#

73.

75.

Simplify within absolute value bars. Take absolute value. Evaluate power. Multiply. Add.

Ð&  È%ÑÐ## Ñ &  " Ð&  #ÑÐ%Ñ œ ' Ð$ÑÐ%Ñ œ ' "# œ ' œ # #Ð&Ñ  Ð$ÑÐ#Ñ )  $#  " "!  ' œ )  *  " % œ "" % œ !

79.

81. Multiply. Add. Subtract.

Multiply. Work inside parenthesesÞ Divide.

Evaluate root and power; subtract. Work inside parentheses. Multiply.

83.

85.

œ

"#  Ð#(Ñ #&'  Ð")Ñ

œ

"& "& œ #$) #$)

87.

AC  )B œ %Ð "# Ñ  )Ð $% Ñ œ#' œ)

89.

BC  C% œ  $% Ð "# Ñ  Ð "# Ñ% œ

Evaluate power; multiply. Add.

91.

93.

Work in numerator and denominator separately. 95.

Write in lowest terms.

" "' ' "  "'  "'

& œ  "'

A  #B  $C  D œ %  #Ð $% Ñ  $Ð "# Ñ  "Þ#& œ %  $#  $#  "Þ#& œ %  "Þ#& œ #Þ(&

Subtract.

&  $Ð#Ñ  ' #!  #" &'' œ " ( œ "

726

#-  +$ #Ð'Ñ  Ð$Ñ$ œ %,  '+ %Ð'%Ñ  'Ð$Ñ

œ  $) 

œ

œ (

%+$  #- œ %Ð$Ñ$  #Ð'Ñ œ %Ð#(Ñ  "# œ "!)  "# œ *'

In Exercises 87–94, A œ %, B œ  $% , C œ "# , and D œ "Þ#&.

Divide.

&  * &  $Œ ' ( *  ""  $ • ( &  $Ð "% ( Ñ  ' œ *  ""  #"

È,  -  + œ È'%  '  Ð$Ñ œ)'$ œ "%  $ œ "(

Since division by ! is undefined, the given expression is undefined.

77.

$+  È, œ $Ð$Ñ  È'% œ $Ð$Ñ  ) œ *  ) œ "

(Ð $% Ñ  *Ð "# Ñ (B  *C œ A % *  #"  #"  ") %  # % œ œ % % % $ $ % $ " $ œ % œ ƒ œ • œ % % " % % "' @ ‚ !Þ&%)&  %)&! ƒ "!!! ‚ $"Þ%% œ Ð"!!,!!! ‚ !Þ&%)&  %)&!Ñ ƒ "!!! ‚ $"Þ%% œ Ð&%,)&!  %)&!Ñ ƒ "!!! ‚ $"Þ%% œ &!,!!! ƒ "!!! ‚ $"Þ%% œ &! ‚ $"Þ%% œ "&(# The owner would pay $"&(# in property taxes.

Review of the Real Number System 97.

@ ‚ !Þ&%)&  %)&! ƒ "!!! ‚ $"Þ%% œ Ð#!!,!!! ‚ !Þ&%)&  %)&!Ñ ƒ "!!! ‚ $"Þ%% œ Ð"!*,(!!  %)&!Ñ ƒ "!!! ‚ $"Þ%% œ "!%,)&! ƒ "!!! ‚ $"Þ%% œ "!%Þ)& ‚ $"Þ%% œ $#*'Þ%)% ¸ $#*'

N4. (a) $Ð>  %Ñ  >  "& œ $>  "#  >  "& œ $>  >  "#  "& œ %>  #( (b) &BÐ'CÑ œ Ò&BÐ'ÑÓC œ Ò&ÐB • 'ÑÓC œ Ò&Ð'BÑÓC œ ÒÐ& • 'ÑBÓC œ Ð$!BÑC œ $!ÐBCÑ œ $!BC

The owner would pay $$#*' in property taxes. 99.

number of oz ‚ % alcohol ‚ !Þ!(& ƒ body weight in lb  hr of drinking ‚ !Þ!"& œ $' ‚ %Þ! ‚ !Þ!(& ƒ "$&  $ ‚ !Þ!"& œ "%% ‚ !Þ!(& ƒ "$&  $ ‚ !Þ!"& œ "!Þ) ƒ "$&  $ ‚ !Þ!"& œ !Þ!)  $ ‚ !Þ!"& œ !Þ!)  !Þ!%& œ !Þ!$&

101. "Þ*!*B  $(*" (a) "Þ*!*Ð"**'Ñ  $(*" ¸ $"*Þ% billion

(d) $%#Þ$ billion is more than twice $"*Þ% billion, so the amount spent on pets more than doubled from 1996 to 2008.

4

Properties of Real Numbers

4 Now Try Exercises N1. (a) #Ð$B  CÑ œ #Ò$B  ÐCÑÓ œ #Ð$BÑ  Ð#ÑÐCÑ œ 'B  #C

1.

The identity element for addition is ! since, for any real number +, +  ! œ !  + œ +. Choice B is correct.

3.

The additive inverse of + is + since, for any real number +, +  Ð+Ñ œ ! and +  + œ !Þ Choice A is correct.

5.

The multiplication property of ! states that the product of ! and any real number is ! .

7.

The associative property is used to change the grouping of three terms or factors.

9.

When simplifying an expression, only like terms can be combined.

11.

Using the distributive property, #Ð7  :Ñ œ #7  #:.

13.

Using the distributive property,

15.

Using the second form of the distributive property,

(b) Use the second form of the distributive property. %5  "#5 œ Ò%  Ð"#ÑÓ5 œ )5 N2. (a) (B  B œ (B  "B œ Ð(  "ÑB œ )B

Identity property Distributive property

(b) Ð&:  $;Ñ œ "Ð&:  $;Ñ œ "Ð&:Ñ  Ð"ÑÐ$;Ñ

Identity prop. Distributive property

"#ÐB  CÑ œ "#cB  ÐCÑd œ "#ÐBÑ  Ð"#ÑÐCÑ œ "#B  "#C.

&5  $5 œ Ð&  $Ñ5 œ )5. 17.

(<  *< œ (<  Ð* M M œ :>

3.

J  $# œ *& G & &ˆ* ‰ * ÐJ  $#Ñ œ * & G & * ÐJ

13.

œ #P  #[ œ #P #P œ # œ P, or P œ

T [ #

(a) Solve for Z œ P[ L for [ . Z œ P[ L Z P[ L œ PL PL Z Z œ [ , or [ œ PL PL

T œ "# ,2 #T œ #ˆ "# ,2‰

" ,2 T T œ #" gives us 2 œ " . " , , # # #,

Choice D, 2 œ

Z œ P[ L Z P[ L œ P[ P[ Z Z œ L, or L œ P[ P[ Solve G œ #1< for œ ". Make a table.

Rate (as a decimal) !Þ!' !Þ!$ !Þ!%

Interest !Þ!'Ð"#,!!!Ñ !Þ!$B !Þ!% "#,!!!  B

The last column gives the equation. Interest interest interest  œ Þ at '% at $% at %% !Þ!'Ð"#,!!!Ñ  !Þ!$B œ !Þ!% "#,!!!  B 'Ð"#,!!!Ñ  $B œ % "#,!!!  B (#,!!!  $B œ %),!!!  %B #%,!!! œ B

Multiply by 100.

He should invest $#%,!!! at $%.

Multiply by 100.

He should invest $%!!! at $% and "#,!!!  %!!! œ $)!!! at %%.

51.

"#,!!! B "#,!!!  B

!Þ!$B !Þ!%Ð"#,!!!  BÑ %%!

Interest interest total  œ at $% at %% interest. !Þ!$B  !Þ!%Ð"#,!!!  BÑ œ %%!

Let B œ the amount of additional money to be invested at $%. Use M œ : with > œ ". Make a table. Use the fact that the total return on the two investments is %%. Principal

Interest

The last column gives the equation.

!Þ!%&B !Þ!$Ð#B  "!!!Ñ "!#!

Interest interest total  œ at %Þ&% at $% interest. !Þ!%&B  !Þ!$Ð#B  "!!!Ñ œ "!#!

Let B œ the 2008 cost. Then B  $Þ)%ÐBÑ œ %#Þ*". B  $Þ)Ð!Þ!"ÑÐBÑ œ %#Þ*" "B  !Þ!$)B œ %#Þ*" !Þ*'#B œ %#Þ*" B œ %#Þ*" !Þ*'# ¸ %%Þ'!

Interest

The last column gives the equation.

To the nearest dollar, the cost was $(!#). 45.

Rate (as a Decimal) !Þ!%& !Þ!$ Total p

Check $"#,!!! @ '% œ $(#! and $#%,!!! @ $% œ $(#!; $(#!  $(#! œ $"%%!, which is the same as Ð$"#,!!!  $#%,!!!Ñ @ %%Þ 55.

Let B œ the number of liters of "!% acid solution needed. Make a table. Liters of Solution "! B B  "!

Percent (as a decimal) !Þ!% !Þ"! !Þ!'

Liters of Pure Acid !Þ!%Ð"!Ñ œ !Þ% !Þ"!B !Þ!' B  "!

Linear Equations, Inequalities, and Applications Acid in %% acid in "!% acid in '%  œ solution solution solutionÞ !Þ%  !Þ"!B œ !Þ!' B  "!

61.

Let B œ the amount of $' per lb nuts. Make a table. Pounds of nuts &! B B  &!

!Þ%  !Þ"!B œ !Þ!'B  !Þ' Distributive property !Þ!%B œ !Þ# Subtract 0.06x and 0.4. Bœ& Divide by 0.04. Check %% of "! is !Þ% and "!% of & is !Þ&; !Þ%  !Þ& œ !Þ*, which is the same as '% of Ð"!  &Ñ.

"!!  'B œ &ÐB  &!Ñ "!!  'B œ &B  #&! B œ "&!

Let B œ the number of liters of the #!% alcohol solution. Make a table. Liters of Solution "# B B  "#

Percent (as a decimal) !Þ"# !Þ#! !Þ"%

Liters of Pure Alcohol !Þ"#Ð"#Ñ œ "Þ%% !Þ#!B !Þ"% B  "#

Alcohol in alcohol in alcohol in  œ "#% solution #!% solution "%% solutionÞ "Þ%%  !Þ#!B œ !Þ"% B  "# "%%  #!B œ "%ÐB  "#Ñ "%%  #!B œ "%B  "') 'B œ #% Bœ%

Multiply by 100. Distributive property Subtract 14x and 144. Divide by 6.

He should use "&! lb of $' nuts. Check &! pounds of the $# per lb nuts are worth $"!! and "&! pounds of the $' per lb nuts are worth $*!!; $"!!  $*!! œ $"!!!, which is the same as Ð&!  "&!Ñ pounds worth $& per lb. 63.

We cannot expect the final mixture to be worth more than the more expensive of the two ingredients. Answers will vary.

65.

(a) Let B œ the amount invested at &%. )!!  B œ the amount invested at "!%. (b) Let C œ the amount of &% acid used. )!!  C œ the amount of "!% acid used.

66.

%L of #!% alcohol solution are needed.

Organize the information in a table. (a)

Check "#% of "# is "Þ%% and #!% of % is !Þ); "Þ%%  !Þ) œ #Þ#%, which is the same as "%% of Ð"#  %ÑÞ 59.

Percent (as a decimal) " !Þ#& !Þ%!

Gallons of Pure Dye "B œ B !Þ#&Ð%Ñ œ " !Þ%! B  %

Distributive property Subtract 0.4x and 1. Divide by 0.6.

One gallon of pure ("!!%) dye is needed. Check "!!% of " is " and #&% of % is "; "  " œ #, which is the same as %!% of Ð"  %ÑÞ

Percent (as a decimal) !Þ!& !Þ"! !Þ!)(&

Interest !Þ!&B !Þ"!Ð)!!  BÑ !Þ!)(&Ð)!!Ñ

The amount of interest earned at &% and "!% is found in the last column of the table, !Þ!&B and !Þ"!Ð)!!  BÑ. (b)

Write the equation from the last column in the table. B  " œ !Þ%ÐB  %Ñ B  " œ !Þ%B  "Þ' !Þ'B œ !Þ' Bœ"

Principal B )!!  B )!!

Let B œ the amount of pure dye used (pure dye is "!!% dye). Make a table. Gallons of Solution B % B%

Total Cost #Ð&!Ñ œ "!! 'B & B  &!

The total value of the $# per lb nuts and the $' per lb nuts must equal the value of the $& per lb nuts.

Five liters of the "!% solution are needed.

57.

Cost per lb $# $' $&

Liters of Solution C )!!  C )!!

Percent (as a decimal) !Þ!& !Þ"! !Þ!)(&

Liters of Pure Acid !Þ!&C !Þ"!Ð)!!  CÑ !Þ!)(&Ð)!!Ñ

The amount of pure acid in the &% and "!% mixtures is found in the last column of the table, !Þ!&C and !Þ"!Ð)!!  CÑ. 67.

Refer to the tables for Exercise 66. In each case, the last column gives the equation. (a) !Þ!&B  !Þ"!Ð)!!  BÑ œ !Þ!)(&Ð)!!Ñ (b) !Þ!&C  !Þ"!Ð)!!  CÑ œ !Þ!)(&Ð)!!Ñ

747

Linear Equations, Inequalities, and Applications 68.

In both cases, multiply by "!,!!! to clear the decimals. (a)

!Þ!&B  !Þ"!Ð)!!  BÑ œ !Þ!)(&Ð)!!Ñ &!!B  "!!!Ð)!!  BÑ œ )(&Ð)!!Ñ &!!B  )!!,!!!  "!!!B œ (!!,!!! &!!B œ "!!,!!! B œ #!!

Jack invested $#!! at &% and )!!  B œ )!!  #!! œ $'!! at "!%. (b)

!Þ!&C  !Þ"!Ð)!!  CÑ œ !Þ!)(&Ð)!!Ñ &!!C  "!!!Ð)!!  CÑ œ )(&Ð)!!Ñ &!!C  )!!,!!!  "!!!C œ (!!,!!! &!!C œ "!!,!!! C œ #!!

Jill used #!! L of &% acid solution and )!!  C œ )!!  #!! œ '!! L of "!% acid solution. (c) The processes used to solve Problems A and B were virtually the same. Aside from the variables chosen, the problem information was organized in similar tables and the equations solved were the same. 69.

. œ ; < œ &!, > œ % T œ +  ,  - ; , œ "$, - œ "%, T œ %' %' œ +  "$  "% %' œ +  #( "* œ +

4

Further Applications of Linear Equations

4 Now Try Exercises N1. Let B œ the number of dimes. Then &#  B œ the number of nickels. Number of Coins B &#  B

Denomination !Þ"! !Þ!& Total p

Value !Þ"!B !Þ!&Ð&#  BÑ $Þ(!

Multiply the number of coins by the denominations, and add the results to get $Þ(!. !Þ"!B  !Þ!&Ð&#  BÑ œ $Þ(! "!B  &Ð&#  BÑ œ $(! "!B  #'!  &B œ $(! &B œ ""! B œ ##

Multiply by 100.

He has ## dimes and &#  ## œ $! nickels.

748

N2. Let B œ the amount of time needed for the trains to be $)(Þ& km apart. Make a table. Use the formula . œ , that is, find each distance by multiplying rate by time. First Train Second Train Total

Rate )! (&

Time B B

Distance )!B (&B $)(Þ&

The total distance traveled is the sum of the distances traveled by each train, since they are traveling in opposite directions. This total is $)(Þ& km. )!B  (&B œ $)(Þ& "&&B œ $)(Þ& B œ $)(Þ& "&& œ #Þ& The trains will be $)(Þ& km apart in #Þ& hr. Check The first train traveled )!Ð#Þ&Ñ œ #!! km. The second train traveled (&Ð#Þ&Ñ œ ")(Þ& km, for a total of #!!  ")(Þ& œ $)(Þ&, as required. N3. Let B œ the driving rate. Then B  $! œ the bicycling rate.

. œ &!Ð%Ñ œ #!! 71.

Check The number of coins is ##  $! œ &# and the value of the coins is $!Þ"!Ð##Ñ  $!Þ!&Ð$!Ñ œ $$Þ(!, as required.

Make a table. Use the formula . œ , that is, find each distance by multiplying rate by time. Rate

Time

Distance

Car

B

" #

" #B

Bike

B  $!

" "# œ

$ #

$ # ÐB

 $!Ñ

The distances are equal. " #B

œ $# ÐB  $!Ñ "B œ $ÐB  $!Ñ B œ $B  *! *! œ #B %& œ B

Multiply by 2.

The distance he travels to work is " #B

œ "# Ð%&Ñ œ ##Þ& miles.

Check The distance he travels to work by bike is $ $ $ # ÐB  $!Ñ œ # Ð%&  $!Ñ œ # Ð"&Ñ œ ##Þ& miles, which is the same as we found above (by car). N4. The sum of the three measures must equal ")!°. B  ÐB  ""Ñ  Ð$B  $'Ñ œ ")! &B  #& œ ")! &B œ #!& B œ %"

Linear Equations, Inequalities, and Applications The sum of the values must equal the total value.

The angles measure %"°, %"  "" œ &#°, and $Ð%"Ñ  $' œ )(°.

"B  #Ð$(  BÑ œ &" B  (%  #B œ &" B  (% œ &" #$ œ B

Check Since %"°  &#°  )(° œ ")!°, the answers are correct.

4 Section Exercises 1.

The total amount is

She has #$ loonies and $(  #$ œ "% toonies.

"%Ð!Þ"!Ñ  "'Ð!Þ#&Ñ œ "Þ%!  %Þ!! œ $&Þ%!. 3.

Use . œ , or < œ .> . Substitute $!! for . and "! for >.  "# Ñ '&> œ ')>  $% $> œ $% > œ $% or "" "$ $

Rate B& B

&!ÐB  "# Ñ

Check Anne travels '!Ð "# Ñ œ $! miles. Johnny travels &!Ð "#  "# Ñ œ &! miles. The sum of the distances is )! miles, as required.

The distances are equal.

25.

" #

ÐB  $!Ñ  Ð#B  "#!Ñ  Ð "# B  "&Ñ œ ")!

Time > >

B

Distance '!B

'!B  &!ÐB  "# Ñ œ )! '!B  &!B  #& œ )! ""!B œ && && B œ ""! œ

Let > œ Mulder's time. Then >  "# œ Scully's time. Mulder

&!

Time B

The total distance is )!.

hr.

Rate '&

Rate '!

They will meet & #

Check Each steamer traveled ##Ð#Þ&Ñ œ && miles for a total of #Ð&&Ñ œ ""! miles, as required. 23.

Let B œ Anne's time. Then B  "# œ Johnny's time. Anne

The total distance traveled is the sum of the distances traveled by each steamer, since they are traveling in opposite directions. This total is ""! mi. ##>  ##> œ ""! %%> œ ""! > œ ""! %% œ

Subtract 3.6x.

Distance $Þ'ÐB  &Ñ %B

With B œ *!, the three angle measures become

and 31.

Ð*!  $!Ñ° œ '!°, c#Ð*!Ñ  "#!d° œ '!°, Ò "# Ð*!Ñ  "&Ó° œ '!°.

The sum of the measures of the three angles of a triangle is ")!°. Ð$B  (Ñ  Ð*B  %Ñ  Ð%B  "Ñ œ ")! "'B  % œ ")! "'B œ "(' B œ "" With B œ "", the three angle measures become

and

Ð$ • ""  (Ñ° œ %!°, Ð* • ""  %Ñ° œ *&°, Ð% • ""  "Ñ° œ %&°Þ

Linear Equations, Inequalities, and Applications 33.

If my current age is &", in "! years I will be

The sum of the measures of the angles of a triangle is ")!°.

&"  "! œ '" years old.

B  #B  '! œ ")! $B  '! œ ")! $B œ "#! B œ %!

45.

The measures of the unknown angles are %!° and #B œ )!°Þ 34.

Two angles which form a straight line add to ")!°, so ")!°  '!° œ "#!°Þ The measure of the unknown angle is "#!°Þ

35.

The sum of the measures of the unknown angles in Exercise 33 is %!°  )!° œ "#!°. This is equal to the measure of the angle in Exercise 34.

36.

is equal to the measure of angle 37.

o o o

The sum of the measures of angles

1

and

B  ÐB  #Ñ œ ÐB  %Ñ  #' #B  # œ B  $! B œ #) The three consecutive even integers are #), $!, and $#. 47.

2

3 .

Vertical angles have equal measure.

The three consecutive odd integers are #", #$, and #&.

The angles are both "##°.

41.

The sum of the two angles is *!°.

49.

Ð%ß ∞Ñ is equivalent to B  %.

Ð&B  "Ñ  #B œ *! (B  " œ *! (B œ *" B œ "$

51.

Ð#ß 'Ñ is equivalent to #  B  '.

The measures of the two angles are c&Ð"$Ñ  "d° œ '%° and c#Ð"$Ñd° œ #'°.

Let B œ the first consecutive integer. Then B  " will be the second consecutive integer, and B  # will be the third consecutive integer.

The sum of the first and twice the second is "( more than twice the third. B  #ÐB  "Ñ œ #ÐB  #Ñ  "( B  #B  # œ #B  %  "( $B  # œ #B  #" B œ "* Since B œ "*, B  " œ #!, and B  # œ #". The three consecutive integers are "*, #!, and #". 43.

Let B œ the least odd integer. Then B  # and B  % are the next two odd consecutive integers. The sum of the least integer and middle integer is "* more than the greatest integer. B  ÐB  #Ñ œ ÐB  %Ñ  "* #B  # œ B  #$ B œ #"

)B  # œ (B  "( B œ "& ) • "&  # œ "## and ( • "&  "( œ "##Þ 39.

Let B œ the least even integer. Then B  # and B  % are the next two even consecutive integers. The sum of the least integer and middle integer is #' more than the greatest integer.

Let B œ my current age. Then B  " will be my age next year. The sum of these ages will be "!$ years. B  ÐB  "Ñ œ "!$ #B  " œ "!$ #B œ "!# B œ &"

Summary Exercises on Solving Applied Problems 1.

Let B œ the width of the rectangle. Then B  $ is the length of the rectangle. If the length were decreased by # inches and the width were increased by " inch, the perimeter would be #% inches. Use the formula T œ #P  #[ , and substitute #% for T , ÐB  $Ñ  # or B  " for P, and B  " for [ . T œ #P  #[ #% œ #ÐB  "Ñ  #ÐB  "Ñ #% œ #B  #  #B  # #% œ %B  % #! œ %B &œB The width of the rectangle is & inches, and the length is &  $ œ ) inches. 751

Linear Equations, Inequalities, and Applications 3.

Check John traveled '!Ð&Ñ œ $!! miles and Pat traveled #)Ð&Ñ œ "%! miles; $!!  "%! œ %%!, as required.

Let B œ the regular price of the item. The sale price after a %'% (or !Þ%') discount was $%'Þ*(, so an equation is B  !Þ%'B œ %'Þ*(. !Þ&%B œ %'Þ*( B œ %'Þ*( !Þ&% ¸ )'Þ*)

11.

Liters of Solution #! B #!  B

To the nearest cent, the regular price was $)'Þ*). 5.

Let B œ the amount invested at %%. Then #B is the amount invested at &%. Use M œ : with > œ " yr. Make a table. Principal B #B

Rate (as a decimal) !Þ!% !Þ!& Total p

!Þ!%B !Þ!&Ð#BÑ œ !Þ"!B ""#

#!!  &B œ )Ð#!  BÑ #!!  &B œ "'!  )B %! œ $B B œ %! or "$ "$ $

Interest interest total  œ at %% at &% interest. !Þ!%B  !Þ"!B œ ""#

13.

Time > >

Distance '!> #)>

The total distance is %%! miles. '!>  #)> œ %%! ))> œ %%! >œ& It will take & hours for John and Pat to meet.

752

Value

& "! o Totals p

&B "!Ð"#'  BÑ )%!

&B  "!Ð"#'  BÑ œ )%! &B  "#'!  "!B œ )%! &B œ %#! B œ )% There are )% $& bills and "#'  )% œ %# $"! bills.

Wade scored #$)' points and James scored #$)'  "$' œ ##&! points.

Rate '! #)

Denomination

The sum of the values must equal the total value.

B  ÐB  "$'Ñ œ %'$' #B  "$' œ %'$' #B œ %((# B œ #$)'

John Pat

%! $ Ñ.

Let B œ the number of $& bills. Then "#'  B is the number of $"! bills. Number of Bills B "#'  B "#'

Let B œ the number of points scored by Wade in the 2008–2009 season. Then B  "$' œ the number of points scored by James in the 2007– 2008 season. The total number of points scored by both was %'$'.

Let > œ the time it will take until John and Pat meet. Use . œ and make a table.

Multiply by 100.

# Check "!% of #! is # and &% of %! $ is $ ; # ) #  $ œ $ , which is the same as )% of Ð#! 

Multiply by 100.

Check $)!! @ %% œ $$# and $"'!! @ &% œ $)!; $$#  $)! œ $""#

9.

Liters of Pure Drug #!Ð!Þ"!Ñ œ # !Þ!&B !Þ!) #!  B

The pharmacist should add "$ "$ L.

$)!! is invested at %% and #Ð$)!!Ñ œ $"'!! at &%.

7.

Percent (as a decimal) !Þ"! !Þ!& !Þ!)

Drug drug drug  œ in "!% in &% in )%Þ #  !Þ!&B œ !Þ!) #!  B

Interest

The last column gives the equation.

%B  "!B œ "",#!! "%B œ "",#!! B œ )!!

Let B œ the number of liters of the &% drug solution.

Check The number of bills is )%  %# œ "#' and the value of the bills is $&Ð)%Ñ  $"!Ð%#Ñ œ $)%!, as required. 15.

The sum of the measures of the three angles of a triangle is ")!°. B  Ð'B  &!Ñ  ÐB  "!Ñ œ ")! )B  '! œ ")! )B œ #%! B œ $! With B œ $!, the three angle measures become Ð' • $!  &!Ñ° œ "$!°, Ð$!  "!Ñ° œ #!°, and $!°Þ

Linear Equations, Inequalities, and Applications 17.

Let B œ the least integer. Then B  " is the middle integer and B  # is the greatest integer. "The sum of the least and greatest of three consecutive integers is $# more than the middle integer" translates to B  ÐB  #Ñ œ $#  ÐB  "ÑÞ #B  # œ B  $$ B œ $"

N2. %B  "   &B " B BŸ"

Subtract 4x. Equivalent inequality.

Check Substitute " for B in %B  " œ &B. ? %Ð"Ñ  " œ &Ð"Ñ &œ&

So " satisfies the equality part of   . Choose ! and # as test points.

The three consecutive integers are $", $#, and $$Þ Check The sum of the least and greatest integers is $"  $$ œ '%, which is the same as $# more than the middle integer. 19.

Let B œ the length of the shortest side. Then #B is the length of the middle side and $B  # is the length of the longest side. The perimeter is $% inches. Using T œ +  ,  gives us B  #B  Ð$B  #Ñ œ $%Þ 'B  # œ $% 'B œ $' Bœ' The lengths of the three sides are ' inches, #Ð'Ñ œ "# inches, and $Ð'Ñ  # œ "' inches. Check The sum of the lengths of the three sides is '  "#  "' œ $% inches, as required.

5

Linear Inequalities in One Variable

True

%B  "   &B Let x œ 0. ?

%Ð!Ñ  "   &Ð!Ñ " ! True ! is in the solution set. Let x œ 2. ?

%Ð#Ñ  "   &Ð#Ñ *   "! # is not in the solution set.

False

The check confirms that Ð∞, "Ó is the solution set.

N3. (a) )B   %! )B %!   ) ) B   &

Divide by 8 > 0; do not reverse the symbol.

Check that the solution set is the interval c&ß ∞ .

5 Now Try Exercises N1. B  "!  ( B$

Add 10.

Check Substitute $ for B in B  "! œ (. ? $  "! œ ( ( œ (

Check that the solution set is the interval Ð∞ß $Ñ.

B  "!  ( ?

Divide by 20 < 0; reverse the symbol.

True

This shows that $ is the boundary point. Choose ! and ( as test points.

Let B œ 0.

(b) #!B  '! #!B '!  #! #! B$

Let B œ 7. ?

!  "!  ( (  "!  ( "!  ( False $  ( True ! is not in the ( is in the solution set. solution set. The check confirms that Ð$ß ∞Ñ is the solution set.

N4. &  #ÐB  %Ñ Ÿ ""  %B &  #B  ) Ÿ ""  %B #B  "$ Ÿ ""  %B #B  "$ Ÿ "" #B Ÿ # #B # Ÿ # # B Ÿ " Check that the solution set is the interval Ð∞ß "Ó.

753

Linear Equations, Inequalities, and Applications N5.

 "& ÐB  )Ñ  #! "& ÐB  )Ñ‘ Mult. by 20. "&ÐB  #Ñ  "!  %ÐB  )Ñ "&B  $!  "!  %B  $# "&B  #!  %B  $# ""B  #!  $# Subtract 4x. ""B  "# Add 20. "# B   "" Divide by 11.

Step 5 She can belong to the health club for a maximum of * months. (She may belong for less time, as indicated by the inequality B Ÿ *.)

$ " % ÐB  #Ñ  # $ " #! % ÐB  #Ñ  # ‘

Step 6 If Sara belongs for * months, she will spend %!  $&Ð*Ñ œ $&&, the maximum amount. N9. Let B œ the grade he must make on the fourth test. To find the average of the four tests, add them and divide by %. This average must be at least *!, that is, greater than or equal to *!.

Check that the solution set is the interval Ð "# "" ß ∞Ñ.

)#  *(  *$  B   *! % #(#  B   *! % #(#  B   $'! B   ))

N6. "  B  #  $ " B  & Add 2 to each part.

Joel must score at least )) on the fourth test.

Check that the solution set is the interval Ð"ß &Ñ.

Check N7.

#  %B  & $  %B $ %B  % % $ %  B

Ÿ( Ÿ "# Add 5 to each part. "# Divide by 4.   % Reverse inequalities.   $ Reduce.

$ Ÿ

  $% Equivalent inequality.

B

Check that the solution set is the interval Ò$ß  $% Ñ.

Multiply by 4. Subtract 272.

)#  *(  *$  )) $'! œ œ *! % %

A score of )) or more will give an average of at least *!, as required.

5 Section Exercises 1.

BŸ$ In interval notation, this inequality is written Ð∞ß $Ó. The bracket indicates that $ is included. The answer is choice D.

3.

B$ In interval notation, this inequality is written Ð∞ß $Ñ. The parenthesis indicates that $ is not included. The graph of this inequality is shown in choice B.

N8. Step 2 Let B œ the number of months she belongs to the health club. Step 3 She must pay $%!, plus $$&B, to belong to the health club for B months, and this amount must be no more than $$&&. Cost of belonging ðóóñóóò Æ %!  $&B

is no more than ðóóñóóò Æ Ÿ

$&& dollars. ðñò Æ $&&

$ Ÿ B Ÿ $ In interval notation, this inequality is written Ò$ß $Ó. The brackets indicates that $ and $ are included. The answer is choice F.

7.

Since %  !, the student should not have reversed the direction of the inequality symbol when dividing by %. We reverse the inequality symbol only when multiplying or dividing by a negative number. The solution set is Ò"'ß ∞Ñ.

9.

B  %   "# B   "'

Add 4.

Check that the solution set is the interval Ò"'ß ∞Ñ.

Step 4 $&B Ÿ $"& BŸ*

754

5.

Subtract 40. Divide by 35.

Linear Equations, Inequalities, and Applications 11.

$5  "  ## $5  #" 5(

&Ð"%Ñ  ' ? œ) ) '% ? ) œ) )œ)

Subtract 1. Divide by 3.

Check that the solution set is the interval Ð(ß ∞Ñ.

True

This shows that "% is the boundary point. Now test a number on each side of "%. We choose ! and #!. 13.

%B  "' B  %

&B  ' ) )

Divide by 4.

Check that the solution set is the interval Ð∞ß %Ñ.

15.

Let B œ 0. Let B œ 20. &Ð!Ñ  ' ? &Ð#!Ñ  ' ? ) ) ) ) ' *% '  )  ) True ) Ðor "" ) Ñ  ) False ! is in the solution set. #! is not in the solution set.

 $% B   $! Multiply both sides by  %$ and reverse the inequality symbol.  %$ Ð $% BÑ Ÿ  %$ Ð$!Ñ B Ÿ %! Check that the solution set is the interval Ð∞ß %!Ó.

17.

The check confirms that Ð∞ß "%Ñ is the solution set.

23.

"Þ$B   &Þ# Divide both sides by "Þ$, and reverse the inequality symbol. "Þ$B &Þ# Ÿ "Þ$ "Þ$ BŸ%

&B  # Ÿ %) &B Ÿ &! B Ÿ "!

Subtract 2. Divide by 5.

25.

Check that the solution set is the interval Ð∞ß "!Ó.

21.

&B  ' ) ) &B  ' )Œ   )•) ) &B  '  '% &B  (! B  "%

Add 6. Divide by 5.

&B  ' Check Let B œ "% in the equation œ ). )

'B  %   #B )B  %   ! )B   % B   %) œ "#

Add 2x. Add 4. Divide by 8.

Check that the solution set is the interval Ò "# ß ∞Ñ.

27. Multiply by 8.

Add 5. Divide by 2.

Check that the solution set is the interval Ð∞ß  "& # Ñ.

Check that the solution set is the interval Ð∞ß %Ó.

19.

#B  & & % Multiply both sides by %, and reverse the inequality symbol. #B  & %Œ   %Ð&Ñ % #B  &  #! #B  "& B   "& #

B  #ÐB  %Ñ Ÿ $B B  #B  ) Ÿ $B B  ) Ÿ $B ) Ÿ %B # Ÿ B, or B   #

Add x.

Check that the solution set is the interval Ò#ß ∞Ñ.

755

Linear Equations, Inequalities, and Applications 29.

Ð%  œ # hr. . œ "'%Ð#Ñ œ $#) The distance is $#) miles.

Let B œ the number of quarters. Then #B  " is the number of dimes. Number of Coins B #B  "

The distance is &$! miles.

Multiply by 100.

Eric should invest $"!,!!! at '% and $"!,!!!  $%!!! œ $'!!! at %%. 27.

Denomination

The sum of the values equals the total value.

The last column gives the equation.

26.

Let B œ the number of nickels. Then "*  B is the number of dimes.

Denomination

Value

!Þ#& !Þ"! Total p

!Þ#&B !Þ"!Ð#B  "Ñ $Þ&!

31.

Let B œ the time it takes for the trains to be #*( mi apart. Use the formula . œ . Passenger Train Freight Train

Rate '! (&

Time B B

Distance '!B (&B #*(

The total distance traveled is the sum of the distances traveled by each train. 771

Linear Equations, Inequalities, and Applications Check %' mph for " hour œ %' miles and %'  ( œ $* mph for " hour œ $* miles; %'  $* œ )&Þ

'!B  (&B œ #*( "$&B œ #*( B œ #Þ# It will take the trains #Þ# hours before they are #*( miles apart. 32.

35.

Ð$B  (Ñ  Ð%B  "Ñ  Ð*B  %Ñ œ ")! "'B  % œ ")! "'B œ "(' B œ ""

Let B œ the rate of the faster car and B  "& œ the rate of the slower car. Make a table. Rate B B  "&

Faster Car Slower Car

Time # #

Distance #B #ÐB  "&Ñ #$!

The first angle is $Ð""Ñ  ( œ %!°. The second angle is %Ð""Ñ  " œ %&°. The third angle is *Ð""Ñ  % œ *&°. 36.

The total distance traveled is the sum of the distances traveled by each car.

The angle measures are "&Ð*Ñ  "& œ "&!° and $Ð*Ñ  $ œ $!°. 37.

Check #Ð'&Ñ  #Ð&!Ñ œ #$! 33.

Let B œ amount of time spent averaging %& miles per hour. Then %  B œ amount of time at &! mph. First Part Second Part Total

Rate %& &!

Time B %B

Distance %&B &!Ð%  BÑ "*&

38.

39.

Check %& mph for " hour œ %& miles and &! mph for $ hours œ "&! miles; %&  "&! œ "*&Þ Let B œ the average rate for the first hour. Then B  ( œ the average rate for the second hour. Using . œ , the distance traveled for the first hour is BÐ"Ñ, for the second hour is ÐB  (ÑÐ"Ñ, and for the whole trip, )&. B  ÐB  (Ñ œ )& #B  ( œ )& #B œ *# B œ %' The average rate for the first hour was %' mph.

772

&B  %   "" &B   "& Divide by &; reverse the inequality symbol. B Ÿ $ The solution set is Ð∞ß $Ó.

The automobile averaged %& mph for " hour.

34.

#  B' $ Multiply by 3. #B  ") Divide by #; reverse the inequality symbol. B  * The solution set is Ð*ß ∞Ñ.

From the last column: %&B  &!Ð%  BÑ œ "*& %&B  #!!  &!B œ "*& &B œ & Bœ"

The marked angles are supplements which have a sum of ")!°. Ð"&B  "&Ñ  Ð$B  $Ñ œ ")! ")B  ") œ ")! ")B œ "'# Bœ*

#B  #ÐB  "&Ñ œ #$! #B  #B  $! œ #$! %B œ #'! B œ '& The faster car travels at '& km/hr, while the slower car travels at '&  "& œ &! km/hr.

The sum of the angles in a triangle is ")!°.

'B  $  $ % Multiply by %; reverse the inequality symbol. 'B  $  "# 'B  * B  *' œ $# The solution set is Ð $# ß ∞Ñ.

40.

&  Ð'  %BÑ   #B  ( &  '  %B   #B  ( %B  "   #B  ( #B   ' B   $ The solution set is Ò$ß ∞Ñ.

41.

) Ÿ $B  "  "% * Ÿ $B  "& $ŸB & The solution set is Ò$ß &Ñ.

Linear Equations, Inequalities, and Applications 42.

& $ ÐB

 #Ñ  #& ÐB  "Ñ  " #&ÐB  #Ñ  'ÐB  "Ñ  "& Multiply by 15. #&B  &!  'B  '  "& $"B  %%  "& $"B  &* B  &* $"

50. 51.

Let B œ the other dimension of the rectangle. One dimension of the rectangle is ## and the perimeter can be no greater than "#!.

The solution set is Ð'ß *Ñ.

52.

T Ÿ "#! #P  #[ Ÿ "#! #ÐBÑ  #Ð##Ñ Ÿ "#! #B  %% Ÿ "#! #B Ÿ (' B Ÿ $) Let B œ the number of tickets that can be purchased. The total cost of the tickets is $%)B. Including the $&! discount and staying within the available $"'!!, we have

45.

53.

The result, )  "$, is a false statement. There are no real numbers that make this inequality true. The solution set is g.

For Exercises 47–50, let E œ ea, b, c, df, F œ ea, c, e, f f, and G œ ea, e, f, gf. 47. 48. 49.

E ∩ F œ ea, b, c, df ∩ ea, c, e, f f œ ea, cf E ∩ G œ ea, b, c, df ∩ ea, e, f, gf œ ea f

F ∪ G œ ea, c, e, f f ∪ ea, e, f, gf œ ea, c, e, f, gf

B  & or B Ÿ $

The solution set is Ð∞ß $Ó ∪ Ð&ß ∞Ñ.

54.

B   # or B  # The graph of the solution set will be all numbers that are either greater than or equal to # or less than #. All real numbers satisfy these criteria. The solution set is Ð∞ß ∞Ñ.

55.

B%' B  "!

and and

B  $ Ÿ "! BŸ(

The graph of the solution set will be all numbers that are both greater than "! and less than or equal to (. There are no real numbers satisfying these criteria.

The student will pass algebra if any score greater than or equal to '"% on the fifth test is achieved. 46.

B  #  "# B  "%

The graph of the solution set will be all numbers that are either greater than & or less than or equal to $Þ

Let B œ the student's score on the fifth test. The average of the five test scores must be at least (!. The inequality is (&  (*  '%  ("  B   (!. & (&  (*  '%  ("  B   $&! #)*  B   $&! B   '"

and and

The solution set is Ð)ß "%Ñ.

%)B  &! Ÿ "'!!Þ %)B Ÿ "'&! B p $%Þ% The group can purchase $% tickets or fewer (but at least "&).

B  %  "# B)

The graph of the solution set will be all numbers between ) and "%, not including the endpoints.

The other dimension must be $) meters or less. 44.

B  ' and B  *

The graph of the solution set will be all numbers which are both greater than ' and less than *. The overlap is the numbers between ' and *, not including the endpoints.

The solution set is Ð &* $" ß ∞Ñ. 43.

E ∪ G œ ea, b, c, df ∪ ea, e, f, gf œ ea, b, c, d, e, f, gf

The solution set is g. 56.

&B  "   "" &B   "! B Ÿ #

or or

$B  &   #' $B   #" B (

The graph of the solution set will be all numbers that are either less than or equal to # or greater than or equal to (. The solution set is Ð∞ß #Ó ∪ Ò(ß ∞Ñ.

773

Linear Equations, Inequalities, and Applications 57.

Ð$ß ∞Ñ ∩ Ð∞ß %Ñ Ð$ß ∞Ñ includes all real numbers greater than $. Ð∞ß %Ñ includes all real numbers less than %. Find the intersection. The numbers common to both sets are greater than $ and less than %.

65.

#B  ( œ ( #B œ "% Bœ(

$  B  % The solution set is Ð$ß %Ñ. 58.

66.

Ð∞ß 'Ñ ∩ Ð∞ß #Ñ Ð∞ß 'Ñ includes all real numbers less than '. Ð∞ß #Ñ includes all real numbers less than #. Find the intersection. The numbers common to both sets are less than #.

Ð%ß ∞Ñ ∪ Ð*ß ∞Ñ Ð%ß ∞Ñ includes all real numbers greater than %. Ð*ß ∞Ñ includes all real numbers greater than *. Find the union. The numbers in the first set, the second set, or in both sets are all the real numbers that are greater than %.

Ð"ß #Ñ ∪ Ð"ß ∞Ñ Ð"ß #Ñ includes the real numbers between " and #, not including " and #. Ð"ß ∞Ñ includes all real numbers greater than ". Find the union. The numbers in the first set, the second set, or in both sets are all real numbers greater than ". The solution set is Ð"ß ∞Ñ.

61.

kBk œ ( Í B œ ( or B œ ( The solution set is Ö(ß (×.

62.

67.

or

%B  # œ % %B œ ' B œ  '%

" #

or

B œ  $#

k$B  "k œ kB  #k

" #

or

or

$B  " œ ÐB  #Ñ $B  " œ B  # %B œ $ B œ 

%$The solution set is ˜ $% ß "# ™. 68.

k#B  "k œ k#B  $k

#B  " œ #B  $ or " œ $ False or

No solution

The solution set is ˜ "# ™. 69.

#B  " œ Ð#B  $Ñ #B  " œ #B  $ %B œ # B œ  #% œ  "#

kBk  "% Í "%  B  "%

The solution set is Ð"%ß "%Ñ. 70.

or or

B  # œ * B œ ""

kB  'k Ÿ (

"$   B

k$B  (k œ )

" Ÿ B 71.

or or

kB  %k œ "#

Subtract 6. Multiply by 1.   " Reverse inequalities. Ÿ "$ Equivalent inequality

The solution set is Ò"ß "$Ó.

$B  ( œ ) $B œ " B œ  "$

k#B  &k Ÿ "

" Ÿ #B  & Ÿ " ' Ÿ #B Ÿ % $ Ÿ B Ÿ #

The solution set is ˜ "$ ß &™.

774

%B  # œ % %B œ # B œ #%

( Ÿ B  ' Ÿ ( "$ Ÿ B Ÿ"

$B  ( œ ) $B œ "& Bœ&

64.

k%B  #k  ( œ $ k%B  #k œ %

Bœ

The solution set is e""ß (f.

63.

or

$B  " œ B  # #B œ "

kB  #k œ * B#œ* Bœ(

#B  ( œ ( #B œ ! Bœ!

The solution set is ˜ $# ß "# ™.

The solution set is Ð%ß ∞Ñ. 60.

or

The solution set is e!ß (f.

Bœ

The solution set is Ð∞ß #Ñ. 59.

k#B  (k  % œ "" k#B  (k œ (

The solution set is Ò$ß #Ó. 72.

kB  "k   $

Since the absolute value of an expression can never be negative, there are no solutions for this equation.

Since the absolute value of an expression is always nonnegative (positive or zero), the inequality is true for any real number B.

The solution set is g.

The solution set is Ð∞ß ∞Ñ.

Linear Equations, Inequalities, and Applications 73.

[5]

&  Ð'  %BÑ  #B  & &  '  %B  #B  & "  %B  #B  & #B  % B  #

80.

[4] Let B œ the first consecutive integer. Then B  " œ the second consecutive integer and B  # œ the third consecutive integer. The sum of the first and third integers is %( more than the second integer, so an equation is B  ÐB  #Ñ œ %(  ÐB  "Ñ. #B  # œ %)  B B œ %'

The solution set is Ð#ß ∞Ñ. 74.

[2]

Solve +5  ,> œ '< for 5Þ +5 œ '<  ,> '<  ,> 5œ +

75.

[6]

B  $ and B   #

Then B  " œ %(, and B  # œ %)Þ The integers are %', %(, and %). 81.

The real numbers that are common to both sets are the numbers greater than or equal to # and less than $.

%B  # $B  " B'  œ % ) "' Clear fractions by multiplying by the LCD, "'Þ %Ð%B  #Ñ  #Ð$B  "Ñ œ B  ' "'B  )  'B  # œ B  ' ##B  ' œ B  ' #"B œ ! Bœ!

The solution set is e!f. 77.

[7]

k$B  'k   !

82.

78.

[5]

[1]

83.

[7]

84.

[5]

Divide by 5 < 0; reverse the symbol.

[2] Use the formula L.

Z œ P[ L , and solve for

Z P[ L œ P[ P[ Z Z œ L , or L œ P[ P[ Substitute "Þ& for [ , & for P, and (& for Z . Lœ

(& (& œ œ "! &Ð"Þ&Ñ (Þ&

The height of the box is "! ft.

kB  $k Ÿ "$

The solution set is Ò"'ß "!Ó.

The solution set is Ð∞ß #Ó. 79.

!Þ!&B  !Þ!$Ð"#!!  BÑ œ %# &B  $Ð"#!!  BÑ œ %#!! Multiply by 100. &B  $'!!  $B œ %#!! #B œ '!! B œ $!!

"$ Ÿ B  $ Ÿ "$ "' Ÿ B Ÿ "!

&B   "! BŸ#

or

$B  # œ & $B œ ( B œ  ($

The solution set is e$!!f.

The absolute value of an expression is always nonnegative, so the inequality is true for any real number 5 . The solution set is Ð∞ß ∞Ñ.

or

The solution set is ˜ ($ ß "™.

The solution set is Ò#ß $Ñ. [1]

k$B  #k  % œ * k$B  #k œ &

$B  # œ & $B œ $ Bœ"

# Ÿ B  $

76.

[7]

$ % ÐB

 #Ñ  "$ Ð&  #BÑ  # *ÐB  #Ñ  %Ð&  #BÑ  #% Multiply by 12. *B  ")  #!  )B  #% "(B  $)  #% "(B  "% B  "% "(

The solution set is Ð∞ß "% "( Ñ. 85.

[5]

%  $  #B  * (  #B  ' ( B  $ # $  B  (#

Subtract 3. Divide by 2. Reverse inequalities. Equivalent inequality

The solution set is Ð$ß (# Ñ.

775

Linear Equations, Inequalities, and Applications 86.

87.

[5]

!Þ$B  #Þ"ÐB  %Ñ Ÿ 'Þ' $B  #"ÐB  %Ñ Ÿ '' Multiply by 10. $B  #"B  )% Ÿ '' ")B  )% Ÿ '' ")B Ÿ ") BŸ" The solution set is Ð∞ß "Ó.

91.

Liters of Solution B "! B  "!

Percent (as a decimal) !Þ#! !Þ&! !Þ%!

!Þ#!B  & œ !Þ%!ÐB  "!Ñ !Þ#!B  & œ !Þ%!B  % " œ !Þ#!B &œB

Multiply by 5.

& L of the #!% solution should be used. 92.

[7]

kB  "k œ k#B  $k

B  " œ #B  $

or

The measure of the angle is )!°. 88.

Any amount greater than or equal to $""!! will qualify the employee for the pension plan. 89.

[7]

or or

93.

94.

[6]

$B B  œ$ & # 'B  &B œ $! B œ $!

Multiply by 10.

The solution set is e$!f.

[7]

kB  $k Ÿ "

The solution set is Ò%ß #Ó. &B  "  "% &B  "$ B   "$ &

95.

The solution set is Ð∞ß ∞Ñ.

[7]

k$B  (k œ %

$B  ( œ % $B œ "" B œ "" $

or or

$B  ( œ % $B œ $ Bœ"

™ The solution set is ˜"ß "" $ .

B   # or B  %

The solution set includes all numbers either greater than or equal to # or all numbers less than %. This is the union and is the set of all real numbers.

776

[1]

" Ÿ B  $ Ÿ " % Ÿ B Ÿ #

The solution set is Ð∞ß  "$ & Ñ ∪ Ð$ß ∞Ñ. 90.

B  " œ Ð#B  $Ñ B  " œ #B  $ $B œ # B œ  #$

The solution set is ˜%ß  #$ ™.

k&B  "k  "%

&B  "  "% &B  "& B$

or

% œ B

[5] Let B œ the employee's earnings during the fifth month. The average of the five months must be at least $"!!!. *!!  "#!!  "!%!  ('!  B   "!!! & *!!  "#!!  "!%!  ('!  B   &!!! $*!!  B   &!!! B   ""!!

Liters of Mixture !Þ#!B !Þ&!Ð"!Ñ œ & !Þ%!ÐB  "!Ñ

From the last column:

[4] Let B œ the angle. Then *!  B is its complement and ")!  B is its supplement. The complement of an angle measures "!° less than one-fifth of its supplement. *!  B œ "& Ð")!  BÑ  "! %&!  &B œ ")!  B  &! %&!  &B œ "$!  B $#! œ %B )! œ B

[3] Let B œ the number of liters of the #!% solution. Then B  "! is the number of liters of the resulting %!% solution.

96.

[1]

&Ð#B  (Ñ œ #Ð&B  $Ñ "!B  $& œ "!B  ' $& œ ' False

This equation is a contradiction. The solution set is g.

Linear Equations, Inequalities, and Applications 97.

[7] (a)

k&B  $k  5

If 5  !, then k&B  $k would be less than a negative number. Since the absolute value of an expression is always nonnegative (positive or zero), the solution set is g.

Test 1.

(b) k&B  $k  5

If 5  !, then k&B  $k would be greater than a negative number. Since the absolute value of an expression is always nonnegative (positive or zero), the solution set is the set of all real numbers, Ð∞ß ∞Ñ.

The solution set is e"*f. 2.

(c) k&B  $k œ 5

If 5  !, then k&B  $k would be equal to a negative number. Since the absolute value of an expression is always nonnegative (positive or zero), the solution set is g. 98.

[6]

B  ' and B  )

$Ð#B  #Ñ  %ÐB  'Ñ œ $B  )  B 'B  '  %B  #% œ %B  ) #B  $! œ %B  ) #B œ $) B œ "*

!Þ!)B  !Þ!'ÐB  *Ñ œ "Þ#% )B  'ÐB  *Ñ œ "#% Multiply by 100. )B  'B  &% œ "#% "%B  &% œ "#% "%B œ (! Bœ&

The solution set is e&f. 3.

The graph of the solution set is all numbers both greater than ' and less than ). This is the intersection. The elements common to both sets are the numbers between ' and ), not including the endpoints. The solution set is Ð'ß )Ñ.

B' B% B#  œ "! "& ' $ÐB  'Ñ  #ÐB  %Ñ œ &ÐB  #Ñ Multiply by 30. $B  ")  #B  ) œ &B  "! &B  "! œ &B  "! True

This is an identity. The solution set is Öall real numbers×.

99.

[6]

&B  "   "" &B   "! B Ÿ #

or or

$B  &   #' $B   #" B (

The graph of the solution set is all numbers either less than or equal to # or greater than or equal to (. This is the union. The solution set is Ð∞ß #Ó ∪ Ò(ß ∞Ñ.

4.

(a) $B  Ð#  BÑ  %B  # œ )B  $ $B  #  B  %B  # œ )B  $ )B œ )B  $ ! œ $ False The false statement indicates that the equation is a contradiction. The solution set is g. (b)

100. [6] (a) All states have less than $ million female workers. The set of states with more than $ million male workers is ÖIllinois×. Illinois is the only state in both sets, so the set of states with less than $ million female workers and more than $ million male workers is ÖIllinois×. (b) The set of states with less than " million female workers or more than # million male workers is ÖIllinois, Maine, North Carolina, Oregon, Utah×. (c) It is easy to see that the sum of the female and male workers for each state doesn't exceed ( million, so the set of states with a total of more than ( million civilian workers is Ö×, or g.

B &B B (œ # * $ ' # Multiply each side by the LCD, '. #B  %# œ &B  "#  $B  &% #B  %# œ #B  %# !œ! True

This equation is an identity. The solution set is Öall real numbers×. (c) %Ð#B  'Ñ œ &B  #%  (B )B  #% œ #B  #% #% œ 'B  #% ! œ 'B !œB This is a conditional equation.

The solution set is e!f.

777

Linear Equations, Inequalities, and Applications 5.

Solve Z œ "$ ,2 for 2. Z œ "$ ,2 $Z œ ,2 $Z œ2 ,

6.

Use the formula . œ .

Multiply by 3. Divide by b.

Slower Car Faster Car

Divide by t.

12.

The three angle measures are %!°, %!°, and Ð# • %!  #!Ñ° œ "!!°. 13.

#(,#$# ¸ !Þ(%# $',(#$ (%Þ#% were classified as post offices.

B #),!!!  B $#),!!!

Rate (as a decimal) !Þ!$ !Þ!& o Totals p

Interest

14.

!Þ!$B !Þ!&Ð#),!!!  BÑ $"#%!

From the last column: !Þ!$B  !Þ!&Ð#),!!!  BÑ œ "#%! $B  &Ð#),!!!  BÑ œ "#%,!!! Multiply by 100. $B  "%!,!!!  &B œ "#%,!!! #B œ "',!!! B œ )!!! He invested $)!!! at $% and $#),!!!  $)!!! œ $#!,!!! at &%.

778

%  'ÐB  $Ñ Ÿ #  $ÐB  'Ñ  $B %  'B  ") Ÿ #  $B  ")  $B 'B  "% Ÿ #! 'B Ÿ ' Divide by ', and reverse the inequality symbol. B " The solution set is Ò"ß ∞Ñ.

Let B œ the amount invested at $%. Then #),!!!  B œ the amount invested at &%. Principal

The sum of the three angle measures is ")!°. Ð#B  #!Ñ  B  B œ ")! %B  #! œ ")! %B œ "'! B œ %!

Use M œ T and substitute $##)"Þ#& for M , $$',&!! for T , and " for >.

The rate of interest is 'Þ#&%.

Distance 'B 'ÐB  "&Ñ '$!

The slower car traveled at %& mph, while the faster car traveled at %&  "& œ '! mph.

##)"Þ#& œ $',&!!# @œ >

7.

11.

%  B  "' ( %B  ""# Multiply by 7. Divide by %, and reverse the inequality symbol. B  #) The solution set is ∞ß #) .

15.

"  $B  %  # $  $B ' "B #

Add 4. Divide by 3.

The solution set is Ð"ß #Ñ.

Linear Equations, Inequalities, and Applications 16.

' Ÿ %$ B  # Ÿ # ") Ÿ %B  ' Ÿ ' "# Ÿ %B Ÿ "# $ Ÿ B Ÿ$

23. Multiply by 3. Add 6. Divide by 4.

A.

The solution set is g.

$B  * B  $

B.

$B  * B  $

D.

$B  * B$

25.

$B  * B$

" $

(b) E ∪ F œ e"ß #ß &ß (f ∪ e"ß &ß *ß "#f œ e"ß #ß &ß (ß *ß "#f $B   ' B #

and and

26.

%B Ÿ #%

or

B '

or

%B  #  "! %B  "# B$

The solution set is all numbers less than $ or greater than or equal to '. This is the union. The solution set is Ð∞ß $Ñ ∪ Ò'ß ∞Ñ. 22.

k%B  $k Ÿ (

( Ÿ %B  $ Ÿ ( "! Ÿ %B Ÿ% %  "! Ÿ B Ÿ % %  &#

ŸB

Ÿ"

The solution set is Ò &# ß "Ó.



k$B  #k  " œ ) k$B  #k œ (

$B  # œ ( $B œ * B œ *$ œ $

( $

Equivalent inequality

or or or

$B  # œ ( $B œ & B œ  &$

The solution set is ˜ &$ ß $™.

27.

k$  &Bk œ k#B  )k $  &B œ #B  ) (B œ &

B%& B*

The solution set is all numbers both greater than or equal to # and less than *. This is the intersection. The numbers common to both sets are between # and *, including # but not *. The solution set is Ò#ß *Ñ.

B

The solution set is Ð "$ ß ($ Ñ.

The minimum score must be )# to guarantee a B. (a) E ∩ F œ e"ß #ß &ß (f ∩ e"ß &ß *ß "#f œ e"ß &f

k$B  %k  %  " k$B  %k  $ $  $B  %  $ (  $B  " (  B  "$ Reverse inequalities. $

Let B œ the score on the fourth test. )$  ('  (*  B   )! % #$)  B   )! % #$)  B   $#! B   )#

21.

k(  Bk Ÿ "

Since the absolute value of an expression is always nonnegative (positive or zero), the inequality is false for any real number B.

Thus, inequality C is equivalent to B  $.

20.

or

For each inequality, divide both sides by $ and reverse the direction of the inequality symbol.

C.

19.

&  'B  "# 'B  "( B  "( '

The solution set is Ð∞ß  (' Ñ ∪ Ð "( ' ß ∞Ñ. 24.

18.

or

&  'B  "# 'B  ( B   ('

The solution set is c$ß $d.

17.

k&  'Bk  "#

B œ  &(

or

or

$  &B œ Ð#B  )Ñ $  &B œ #B  ) $B œ "" B œ "" $

™ The solution set is ˜ &( ß "" $ . 28.

(a) k)B  &k  5

If 5  !, then k)B  &k would be less than a negative number. Since the absolute value of an expression is always nonnegative (positive or zero), the solution set is g. (b) k)B  &k  5

If 5  !, then k)B  &k would be greater than a negative number. Since the absolute value of an expression is always nonnegative (positive or zero), the solution set is the set of all real numbers, Ð∞ß ∞Ñ. (c) k)B  &k œ 5

If 5  !, then k)B  &k would be equal to a negative number. Since the absolute value of an expression is always nonnegative (positive or zero), the solution set is g. 779

780

The B-intercept is Ð%ß !Ñ.

GRAPHS, LINEAR EQUATIONS, AND FUNCTIONS 1

To find the C-intercept, let B œ !. B  #C œ % !  #C œ % #C œ % C œ #

The Rectangular Coordinate System

1 Now Try Exercises

The C-intercept is Ð!ß #Ñ.

N1. #B  C œ % To complete the ordered pairs, substitute the given value of B or C in the equation. For Ð!ß ****Ñ, let B œ !.

Plot the intercepts, and draw the line through them.

#B  C œ % #Ð!Ñ  C œ % C œ % C œ % The ordered pair is Ð!ß %Ñ. For Ð****ß !Ñ let C œ !. #B  C œ % #B  ! œ % #B œ % Bœ#

N3. C œ # In standard form, the equation is !B  C œ #. Every value of B leads to C œ #, so the C-intercept is Ð!ß #Ñ. There is no B-intercept. The graph is the horizontal line through Ð!ß #Ñ.

The ordered pair is Ð#ß !Ñ. For Ð%ß ****Ñ, let B œ %. #B  C œ % #Ð%Ñ  C œ % )C œ% C œ % Cœ% The ordered pair is Ð%ß %Ñ. For Ð****ß #Ñ, let C œ #. #B  C œ % #B  # œ % #B œ ' Bœ$

N4. B  $ œ ! In standard form, the equation is B  !C œ $. Every value of C leads to B œ $, the B-intercept is Ð$ß !Ñ. There is no C-intercept. The graph is the vertical line through Ð$ß !Ñ.

The ordered pair is Ð$ß #Ñ. The completed table follows. B ! # % $

C % ! % #

N2. B  #C œ % To find the B-intercept, let C œ !. B  #C œ % B  #Ð!Ñ œ % Bœ%

N5. #B  $C œ ! To find the B-intercept, let C œ !. #B  $Ð!Ñ œ ! #B œ ! Bœ! Since the B-intercept is Ð!ß !Ñ, the C-intercept is also Ð!ß !Ñ. Find another point. Let B œ $.

From Chapter 3 of Student’s Solutions Manual for Intermediate Algebra, Eleventh Edition. Margaret L. Lial, John Hornsby, Terry McGinnis. Copyright © 2012 by Pearson Education, Inc. Publishing as Addison-Wesley. All rights reserved.

781

Graphs, Linear Equations, and Functions (e) The point Ð$ß !Ñ is located on the B-axis, so it does not belong to any quadrant.

#Ð$Ñ  $C œ ! '  $C œ ! $C œ ' C œ # This gives the ordered pair Ð$ß #Ñ. Plot Ð$ß #Ñ and Ð!ß !Ñ and draw the line through them.

(f) The point Ð!ß !Þ&Ñ is located on the C-axis, so it does not belong to any quadrant. 11.

(a) If BC  !, then both B and C have the same sign. ÐBß CÑ is in quadrant I if B and C are positive. ÐBß CÑ is in quadrant III if B and C are negative. (b) If BC  !, then B and C have different signs. ÐBß CÑ is in quadrant II if B  ! and C  !. ÐBß CÑ is in quadrant IV if B  ! and C  !. (c) If BC  !, then B and C have different signs. ÐBß CÑ is in either quadrant II or IV. (See part (b).)

N6. By the midpoint formula, the midpoint of the segment with endpoints Ð#ß &Ñ and Ð%ß (Ñ is Œ

#  Ð%Ñ &  ( # # ß  œ Œ ß  œ Ð"ß "ÑÞ # # # #

1 Section Exercises 1.

(d) If BC  !, then B and C have the same sign. ÐBß CÑ is in either quadrant I or III. (See part (a).) For Exercises 13–22, see the rectangular coordinate system after Exercise 21. 13.

To plot Ð#ß $Ñ, go two units from zero to the right along the B-axis, and then go three units up parallel to the C-axis.

(b) The dot above the year 2007 appears to be at about "&!, so the revenue in 2007 was $"&! billion.

15.

To plot Ð$ß #Ñ, go three units from zero to the left along the B-axis, and then go two units down parallel to the C-axis.

(c) The ordered pair is Ð#!!(ß "&!ÑÞ

17.

To plot Ð!ß &Ñ, do not move along the B-axis at all since the B-coordinate is !. Move five units up along the C-axis.

19.

To plot Ð#ß %Ñ, go two units from zero to the left along the B-axis, and then go four units up parallel to the C-axis.

21.

To plot Ð#ß !Ñ, go two units to the left along the B-axis. Do not move up or down since the C-coordinate is !.

23.

C œB% (a) To complete the table, substitute the given values for B and C in the equation. For B œ !: C œ B  % C œ!% C œ % Ð!ß %Ñ

(a) B represents the year; C represents the higher education aid in billions of dollars.

(d) In 1997, federal tax revenues were about $(& billion. 3.

The point with coordinates Ð!ß !Ñ is called the origin of a rectangular coordinate system.

5.

The B-intercept is the point where a line crosses the B-axis. To find the B-intercept of a line, we let C equal ! and solve for B . The C-intercept is the point where a line crosses the C-axis. To find the C-intercept of a line, we let B equal ! and solve for C .

7.

To graph a straight line, we must find a minimum of two points. A third point is sometimes found to check the accuracy of the first two points.

9.

(a) The point Ð"ß 'Ñ is located in quadrant I, since the B- and C-coordinates are both positive. (b) The point Ð%ß #Ñ is located in quadrant III, since the B- and C-coordinates are both negative. (c) The point Ð$ß 'Ñ is located in quadrant II, since the B-coordinate is negative and the C-coordinate is positive. (d) The point Ð(ß &Ñ is located in quadrant IV, since the B-coordinate is positive and the C-coordinate is negative.

782

For B œ ": C œ B  % C œ"% C œ $

Ð"ß $Ñ

Graphs, Linear Equations, and Functions In a similar manner, substitute B œ #, $, and % to get C œ #, ", and !. B ! " # $ %

For C œ !:

C % $ # " !

B  #C œ & B  #Ð!Ñ œ & B!œ& Bœ&

For B œ #: B  #C œ & #  #C œ & #C œ $ C œ $#

(b) Plot the ordered pairs and draw the line through them.

For C œ #:

Ð&ß !Ñ

Ð#ß $# Ñ

B  #C œ & B  #Ð#Ñ œ & B%œ& Bœ"

"ß #

(b) Plot the ordered pairs and draw the line through them.

25.

BC œ$ (a) To complete the table, substitute the given values for B and C in the equation. For B œ !: B  C œ $ !C œ$ C œ $ Ð!ß $Ñ For C œ !: B  C œ $ B!œ$ Bœ$ For B œ &: B  C œ $ &C œ$ C œ # Cœ# For B œ #: B  C œ $ #C œ$ C œ " C œ "

29.

%B  &C œ #! (a) For B œ !:

Ð$ß !Ñ

For C œ !: Ð&ß #Ñ

For B œ #: Ð#ß "Ñ

(b) Plot the ordered pairs and draw the line through them. For C œ $:

27.

B  #C œ & (a) To complete the table, substitute the given values for B or C in the equation. For B œ !: B  #C œ & !  #C œ & #C œ & C œ &# Ð!ß &# Ñ

%B  &C œ #! %Ð!Ñ  &C œ #! &C œ #! C œ %

Ð!ß %Ñ

%B  &C œ #! %B  &Ð!Ñ œ #! %B œ #! Bœ&

Ð&ß !Ñ

%B  &C œ #! %Ð#Ñ  &C œ #! )  &C œ #! &C œ "# C œ  "# &

Ð#ß  "# & Ñ

%B  &C œ #! %B  &Ð$Ñ œ #! %B  "& œ #! %B œ & B œ &%

Ð &% ß $Ñ

(b) Plot the ordered pairs and draw the line through them.

783

Graphs, Linear Equations, and Functions 31.

C œ #B  $ (a) B #B ! ! " # # % $ '

The B-intercept is Ð'ß !Ñ. To find the C-intercept, let B œ !.

C œ #B  $ $ " " $

B  $C œ ' !  $C œ ' $C œ ' C œ #

(b) Notice that as the value of B increases by ", the value of C decreases by #.

33.

(a) The C-values are %, $, #, ", and !. They increase by " unit.

The C-intercept is Ð!ß #Ñ. Plot the intercepts and draw the line through them.

39.

(b) The C-values are $, ", ", and $. They decrease by # units.

&B  'Ð!Ñ œ "! &B œ "! B œ #

(c) It appears that the C-value increases (or decreases) by the value of the coefficient of B. So for C œ #B  %, a conjecture is "For every increase in B by " unit, C increases by # units." 35.

&B  'C œ "! To find the B-intercept, let C œ !.

The B-intercept is Ð#ß !Ñ. To find the C-intercept, let B œ !. &Ð!Ñ  'C œ "! 'C œ "! & C œ  "! ' œ $

#B  $C œ "# To find the B-intercept, let C œ !. #B  $C œ "# #B  $Ð!Ñ œ "# #B œ "# Bœ'

The C-intercept is Ð!ß  &$ ÑÞ Plot the intercepts and draw the line through them.

The B-intercept is Ð'ß !ÑÞ To find the C-intercept, let B œ !. #B  $C œ "# #Ð!Ñ  $C œ "# $C œ "# Cœ% The C-intercept is Ð!ß %Ñ. Plot the intercepts and draw the line through them.

41.

# $B

 $C œ ( To find the B-intercept, let C œ !. # $B

 $Ð!Ñ œ ( # $B

œ(

Bœ

$ #

•( œ

#" #

The B-intercept is Ð #" # ß !Ñ. To find the C-intercept, let B œ !. 37.

B  $C œ ' To find the B-intercept, let C œ !. B  $C œ ' B  $Ð!Ñ œ ' B!œ' Bœ'

784

# $

!  $C œ ( $C œ ( C œ  ($

The C-intercept is Ð!ß  ($ Ñ. Plot the intercepts and draw the line through them.

Graphs, Linear Equations, and Functions Since every point of the line has C-coordinate #, the C-intercept is Ð!ß #Ñ. Draw the horizontal line through Ð!ß #Ñ.

43.

Cœ& This is a horizontal line. Every point has C-coordinate &, so no point has C-coordinate !. There is no B-intercept. Since every point of the line has C-coordinate &, the C-intercept is Ð!ß &Ñ. Draw the horizontal line through Ð!ß &Ñ.

51.

B  &C œ ! To find the B-intercept, let C œ !. B  &C œ ! B  &Ð!Ñ œ ! Bœ!

45.

47.

The B-intercept is Ð!ß !Ñ, and since B œ !, this is also the C-intercept. Since the intercepts are the same, another point is needed to graph the line. Choose any number for C, say C œ ", and solve the equation for B.

Bœ# This is a vertical line. Every point has B-coordinate #, so the B-intercept is Ð#ß !Ñ. Since every point of the line has B-coordinate #, no point has B-coordinate !. There is no C-intercept. Draw the vertical line through Ð#ß !Ñ.

B  % œ ! ÐB œ %Ñ This is a vertical line. Every point has B-coordinate %, so the B-intercept is Ð%ß !Ñ. Since every point of the line has B-coordinate %, no point has B-coordinate !. There is no C-intercept. Draw the vertical line through Ð%ß !Ñ.

B  &C œ ! B  &Ð"Ñ œ ! Bœ& This gives the ordered pair Ð&ß "Ñ. Plot Ð&ß "Ñ and Ð!ß !Ñ, and draw the line through them.

53.

#B œ $C If B œ !, then C œ !, so the B- and C-intercepts are Ð!ß !Ñ. To get another point, let B œ $. #Ð$Ñ œ $C #œC Plot Ð$ß #Ñ and Ð!ß !Ñ, and draw the line through them.

49.

C  # œ ! ÐC œ #Ñ This is a horizontal line. Every point has C-coordinate #, so no point has C-coordinate !. There is no B-intercept.

785

Graphs, Linear Equations, and Functions &B œ) # &  B œ "' B œ ""

 #$ C œ B If B œ !, then C œ !, so the B- and C-intercepts are Ð!ß !Ñ. To get another point, let C œ $.

55.

 #$ Ð$Ñ œ B # œ B

)C œ# # )C œ% C œ %

Thus, the endpoint U is Ð""ß %Ñ.

Plot Ð#ß $Ñ and Ð!ß !Ñ, and draw the line through them.

71.

midpoint of T Ð"Þ&ß "Þ#&Ñ and UÐBß CÑ œ Q Ð$ß "Ñ "Þ&  B "Þ#&  C ß Œ  œ Ð$ß "Ñ # # The B- and C-coordinates must be equal. "Þ&  B "Þ#&  C œ$ œ" # # "Þ&  B œ ' "Þ#&  C œ # B œ %Þ& C œ !Þ(& Thus, the endpoint U is Ð%Þ&ß !Þ(&Ñ.

By the Midpoint Formula, the midpoint of the segment with endpoints Ð)ß %Ñ and Ð#ß 'Ñ is

57. Œ

73.

The graph goes through the point Ð#ß !Ñ which satisfies only equations B and C. The graph also goes through the point Ð!ß $Ñ which satisfies only equations A and B. Therefore, the correct equation is B.

75.

The screen on the right is more useful because it shows the intercepts.

77.

We need to solve the given equation for C.

)  Ð#Ñ %  Ð'Ñ "! # ß ß œŒ  œ Ð&ß "ÑÞ # # # # By the Midpoint Formula, the midpoint of the segment with endpoints Ð$ß 'Ñ and Ð'ß $Ñ is

59.

$  ' '  $ * $ * $ ß Œ œŒ ß  œ Œ ß  Þ # # # # # #

&B  #C œ "! #C œ &B  "! C œ  &# B  &

By the Midpoint Formula, the midpoint of the segment with endpoints Ð*ß $Ñ and Ð*ß )Ñ is

61.

Graph Y" œ #Þ&X  & in a standard viewing window.

*  * $  ) ! "" "" ß Œ  œ Œ ß  œ Œ!ß Þ # # # # # By the Midpoint Formula, the midpoint of the segment with endpoints Ð#Þ&ß $Þ"Ñ and Ð"Þ(ß "Þ$Ñ is

63.

Œ

#Þ&  "Þ( $Þ"  Ð"Þ$Ñ %Þ# "Þ) ß œŒ ß  œ #Þ"ß !Þ* Þ # # # # By the Midpoint Formula, the midpoint of the segment with endpoints ˆ "# ß "$ ‰ and ˆ $# ß &$ ‰ is

65.

 $#  # ß " #

67.  69.

" $

% '  &$ # # # $ œ ß  œ Œ ß  œ "ß " Þ   # # # # #

By the Midpoint Formula, the midpoint of the " ‰ segment with endpoints ˆ "$ ß #( ‰ and ˆ "# ß "% is  "$

ˆ "# ‰

 #

ß

# (

 #

" "%

& &  œ  # ß #  œ Œ "# ß #) Þ  '&

79. 81. 83.

& "%

midpoint of T Ð&ß )Ñ and UÐBß CÑ œ Q Ð)ß #Ñ &B )C ß Œ  œ Ð)ß #Ñ # # The B- and C-coordinates must be equal.

786

Subtract 5x. Divide by 2.

85.

'# % œ œ# &$ # &  Ð&Ñ &  & ! œ œ œ! $# " " %B  C œ "! C œ %B  "! Subtract 4x. C œ %B  "! Multiply by 1. $B  %C œ "# %C œ $B  "# "#  $B , Cœ % $ or C œ  % B  $

Subtract 3x. Divide by 4.

Graphs, Linear Equations, and Functions

2

From Ð%ß "Ñ, move down # units and then $ units to the right to Ð"ß "Ñ. Draw the line through the two points.

The Slope of a Line

2 Now Try Exercises N1. If B" ß C" œ Ð#ß 'Ñ and B# ß C# œ Ð$ß &Ñ, then 7œ

C#  C " &  Ð'Ñ "" "" œ œ œ . B#  B " $  # & &

The slope is  "" & . N2.

To find the slope of the line with equation $B  (C œ #", first find the intercepts. The B-intercept is Ð(ß !Ñ, and the C-intercept is Ð!ß $Ñ. The slope is then $  ! $ $ œ œ . !( ( (

7œ The slope is $( .

N6. Find the slope of each line. The line through Ð#ß &Ñ and Ð%ß )Ñ has slope 7" œ

The line through Ð#ß !Ñ and Ð"ß #Ñ has slope 7# œ

N3. (a) To find the slope of the line with equation C  ' œ !, select two different points on the line, such as Ð!ß 'Ñ and Ð#ß 'Ñ, and use the slope formula.

)& $ œ . %# #

#  ! # # œ œ Þ "  # $ $

The slopes are not the same, so the lines are not parallel. N7. Solve each equation for C. B  #C œ ( #C œ B  ( C œ  "# B  (&

#B  C œ % C œ #B  % C œ #B  %

The slope is !.

Ðso 7" œ  "# Ñ

Ðso 7# œ #Ñ

(b) To find the slope of the line with equation

Since 7" 7# œ Ð "# ÑÐ#Ñ œ ", the lines are perpendicular.

7œ

'' ! œ œ! #! #

B œ %,

N8. Solve each equation for C.

select two different points on the line, such as Ð%ß !Ñ and Ð%ß $Ñ, and use the slope formula. 7œ

$! $ œ %% !

Since division by zero is undefined, the slope is undefined. N4. Solve the equation for C. &B  %C œ ( %C œ &B  ( C œ &% B  (%

Subtract 5x. Divide by 4.

The slope is given by the coefficient of B, so the slope is &% . N5. Through Ð%ß "Ñ; 7 œ  #$ Locate the point Ð%ß "Ñ on the graph. Use the slope formula to find a second point on the line. 7œ

change in C # œ change in B $

#B  C œ % C œ #B  % C œ #B  % (so 7" œ #)

#B  C œ ' C œ #B  ' (so 7# œ #)

Since 7" Á 7# , the lines are not parallel. Since 7" 7# œ #Ð#Ñ œ %, the lines are not perpendicular either. Therefore, the answer is neither. N9. (a) B" ß C" œ Ð#!!!ß '*!Ñ and B# ß C# œ Ð#!!#ß )#)Ñ. C#  C" B#  B" )#)  '*! œ #!!#  #!!! "$) œ œ '* #

average rate of change œ

The average rate of change is '* hr per yr. (b) It is greater than &), which is the average rate of change from 2000 to 2005 found in Example 9.

787

Graphs, Linear Equations, and Functions N10. B" ß C" œ Ð#!!!ß #)$)Ñ and B# ß C# œ Ð#!!)ß "))&Ñ.

17.

C#  C" B#  B" "))&  #)$) œ #!!)  #!!! *&$ œ œ ""*Þ"#& )

average rate of change œ

Thus, the average rate of change in sales of digital camcorders in the United States from 2000 to #!!8 was about $""* million per year.

19. 21.

23.

Choices A, !Þ$, B, correct. 3.

5.

D,

$! "!! ,

change in vertical position change in horizontal position "& feet $œ change in horizontal position $ ‚ change œ "& "& change œ œ& $

25.

So the change in horizontal position is & feet.

27.

(b) The slope is positive, so the line rises.

(b) The slope is positive, so the line rises. 29.

788

slope of GH œ

11.

$ rise œ œ" slope of IJ œ $ run

13.

slope of EJ œ

15.

7œ

$ rise œ œ " $ run

'# % œ œ# &$ #

(a) Let B" ß C" œ Ð#ß %Ñ and B# ß C# œ Ð%ß %Ñ. Then 7œ

C#  C " %% ! œ œ œ !. B#  B " %  # '

The slope is !. (b) The slope is zero, so the line is horizontal. 31.

( rise œ , which is undefined. ! run

9.

C#  C " '" & œ œ . B#  B " #  Ð%Ñ '

The slope is &' .

To get to F from E, we must go up # units and move right " unit. Thus, # rise œ œ #. " run

(a) Let B" ß C" œ Ð%ß "Ñ and B# ß C# œ Ð#ß 'Ñ. Then 7œ

(d) Graph B is the only graph that indicates that sales fell during the first two quarters.

slope of EF œ

C#  C " &  Ð$Ñ ) œ œ œ ). B#  B " "  Ð#Ñ "

The slope is ).

(b) Graph A indicates that sales leveled off during the fourth quarter.

7.

(a) Let B" ß C" œ Ð#ß $Ñ and B# ß C# œ Ð"ß &Ñ. Then 7œ

(a) Graph C indicates that sales leveled off during the second quarter.

(c) Graph D indicates that sales rose sharply during the first quarter, and then fell to the original level during the second quarter.

(a) "The line has positive slope" means that the line goes up from left to right. This is line B.

(d) "The line has undefined slope" means that there is no horizontal change; that is, the line is vertical. This is line D.

and G, $!%, are all

slope œ

&  Ð&Ñ &  & ! œ œ œ! $# " " $) & & 7œ œ œ , which is #  Ð#Ñ #  # ! undefined. 7œ

(c) "The line has slope !" means that there is no vertical change; that is, the line is horizontal. This is line A.

change in vertical position change in horizontal position $! feet œ "!! feet

slope œ

$ "! ,

%  Ð"Ñ %" & œ œ $  Ð&Ñ $  & #

(b) "The line has negative slope" means that the line goes down from left to right. This is line C.

2 Section Exercises 1.

7œ

(a) Let B" ß C" œ Ð#ß #Ñ and B# ß C# œ Ð%ß "Ñ. Then 7œ

C#  C " "  # $ " œ œ œ . B#  B " %  Ð#Ñ ' #

The slope is  "# . (b) The slope is negative, so the line falls. 33.

(a) Let B" ß C" œ Ð&ß $Ñ and B# ß C# œ Ð&ß #Ñ. Then

Graphs, Linear Equations, and Functions 7œ

C#  C " #  Ð$Ñ & œ œ . B#  B " && !

47.

To find the slope of &B  #C œ "!,

The slope is undefined.

first find the intercepts. Replace C with ! to find that the B-intercept is Ð#ß !Ñ; replace B with ! to find that the C-intercept is Ð!ß &Ñ. The slope is then

(b) The slope is undefined, so the line is vertical. 35.

(a) Let B" ß C" œ Ð"Þ&ß #Þ'Ñ and B# ß C# œ Ð!Þ&ß $Þ'Ñ. Then 7œ

C#  C " $Þ'  #Þ' " œ œ œ ". B#  B" !Þ&  "Þ& "

7œ

The slope is ".

To sketch the graph, plot the intercepts and draw the line through them.

(b) The slope is negative, so the line falls. 37.

&  ! & & œ œ . !# # #

Let B" ß C" œ Ð "' ß "# Ñ and B# ß C# œ Ð &' ß *# Ñ. Then C#  C " œ B#  B "

7œ

* # & '

 

" # " '

œ

) # % '

œ % • $# œ '.

The slope is '. 39.

& Let B" ß C" œ Ð #* ß ") Ñ and " B# ß C# œ Ð ") ß  *& Ñ. Then

7œ

C#  C " œ B#  B " œ

49.

& "&  &*  ")  ") œ " # & ")  Ð * Ñ ") ")  "& ") • & œ $.

C œ %B, replace B with ! and then B with " to get the ordered pairs !ß ! and "ß % , respectively. (There are other possibilities for ordered pairs.) The slope is then

The slope is $. 41.

The points shown on the line are Ð$ß $Ñ and "ß # . The slope is 7œ

43.

45.

7œ

#  $ & & œ œ . "  Ð$Ñ # #

%! % œ œ %. "! "

To sketch the graph, plot the two points and draw the line through them.

The points shown on the line are Ð$ß $Ñ and $ß $ . The slope is 7œ

In the equation

$  $ ' , which is undefined. œ $$ !

To find the slope of B  #C œ %, first find the intercepts. Replace C with ! to find that the B-intercept is Ð%ß !Ñ; replace B with ! to find that the C-intercept is Ð!ß #Ñ. The slope is then 7œ

#! # " œ œ . !% % #

To sketch the graph, plot the intercepts and draw the line through them.

51.

B$œ!

ÐB œ $Ñ

The graph of B œ $ is the vertical line with B-intercept Ð$ß !Ñ. The slope of a vertical line is undefined.

789

Graphs, Linear Equations, and Functions 53.

C œ & The graph of C œ & is the horizontal line with C-intercept Ð!ß &Ñ. The slope of a horizontal line is !.

55.

#C œ $

61.

Locate Ð"ß #Ñ. Then use 7 œ $ œ $" to go $ units up and " unit right to Ð!ß "Ñ. Draw the line through Ð"ß #Ñ and Ð!ß "Ñ.

63.

Locate Ð#ß &Ñ. A slope of ! means that the line is horizontal, so C œ & at every point. Draw the horizontal line through Ð#ß &Ñ.

65.

Locate Ð$ß "Ñ. Since the slope is undefined, the line is vertical. The B-value of every point is $. Draw the vertical line through Ð$ß "Ñ.

67.

If a line has slope  %* , then any line parallel to it has slope  %* (the slope must be the same), and

ÐC œ $# Ñ

The graph of C œ $# is the horizontal line with C-intercept Ð!ß $# Ñ. The slope of a horizontal line is !.

57.

To graph the line through %ß # with slope 7 œ "# , locate Ð%ß #Ñ on the graph. To find a second point, use the definition of slope. 7œ

change in C " œ change in B #

From %ß # , go up " unit. Then go # units to the right to get to Ð#ß $Ñ. Draw the line through %ß # and #ß $ .

59.

To graph the line through !ß # with slope 7 œ  #$ , locate the point Ð!ß #Ñ on the graph. To find a second point on the line, use the definition of slope, writing  #$ as # $ . 7œ

change in C # œ change in B $

From !ß # , move # units down and then $ units to the right. Draw a line through this second point and !ß # . (Note that the slope could also be # written as $ . In this case, move # units up and $ units to the left to get another point on the same line.)

790

any line perpendicular to it has slope must be the negative reciprocal). 69.

* %

(the slope

The slope of the line through "&ß * and "#ß ( is 7œ

(  * "' "' . œ œ "#  "& $ $

The slope of the line through )ß % and &ß #! is 7œ

#!  Ð%Ñ "' "' . œ œ &) $ $

Graphs, Linear Equations, and Functions Since the slopes are equal, the two lines are parallel. 71.

and B  %C œ ( %B  C œ $ Solve the equations for C. %C œ B  ( C œ %B  $ " ( C œ %B  % C œ %B  $

87.

75.

and $B  %C œ # %B  $C œ ' Solve the equations for C. $C œ %B  ' %C œ $B  # C œ %$ B  # C œ $% B  "# The slopes are %$ and $% . The lines are neither parallel nor perpendicular. Bœ'

and

'Bœ)

89.

79.

81.

%!&  %%$ $) œ ¸ & #!!(  #!!! (

The average rate of change is & theaters per year. (b) The negative slope means that the number of drive-in theaters decreased by an average of & each year from 2000 to 2007. 91.

Use Ð#!!$ß "&*!Ñ and Ð#!!'ß &(!&ÑÞ average rate of change œ

and %B  C œ ! &B  ) œ #C Solve the equations for C. & C œ %B #B  % œ C & The slopes are % and # . The lines are neither parallel nor perpendicular. and #B œ C  $ #C  B œ $ Solve the equations for C. #B  $ œ C #C œ B  $ C œ  "# B  $# " The slopes, # and  # , are negative reciprocals of one another, so the lines are perpendicular.

(a) Use Ð#!!!ß %%$Ñ and Ð#!!(ß %!&Ñ. 7œ

The second equation can be simplified as B œ #. Both lines are vertical lines, so they are parallel. 77.

#(!Þ*  #!(Þ$ '$Þ' œ œ #"Þ# #!!)  #!!& $

(b) The number of subscribers increased by an average of #"Þ# million each year from 2005 to 2008.

The slopes,  "% and %, are negative reciprocals of one another, so the lines are perpendicular. 73.

(a) 7 œ

&(!&  "&*! %""& œ ¸ "$("Þ'( #!!'  #!!$ $

The average rate of change in sales is about $"$("Þ'( million per year, that is, the sales of plasma TVs in the United States increased by an average of $"$("Þ'( million each year from 2003 to 2006. 93.

Label the points as shown in the figure.

Use the points Ð!ß #!Ñ and Ð%ß %Ñ. average rate of change œ

change in C %  #! "' œ œ œ % change in B %! %

The average rate of change is $%!!! per year, that is, the value of the machine is decreasing $%!!! each year during these years. 83.

We can see that there is no change in the percent of pay raise. Thus, the average rate of change is !% per year, that is, the percent of pay raise is not changing—it is $% each year during these years.

85.

Let C be the vertical rise. Since the slope is the vertical rise divided by the horizontal run, C . !Þ"$ œ "&! Solving for C gives C œ !Þ"$Ð"&!Ñ œ "*Þ&. The vertical rise could be a maximum of "*Þ& ft.

In order to determine whether ABCD is a parallelogram, we need to show that the slope of AB equals the slope of CD and that the slope of AD equals the slope of BC. *  Ð"Ñ ) œ œ% "$  Ð""Ñ # '  Ð#Ñ ) Slope of CD œ œ œ% %# # '  Ð"Ñ ( Slope of AD œ œ %  Ð""Ñ "& #  Ð*Ñ ( Slope of BC œ œ #  Ð"$Ñ "& Thus, the figure is a parallelogram. Slope of AB œ

791

Graphs, Linear Equations, and Functions 95.

For EÐ$ß "Ñ and FÐ'ß #Ñ, the slope of AB is 7œ

96.

For FÐ'ß #Ñ and GÐ*ß $Ñ, the slope of BC is 7œ

97.

99.

$# " œ . *' $

Linear Equations in Two Variables

3 Now Try Exercises N1. Slope #$ ; C-intercept Ð!ß "Ñ Here 7 œ #$ and , œ ". Substitute these values into the slope-intercept form. C œ 7B  , C œ #$ B  "

For EÐ$ß "Ñ and GÐ*ß $Ñ, the slope of AC is 7œ

98.

#" " œ . '$ $

3

$" # " œ œ . *$ ' $

The slope of AB œ slope of BC œ slope of AC œ "$ . For EÐ"ß #Ñ and FÐ$ß "Ñ, the slope of AB is 7œ

"  Ð#Ñ " œ . $" #

N2. %B  $C œ ' Solve the equation for C. $C œ %B  ' C œ  %$ B  # Plot the C-intercept Ð!ß #Ñ. The slope can be % % interpreted as either % $ or $ . Using $ , move from Ð!ß #Ñ down % units and to the right $ units to locate the point Ð$ß #Ñ. Draw a line through the two points.

For FÐ$ß "Ñ and GÐ&ß !Ñ, the slope of BC is 7œ

!  Ð"Ñ " œ . &$ #

For EÐ"ß #Ñ and GÐ&ß !Ñ, the slope of AC is 7œ

!  Ð#Ñ # " œ œ . &" % #

Since the three slopes are the same, the three points are collinear. 100. For EÐ!ß 'Ñ and FÐ%ß &Ñ, the slope of AB is 7œ

&  ' "" "" œ œ . %! % %

For FÐ%ß &Ñ and GÐ#ß "#Ñ, the slope of BC is "#  Ð&Ñ "( "( 7œ œ œ . #  % ' ' Since these two slopes are not the same, the three points are not collinear. 101. $B  #C œ ) #C œ $B  ) C œ  $# B  % 103. C  # œ %ÐB  $Ñ C  # œ %B  "# C œ %B  "% 105. C  Ð"Ñ œ  "# ÒB  Ð#ÑÓ C  " œ  "# ÐB  #Ñ #ÐC  "Ñ œ "ÐB  #Ñ #C  # œ B  # B  #C œ %

792

N3. Through Ð&ß $Ñ; slope œ 7 œ  "& Use the point-slope form with B" ß C" œ Ð&ß $Ñ and 7 œ  "& . C  C" œ 7 B  B " C  Ð$Ñ œ  "& ÐB  &Ñ C  $ œ  "& ÐB  &Ñ C  $ œ  "& B  " C œ  "& B  #

Subtract 3.

N4. (a) Through Ð%ß %Ñ; 7 undefined This is a vertical line since the slope is undefined. A vertical line through the point Ð+ß ,Ñ has equation B œ +. Here the B-coordinate is %, so the equation is B œ %. (b) Through Ð%ß %Ñ; 7 œ ! Since the slope is !, this is a horizontal line. A horizontal line through the point Ð+ß ,Ñ has equation C œ , . Here the C-coordinate is %, so the equation is C œ %. N5. Through Ð$ß %Ñ and Ð#ß "Ñ 7œ

"  Ð%Ñ $ $ œ œ #  $ & &

Let B" ß C" œ Ð$ß %Ñ.

Graphs, Linear Equations, and Functions N9. (a) Use Ð#ß ")$Ñ and Ð'ß #&"Ñ.

C  C" œ 7ÐB  B" Ñ C  Ð%Ñ œ  $& ÐB  $Ñ C  % œ  $& ÐB  $Ñ &C  #! œ $B  * $B  &C œ ""

7œ Multiply by 5. Standard form

Use the point-slope form with ÐB" ß C" Ñ œ Ð#ß ")$Ñ. C  C" œ 7 B  B " C  ")$ œ "(ÐB  #Ñ C  ")$ œ "(B  $% C œ "(B  "%*

N6. (a) Through Ð'ß "Ñ; parallel to the line $B  &C œ ( Find the slope of the given line. $B  &C œ ( &C œ $B  ( C œ $& B  (&

(b) For 2011, B œ #!""  #!!! œ "". C œ "(B  "%* C œ "(Ð""Ñ  "%* œ ")(  "%* œ $$'

The slope is $& , so a line parallel to it also has slope $ $ & . Use 7 œ & and B" ß C" œ Ð'ß "Ñ in the point-slope form.

C œ $& B  C œ $& B 

") & ") & #$ &



Let x œ 11.

Using the equation from part (a), we estimate the retail spending on prescription drugs in 2011 to be $$$' billion.

C  C" œ 7ÐB  B" Ñ C  Ð"Ñ œ $& ÐB  'Ñ C  " œ $& B 

#&"  ")$ ') œ œ "( '# %

& &

(b) Through Ð'ß "Ñ; perpendicular to $B  &C œ (

3 Section Exercises 1.

Choice A, $B  #C œ &, is in the form EB  FC œ G with E   ! and integers E, F , and G having no common factor (except ").

The slope of $B  &C œ ( is $& . The negative reciprocal of $& is  &$ , so the slope of the line through Ð'ß "Ñ is  &$ .

3.

Choice A, C œ 'B  #, is in the form C œ 7B  , .

5.

C  C" œ 7ÐB  B" Ñ C  Ð"Ñ œ  &$ ÐB  'Ñ

C  # œ $ÐB  %Ñ C  # œ $B  "# $B  C œ "!

7.

C œ #B  $ This line is in slope-intercept form with slope 7 œ # and C-intercept Ð!ß ,Ñ œ Ð!ß $Ñ. The only graph with positive slope and with a positive C-coordinate of its C-intercept is A.

9.

C œ #B  $ This line is in slope-intercept form with slope 7 œ # and C-intercept Ð!ß ,Ñ œ Ð!ß $Ñ. The only graph with negative slope and with a negative C-coordinate of its C-intercept is C.

11.

C œ #B This line has slope 7 œ # and C-intercept Ð!ß ,Ñ œ Ð!ß !Ñ. The only graph with positive slope and with C-intercept Ð!ß !Ñ is H.

13.

Cœ$ This line is a horizontal line with C -intercept Ð!ß $Ñ. Its C-coordinate is positive. The only graph that has these characteristics is B.

15.

7 œ &; , œ "& Substitute these values in the slope-intercept form.

C  " œ  &$ B  "! C œ  &$ B  * N7. Since the price you pay is $)& per month plus a flat fee of $"!!, an equation for B months is C œ )&B  "!!. N8. (a) Use Ð!ß #!(*Ñ for 2004 and Ð%ß #$(#Ñ for 2008. 7œ

#$(#  #!(* #*$ œ œ ($Þ#& %! % C œ 7B  , C œ ($Þ#&B  #!(*

(b) For 2009, B œ #!!*  #!!% œ &. C œ ($Þ#&B  #!(* C œ ($Þ#&Ð&Ñ  #!(* C œ $''Þ#&  #!(* C œ #%%&Þ#&

Let x = 5.

According to the model, average tuition and fees for in-state students at public two-year colleges in 2009 were about $#%%&.

Standard form

C œ 7B  , C œ &B  "& 793

Graphs, Linear Equations, and Functions 17.

7 œ  #$ ; , œ %& Substitute these values in the slope-intercept form. C œ 7B  , C œ  #$ B  %&

19.

Slope "; C-intercept Ð!ß "Ñ Here, 7 œ " and , œ ". Substitute these values in the slope-intercept form.

(d)

31.

C œ 7B  , C œ "B  ", or C œ B  " 21.

%B  &C œ #! &C œ %B  #! C œ %& B  %

Slope #& ; C-intercept Ð!ß &Ñ Here, 7 œ #& and , œ &. Substitute these values in the slope-intercept form.

(b) The slope is the coefficient of B, %& Þ

C œ 7B  , C œ #& B  & 23.

(c) The C-intercept is the point Ð!ß ,Ñ, or Ð!ß %Ñ.

To get to the point Ð$ß $Ñ from the C-intercept Ð!ß "Ñ, we must go up # units and to the right $ units, so the slope is #$ . The slope-intercept form is (d)

C œ #$ B  ". 25.

To get to the point Ð$ß "Ñ from the C-intercept Ð!ß #Ñ, we must go up $ units and to the left $ $ units, so the slope is $ œ ". The slopeintercept form is

33.

C œ "B  #, or C œ B  #. 27.

%B  &C œ #! (a) Solve for C to get the equation in slopeintercept form.

B  #C œ % (a) Solve for C to get the equation in slopeintercept form.

B  C œ % (a) Solve for C to get the equation in slopeintercept form.

B  #C œ % #C œ B  % C œ  "# B  #

B  C œ % C œB%

(b) The slope is the coefficient of B,  "# Þ (c) The C-intercept is the point Ð!ß ,Ñ, or Ð!ß #Ñ.

(b) The slope is the coefficient of B, ". (c) The C-intercept is the point Ð!ß ,Ñ, or Ð!ß %Ñ.

(d) (d) 35. 29.

'B  &C œ $! (a) Solve for C to get the equation in slopeintercept form. 'B  &C œ $! &C œ 'B  $! C œ  '& B  ' (b) The slope is the coefficient of B,  '& Þ (c) The C-intercept is the point Ð!ß ,Ñ, or Ð!ß 'Ñ.

794

(a) Through Ð&ß )Ñ; slope # Use the point-slope form with B" ß C" œ Ð&ß )Ñ and 7 œ #. C  C" œ 7ÐB  B" Ñ C  ) œ # B  & C  ) œ #B  "! #B  C œ ") (b) Solve the last equation from part (a) for C. #B  C œ ") C œ #B  ")

Graphs, Linear Equations, and Functions 37.

(a) Through Ð#ß %Ñ; slope  $% Use the point-slope form with B" ß C" œ Ð#ß %Ñ and 7 œ  $% .

&C  $% œ (B  "% (B  &C œ #! (B  &C œ #!

C  C" œ 7ÐB  B" Ñ C  % œ  $% cB  Ð#Ñd % C  % œ $ÐB  #Ñ %C  "' œ $B  ' $B  %C œ "! (b) Solve the last equation from part (a) for C.

(b) Solve the last equation from part (a) for C. (B  &C œ #! &C œ (B  #! C œ (& B  %, or C œ "Þ%B  % 45.

Through Ð*ß &Ñ; slope ! A line with slope ! is a horizontal line. A horizontal line through the point ÐBß 5Ñ has equation C œ 5 . Here 5 œ &, so an equation is C œ &.

47.

Through Ð*ß "!Ñ; undefined slope A vertical line has undefined slope and equation B œ - . Since the B-value in Ð*ß "!Ñ is *, the equation is B œ *.

49.

Through Ð $% ß  $# Ñ; slope ! The equation of this line is C œ  $# .

51.

Through Ð(ß )Ñ; horizontal A horizontal line through the point ÐBß 5Ñ has equation C œ 5 , so the equation is C œ ).

53.

Through Ð!Þ&ß !Þ#Ñ; vertical A vertical line through the point Ð5ß CÑ has equation B œ 5 . Here 5 œ !Þ&, so the equation is B œ !Þ&.

55.

(a) $ß % and &ß ) Find the slope.

$B  %C œ "! %C œ $B  "! C œ  $% B  Cœ 39.

 $% B



"! % & #

" #

(a) Through Ð&ß %Ñ; slope Use the point-slope form with B" ß C" œ Ð&ß %Ñ and 7 œ "# . C  C" œ 7ÐB  B" Ñ C  % œ "# cB  Ð&Ñd # C  % œ "ÐB  &Ñ #C  ) œ B  & B  #C œ "$ B  #C œ "$ (b) Solve the last equation from part (a) for C. B  #C œ "$ #C œ B  "$ C œ "# B  "$ #

41.

(a) Through Ð$ß !Ñ; slope % Use the point-slope form with B" ß C" œ Ð$ß !Ñ and 7 œ %. C  C" œ 7ÐB  B" Ñ C  ! œ %ÐB  $Ñ C œ %B  "# %B  C œ "# %B  C œ "# (b) Solve the last equation from part (a) for C. %B  C œ "# C œ %B  "# C œ %B  "#

43.

(a) Through Ð#ß 'Þ)Ñ; slope "Þ% Use the point-slope form with B" ß C" œ Ð#ß 'Þ)Ñ and 7 œ "Þ%. C  C" œ 7ÐB  B" Ñ C  'Þ) œ "Þ%ÐB  #Ñ C  'Þ) œ (& ÐB  #Ñ &ÐC  'Þ)Ñ œ (ÐB  #Ñ

7œ

)% % œ œ# &$ #

Use the point-slope form with B" ß C" œ Ð$ß %Ñ and 7 œ #. C  C" œ 7ÐB  B" Ñ C  % œ #ÐB  $Ñ C  % œ #B  ' #B  C œ # #B  C œ # (b) Solve the last equation from part (a) for C. #B  C œ # C œ #B  # C œ #B  # Note: You could use any of the equations in part (a) to solve for C. In this exercise, choosing C  % œ #B  ' or #B  C œ #, easily leads to the equation C œ #B  #.

795

Graphs, Linear Equations, and Functions 57.

Use the point-slope form with # B" ß C" œ Ð #& ß #& Ñ and 7 œ "$ .

(a) 'ß " and #ß & Find the slope. 7œ

&" % " œ œ #  ' ) #

C "$C 

67.

(a) Through Ð(ß #Ñ; parallel to the graph of the line having equation $B  C œ ) Find the slope of $B  C œ ).

C  C" œ 7ÐB  B" Ñ C  # œ $ÐB  (Ñ C  # œ $B  #" C œ $B  "*

(b) C œ & is already in the slope-intercept form. (a) (ß ' and (ß ) Find the slope.

(b)

69.

(a) Through Ð#ß #Ñ; parallel to B  #C œ "! Find the slope of B  #C œ "!.

The slope is "# , so a line parallel to it also has slope " " # . Use 7 œ # and B" ß C" œ Ð#ß #Ñ in the point-slope form.

(a) Ð "# ß $Ñ and Ð #$ ß $Ñ Find the slope. 7œ

C  C" œ 7ÐB  B" Ñ C  Ð#Ñ œ "# cB  Ð#Ñd

$  Ð$Ñ ! œ ( œ! # " $  # '

C#œ

A line with slope ! is horizontal. A horizontal line through the point ÐBß 5Ñ has equation C œ 5 , so the equation is C œ $.

C#œ Cœ (b)

(b) C œ $ is already in the slope-intercept form. 65.

(a) Ð #& ß #& Ñ and Ð %$ ß #$ Ñ Find the slope. 7œ œ

796

# $



# &

œ

% # $  Ð & Ñ % % "& œ œ #' #' "&

"!  ' "& #!  ' "&

# "$

C œ $B  "* $B  C œ "* $B  C œ "*

#C œ B  "! C œ "# B  &

(b) It is not possible to write B œ ( in slopeintercept form. 63.

% &

The slope is $, so a line parallel to it also has slope $. Use 7 œ $ and B" ß C" œ Ð(ß #Ñ in the pointslope form.

&& ! œ œ! "# " A line with slope ! is horizontal. A horizontal line through the point ÐBß 5Ñ has equation C œ 5 , so the equation is C œ &.

A line with undefined slope is a vertical line. The equation of a vertical line is B œ 5 , where 5 is the common B-value. So the equation is B œ (.

œ #B 

C œ $B  ) C œ $B  )

7œ

)  ' "% Undefined œ (( !

#' &

#B  "$C œ ' "$C œ #B  ' # ' C œ "$ B  "$

(a) #ß & and "ß & Find the slope.

7œ

 Ð #& Ñ‘

(b) Solve the last equation from part (a) for C.

(b) Solve the last equation from part (a) for C.

61.

#  "$ B

#B  "$C œ $! & #B  "$C œ '

C  C" œ 7ÐB  B" Ñ C  " œ  "# ÐB  'Ñ # C  " œ "ÐB  'Ñ #C  # œ B  ' B  #C œ )

59.

œ

"$ÐC  #& Ñ œ #ÐB  #& Ñ

Use the point-slope form with B" ß C" œ Ð'ß "Ñ and 7 œ  "# .

B  #C œ ) #C œ B  ) C œ  "# B  %

# &

71.

" # B# " #B  " " #B  "

" #B

Cœ " #C œ B  # B  #C œ # B  #C œ #

Multiply by 2.

(a) Through Ð)ß &Ñ; perpendicular to #B  C œ ( Find the slope of #B  C œ (. C œ #B  ( C œ #B  (

Graphs, Linear Equations, and Functions The slope of the line is #. Therefore, the slope of the line perpendicular to it is  "# since #Ð "# Ñ œ ". Use 7 œ  "# and B" ß C" œ Ð)ß &Ñ in the point-slope form. C  C" œ 7ÐB  B" Ñ C  & œ  "# ÐB  )Ñ

(b) If B œ &, C œ %"Ð&Ñ  ** œ $!%Þ The ordered pair is Ð&ß $!%Ñ. The cost for a 5-month membership is $$!%. (c) If B œ "#, C œ %"Ð"#Ñ  ** œ &*"Þ The cost for the first year's membership is $&*"Þ 85.

C  & œ  "# B  %

(a) The fixed cost is $$', so that is the value of , . The variable cost is $'!, so

C œ  "# B  * (b)

73.

C œ 7B  , œ '!B  $'Þ

 "# B

Cœ * #C œ B  ") B  #C œ ")

(b) If B œ &, C œ '!Ð&Ñ  $' œ $$'Þ The ordered pair is Ð&ß $$'Ñ. The cost of the plan for & months is $$$'Þ

Multiply by 2.

(a) Through Ð#ß (Ñ; perpendicular to B œ * B œ * is a vertical line so a line perpendicular to it will be a horizontal line. It goes through Ð#ß (Ñ, so its equation is

(c) For a 1-year contract, B œ "#, so C œ '!Ð"#Ñ  $' œ (&'Þ The cost of the plan for " year is $(&'Þ 87.

C œ (Þ

(a) The fixed cost is $$!, so that is the value of , . The variable cost is $', so C œ 7B  , œ 'B  $!Þ

(b) C œ ( is already in standard form. 75.

(b) If B œ &, C œ 'Ð&Ñ  $! œ '!Þ The ordered pair is Ð&ß '!Ñ. It costs $'! to rent the saw for & days.

Distance œ (rate)(time), so C œ %&BÞ B ! & "!

77.

C œ %&B %&Ð!Ñ œ ! %&Ð&Ñ œ ##& %&Ð"!Ñ œ %&!

Ordered Pair Ð!ß !Ñ Ð&ß ##&Ñ Ð"!ß %&!Ñ

Total cost œ (cost/gal)(number of gallons), so C œ $Þ"!BÞ B ! & "!

79.

(c) "$) œ 'B  $! "!) œ 'B "!) Bœ œ ") '

C œ $Þ"!B $Þ"!Ð!Ñ œ ! $Þ"!Ð&Ñ œ "&Þ&! $Þ"!Ð"!Ñ œ $"Þ!!

The saw is rented for ") days. 89.

Ordered Pair Ð!ß !Ñ Ð&ß "&Þ&!Ñ Ð"!ß $"Þ!!Ñ

81.

Ordered Pair Ð!ß !Ñ Ð&ß &&&Ñ Ð"!ß """!Ñ

(b) The year 2007 corresponds to B œ %, so C œ "#*%Þ(Ð%Ñ  $*#" œ $*!**Þ) million. 91.

C  $) œ &Þ#&ÐB  $Ñ C  $) œ &Þ#&B  "&Þ(& C œ &Þ#&B  ##Þ#& (b) The year 2005 corresponds to B œ &, so C œ &Þ#&Ð&Ñ  ##Þ#& œ %)Þ&. This value is greater than the actual value.

(c) If B œ #, C œ ""#Þ&!Ð#Ñ  "# œ #$(Þ The cost of 2 tickets and a parking pass is $#$(Þ

C œ 7B  , œ %"B  **Þ

&*Þ!  $)Þ! #" œ œ &Þ#& ($ %

Now use the point-slope form.

(b) If B œ &, C œ ""#Þ&!Ð&Ñ  "# œ &(%Þ&!Þ The ordered pair is Ð&ß &(%Þ&!Ñ. The cost of 5 tickets and a parking pass is $&(%Þ&!.

(a) The fixed cost is $**, so that is the value of , . The variable cost is $%", so

(a) Use Ð$ß $)Þ!Ñ and Ð(ß &*Þ!Ñ. 7œ

(a) The fixed cost is $"#, so that is the value of , . The variable cost is $""#Þ&!, so C œ 7B  , œ ""#Þ&!B  "#Þ

83.

()!&  $*#" $))% œ ¸ "#*%Þ( $! $

The equation is C œ "#*%Þ(B  $*#". The slope tells us that the sales of digital cameras in the United States increased by $1294.7 million per year from 2003 to 2006.

C œ """BÞ C œ """B """Ð!Ñ œ ! """Ð&Ñ œ &&& """Ð"!Ñ œ """!

(a) Use Ð!ß $*#"Ñ and Ð$ß ()!&Ñ. 7œ

Total cost œ (cost/credit)(number of credits), so

B ! & "!

Let y œ 138.

93.

When C œ !°, F œ $#° , and when C œ "!!°, F œ #"#° . These are the freezing and boiling temperatures for water. 797

Graphs, Linear Equations, and Functions 94.

The two points of the form ÐCß FÑ would be Ð!ß $#Ñ and Ð"!!ß #"#Ñ.

95.

7œ

96.

Let 7 œ

#"#  $# ")! * œ œ "!!  ! "!! & * &

Summary Exercises on Slopes and Equations of Lines 1.

$B  &C œ * &C œ $B  * C œ  $& B  *&

and B" ß C" œ Ð!ß $#Ñ. C  C" œ 7ÐB  B" Ñ F  $# œ *& ÐC  !Ñ

The slope is  $& .

F  $# œ *& C F œ *& C  $# F œ *& C  $#

97.

F  $# œ *& C & * ÐF

98.

Solve the equation for C.

3.

C œ #B  & is already in the slope-intercept form, so the slope is #.

5.

The graph of B  % œ !, or B œ %, is a vertical line with B-intercept Ð%ß !Ñ. The slope of a vertical line is undefined because the denominator equals zero in the slope formula.

7.

Through Ð#ß 'Ñ and Ð%ß "Ñ

 $#Ñ œ C

J œ *& G  $# If G œ $!,

J œ *& Ð$!Ñ  $# œ &%  $# œ )'Þ

(a) The slope is

Thus, when G œ $!°, J œ )'°. 99.

Gœ

& *

7œ

J  $#

When J œ &!,

Use the point-slope form.

G œ &* Ð&!  $#Ñ

C  C" œ 7ÐB  B" Ñ C  ' œ  &' ÒB  Ð#ÑÓ

œ &* Ð")Ñ œ "!Þ Thus, when J œ &!°, G œ "!°.

C  ' œ  &' B 

100. Let J œ G in the equation obtained in Exercise 96.

C œ  &' B  C œ  &' B 

J œ *& G  $# G œ *& G  $# &G œ &Ð *& G  $#Ñ %G œ "'! G œ %!

(b)

Let F = C. Multiply by 5. 9.

&G œ *G  "'! Subtract 9C.

101. #B  &  * #B  % B#

The solution set is Ð∞ß  %$ Ó.

798

Multiply by 6.

(a) Find the slope of #B  &C œ '. &C œ #B  ' C œ #& B  '& The slope of the line is #& . Therefore, the slope of the line perpendicular to it is  &# since # & & & Ð # Ñ œ ". Use 7 œ  # and B" ß C" œ Ð!ß !Ñ in the point-slope form. C  C" œ 7ÐB  B" Ñ C  ! œ  &# ÐB  !Ñ

The solution set is Ð∞ß #Ñ. 103. &  $B   * $B   % B Ÿ  %$

C œ  &' B  "$ $ 'C œ &B  #' &B  'C œ #'

& $ & ") $  $ "$ $

Through Ð!ß !Ñ; perpendicular to #B  &C œ '

Divide by 4.

(The same result may be found by using either form of the equation obtained in Exercise 97.) The Celsius and Fahrenheit temperatures are equal ÐJ œ GÑ at %! degrees.

"' & & œ œ . %  Ð#Ñ ' '

C œ  &# B (b)

C œ  &# B #C œ &B &B  #C œ !

Graphs, Linear Equations, and Functions 11.

The slope of the line through Ð!ß $Ñ and "ß "! is

Through Ð $% ß  (* Ñ; perpendicular to B œ #$ (a) B œ #$ is a vertical line so a line perpendicular to it will be a horizontal line. It goes through Ð $% ß  (* Ñ, so its equation is

7œ

Since we are given the C-intercept, Ð!ß $Ñ, the slope-intercept form is

C œ  (* Þ (b) 13.

C œ  (* *C œ (

(b)

Through Ð%ß #Ñ; parallel to the line through Ð$ß *Ñ and Ð'ß ""Ñ (a) The slope of the line through Ð$ß *Ñ and Ð'ß ""Ñ is ""  * # 7œ œ . '$ $ Use the point-slope form with ÐB" ß C" Ñ œ Ð%ß #Ñ and 7 œ #$ (since the slope of the desired line must equal the slope of the given line). C  C" œ 7ÐB  B" Ñ C  # œ #$ ÒB  Ð%ÑÓ C  # œ #$ ÐB  %Ñ C œ #$ B  C œ #$ B  (b)

15.

C œ (B  $Þ

Multiply by 9.

C  # œ #$ B 

"!  $ ( œ œ (. "  ! "

) $ ) ' $  $ "% $

C œ #$ B  "% $ $C œ #B  "% #B  $C œ "% #B  $C œ "%

4

C œ (B  $ (B  C œ $

Linear Inequalities in Two Variables

4 Now Try Exercises N1. B  #C   % Step 1 Graph the line, B  #C œ %, which has intercepts Ð%ß !Ñ and Ð!ß #Ñ, as a solid line since the inequality involves " Ÿ ". Step 2 Test Ð!ß !Ñ. BC  % ?

!! % ! %

False

Step 3 Since the result is false, shade the region that does not contain Ð!ß !Ñ.

Through Ð%ß )Ñ and Ð%ß "#Ñ (a) The slope is 7œ

"#  Ð)Ñ #! & œ œ . %  % ) #

Use the point-slope form.

N2. $B  C  '

C  C" œ 7ÐB  B" Ñ C  Ð)Ñ œ  &# ÐB  %Ñ C  ) œ  &# B  "! C œ  &# B  # (b)

17.

 &# B

Cœ # #C œ &B  % &B  #C œ %

Multiply by 2.

Through Ð!ß $Ñ and the midpoint of the segment with endpoints Ð#ß )Ñ and Ð%ß "#Ñ

Solve the inequality for C. C  $B  ' C  $B  '

Subtract 3x. Multiply by 1.

Graph the boundary line, C œ $B  ' [which has slope $ and C-intercept Ð!ß 'Ñ], as a dashed line because the inequality symbol is  . Since the inequality is solved for C and the inequality symbol is  , we shade the half-plane above the boundary line.

(a) The midpoint of the segment with endpoints Ð#ß )Ñ and Ð%ß "#Ñ is Œ

#  Ð%Ñ )  "# # #! ß ß  œ "ß "! . œŒ # # # #

799

Graphs, Linear Equations, and Functions The graph of the union is the region that includes all the points in both graphs.

N3. B  C  $ and C Ÿ # Graph B  C œ $, which has intercepts Ð$ß !Ñ and Ð!ß $Ñ, as a dashed line since the inequality involves "  ". Test Ð!ß !Ñ, which yields !  $, a true statement. Shade the region that includes Ð!ß !Ñ.

4 Section Exercises 1.

The boundary of the graph of C Ÿ B  # will be a solid line (since the inequality involves Ÿ ), and the shading will be below the line (since the inequality sign is Ÿ or  ).

3.

The boundary of the graph of C  B  # will be a dashed line (since the inequality involves  ), and the shading will be above the line (since the inequality sign is   or  ).

5.

The graph of EB  FC œ G divides the plane into two regions. In one of these regions, the ordered pairs satisfy EB  FC  G ; in the other, they satisfy EB  FC  G .

7.

BC Ÿ# Graph the line B  C œ # by drawing a solid line (since the inequality involves Ÿ ) through the intercepts #ß ! and !ß # . Test a point not on this line, such as Ð!ß !Ñ.

Graph C œ # as a solid horizontal line through Ð!ß #Ñ. Shade the region below C œ #.

The graph of the intersection is the region common to both graphs.

BC Ÿ# ?

!!Ÿ# !Ÿ#

N4. $B  &C  "& or B  % Graph $B  &C œ "& as a dashed line through its intercepts Ð&ß !Ñ and Ð!ß $Ñ. Test Ð!ß !Ñ, which yields !  "&, a true statement. Shade the region that includes Ð!ß !Ñ.

Graph B œ % as a dashed vertical line through Ð%ß !Ñ. Shade the region to the right of B œ %.

True

Shade the side of the line containing the test point Ð!ß !Ñ.

9.

%B  C  % Graph the line %B  C œ % by drawing a dashed line (since the inequality involves  ) through the intercepts "ß ! and !ß % . Instead of using a test point, we will solve the inequality for C . C  %B  % C  %B  %

800

Graphs, Linear Equations, and Functions Shade the side of the line not containing the test point Ð!ß !Ñ.

Since we have "C  " in the last inequality, shade the region above the boundary line.

11.

B  $C   # Graph the solid line B  $C œ # (since the inequality involves   ) through the intercepts #ß ! and Ð!ß  #$ Ñ. Test a point not on this line such as Ð!ß !Ñ.

17.

?

!  $Ð!Ñ   # !   #

BC ! Graph the line B  C œ !, which includes the points Ð!ß !Ñ and Ð#ß #Ñ, as a dashed line (since the inequality involves  ). Solving the inequality for C gives us C  B, So shade the region above the boundary line.

True

Shade the side of the line containing the test point Ð!ß !Ñ.

19. 13.

#B  $C   ' Graph the solid line #B  $C œ ' (since the inequality involves   ) through the intercepts $ß ! and Ð!ß #Ñ. Test a point not on this line such as Ð!ß !Ñ.

$C Ÿ B C   "$ B Shade the region above the boundary line.

?

#Ð!Ñ  $Ð!Ñ   ' ! '

B  $C Ÿ ! Graph the solid line B  $C œ ! through the points !ß ! and $ß " . Solve the inequality for C.

False

Shade the side of the line not containing the test point Ð!ß !Ñ.

21.

15.

CB Graph the dashed line C œ B through Ð!ß !Ñ and Ð#ß #Ñ. Since we have "C  " in the inequality, shade the region below the boundary line.

&B  $C  "& Graph the dashed line &B  $C œ "& (since the inequality involves  ) through the intercepts $ß ! and Ð!ß &Ñ. Test a point not on this line such as Ð!ß !Ñ. ?

&Ð!Ñ  $Ð!Ñ  "& !  "&

False

801

Graphs, Linear Equations, and Functions 23.

B  C Ÿ " and B   " Graph the solid line B  C œ " through !ß " and "ß ! . The inequality B  C Ÿ " can be written as C Ÿ B  ", so shade the region below the boundary line. Graph the solid vertical line B œ " through Ð"ß !Ñ and shade the region to the right. The required graph is the common shaded area as well as the portions of the lines that bound it.

29.

31.

kBk  $ can be rewritten as $  B  $. The boundaries are the dashed vertical lines B œ $ and B œ $. Since B is between $ and $, the graph includes all points between the lines.

kB  "k  # can be rewritten as #  B  "  # $  B  ".

25.

27.

802

The boundaries are the dashed vertical lines B œ $ and B œ ". Since B is between $ and ", the graph includes all points between the lines.

#B  C   # and C  % Graph the solid line #B  C œ # through the intercepts "ß ! and !ß # . Test Ð!ß !Ñ to get !   #, a false statement. Shade the side of the line not containing Ð!ß !Ñ. To graph C  % on the same axes, graph the dashed horizontal line through Ð!ß %Ñ. Test Ð!ß !Ñ to get !  %, a true statement. Shade the side of the dashed line containing Ð!ß !Ñ. The word "and" indicates the intersection of the two graphs. The final solution set consists of the region where the two shaded regions overlap.

33.

B  C   " or C   # Graph the solid line B  C œ ", which crosses the C-axis at " and the B-axis at ". Use Ð!ß !Ñ as a test point, which yields !   ", a false statement. Shade the region that does not include Ð!ß !Ñ. Now graph the solid line C œ #. Since the inequality is C   #, shade above this line. The required graph of the union includes all the shaded regions, that is, all the points that satisfy either inequality.

B  C  & and C  # Graph B  C œ &, which has intercepts Ð&ß !Ñ and Ð!ß &Ñ, as a dashed line. Test Ð!ß !Ñ, which yields !  &, a true statement. Shade the region that includes Ð!ß !Ñ. Graph C œ # as a dashed horizontal line. Shade the region below C œ #. The required graph of the intersection is the region common to both graphs.

35.

B  #  C or B  " Graph B  # œ C, which has intercepts #ß ! and !ß # , as a dashed line. Test Ð!ß !Ñ, which yields #  !, a false statement. Shade the region that does not include Ð!ß !Ñ. Graph B œ " as a dashed vertical line. Shade the region to the left of B œ ". The required graph of the union includes all the shaded regions, that is, all the points that satisfy either inequality.

Graphs, Linear Equations, and Functions 47.

37.

39.

41.

43.

$B  #C  ' or B  #C  # Graph $B  #C œ ', which has intercepts #ß ! and !ß $ , as a dashed line. Test Ð!ß !Ñ, which yields !  ', a true statement. Shade the region that includes Ð!ß !Ñ. Graph B  #C œ #, which has intercepts #ß ! and !ß " , as a dashed line. Test Ð!ß !Ñ, which yields !  #, a false statement. Shade the region that does not include Ð!ß !Ñ. The required graph of the union includes all the shaded regions, that is, all the points that satisfy either inequality.

(c) The solution set for C  ! is Ð%ß ∞Ñ, since the graph is above the B-axis for these values of B. (a) The B-intercept is Ð$Þ&ß !Ñ, so the solution set for C œ ! is e$Þ&f.

(b) The solution set for C  ! is Ð$Þ&ß ∞Ñ, since the graph is below the B-axis for these values of B. (c) The solution set for C  ! is Ð∞ß $Þ&Ñ, since the graph is above the B-axis for these values of B.

49.

The total daily cost G consists of $&! per worker and $"!! to manufacture " unit, so G œ &!B  "!!C.

50.

Some examples of points in the shaded region are Ð"&!ß %!!!Ñ, Ð"#!ß $&!!Ñ, and Ð")!ß '!!!Ñ. Some examples of points on the boundary are Ð"!!ß &!!!Ñ, Ð"&!ß $!!!Ñ, and Ð#!!ß %!!!Ñ. The corner points are Ð"!!ß $!!!Ñ and Ð#!!ß $!!!Ñ.

51.

ÐBß CÑ Ð"&!ß %!!!Ñ Ð"#!ß $&!!Ñ Ð")!ß '!!!Ñ Ð"!!ß &!!!Ñ Ð"&!ß $!!!Ñ Ð#!!ß %!!!Ñ Ð"!!ß $!!!Ñ Ð#!!ß $!!!Ñ

C Ÿ $B  ' The slope of the boundary line C œ $B  ' is $, and the C-intercept is '. This would be graph A or graph D. The inequality sign is Ÿ , so we want the region on or below the boundary line. The answer is graph A.

(b) The solution set for C  ! is Ð∞ß %Ñ, since the graph is below the B-axis for these values of B.

45.

48.

C Ÿ $B  ' The boundary line, C œ $B  ', has slope $ and C-intercept '. This would be graph B or graph C. Since we want the region less than or equal to $B  ', we want the region on or below the boundary line. The answer is graph C.

(a) The B-intercept is Ð%ß !Ñ, so the solution set for C œ ! is e%f.

"A factory can have no more than #!! workers on a shift, but must have at least "!!" can be translated as B Ÿ #!! and B   "!!. "Must manufacture at least $!!! units" can be translated as C   $!!!.

5

&!B  "!!C œ G &!Ð"&!Ñ  "!!Ð%!!!Ñ œ %!(,&!! &!Ð"#!Ñ  "!!Ð$&!!Ñ œ $&',!!! &!Ð")!Ñ  "!!Ð'!!!Ñ œ '!*,!!! &!Ð"!!Ñ  "!!Ð&!!!Ñ œ &!&,!!! &!Ð"&!Ñ  "!!Ð$!!!Ñ œ $!(,&!! &!Ð#!!Ñ  "!!Ð%!!!Ñ œ %"!,!!! &!Ð"!!Ñ  "!!Ð$!!!Ñ œ $!&,!!! (least value) &!Ð#!!Ñ  "!!Ð$!!!Ñ œ $"!,!!!

52.

The company should use "!! workers and manufacture $!!! units to achieve the least possible cost.

53.

B   ! written in interval notation is Ò!ß ∞ÑÞ

55.

B  " or B  " written in interval notation is Ð∞ß "Ñ ∪ Ð"ß ∞ÑÞ

Introduction to Relations and Functions

5 Now Try Exercises N1. (a) ÖÐ"ß &Ñß Ð$ß &Ñß Ð&ß &Ñ× The relation is a function because for each different B-value there is exactly one C-value. It is acceptable to have different B-values paired with the same C-value. (b) eÐ"ß $Ñß Ð!ß #Ñß Ð"ß 'Ñf The first and last ordered pairs have the same B-value paired with two different C-values (" is paired with both $ and '), so this relation is not a function.

803

Graphs, Linear Equations, and Functions N2. (a) eÐ#ß #Ñß Ð#ß &Ñß Ð%ß )Ñß Ð'ß &Ñf The first two ordered pairs have the same B-value paired with two different C-values (# is paired with both # and &), so this relation is not a function. (b) The domain of this relation is the set of all first components, that is, Ö"ß "!ß "&×. The range of this relation is the set of all second components, that is, Ö$"Þ$*ß $"!Þ$*ß $#!Þ)&×. This relation is a function because for each different first component, there is exactly one second component.

5 Section Exercises 1.

Answers will vary. A function is a set of ordered pairs in which each first component corresponds to exactly one second component. For example, ÖÐ!ß "Ñß Ð"ß #Ñß Ð#ß $Ñß Ð$ß %Ñß á × is a function.

3.

In an ordered pair of a relation, the first element is the independent variable.

5.

eÐ!ß #Ñß Ð#ß %Ñß Ð%ß 'Ñf We can represent this set of ordered pairs by plotting them on a graph.

N3. The arrowheads indicate that the graph extends indefinitely left and right, as well as upward. The domain includes all real numbers, written Ð∞ß ∞Ñ. Because there is a least C -value, #, the range includes all numbers greater than or equal to #, Ò#ß ∞Ñ. N4. A vertical line intersects the graph more than once, so the relation is not a function.

7.

B $ $ #

N5. (a) C œ %B  $ is a function because each value of B corresponds to exactly one value of C. Its domain is the set of all real numbers, Ð∞ß ∞Ñ.

(b) C œ È#B  % is a function because each value of B corresponds to exactly one value of C. Since the quantity under the radical must be nonnegative, the domain is the set of real numbers that satisfy the condition #B  %   ! #B   % B   #.

9.

11.

Therefore, the domain is Ò#ß ∞Ñ. " B# Given any value of B in the domain, we find C by subtracting #, and then dividing the result into ". This process produces exactly one value of C for each value in the domain, so the given equation defines a function. The domain includes all real numbers except those that make the denominator !. We find those numbers by setting the denominator equal to ! and solving for B. (c) C œ

13.

15.

B#œ! Bœ# The domain includes all real numbers except #, written as Ð∞ß #Ñ ∪ Ð#ß ∞Ñ. (d) C  $B  " is not a function because if B œ !, then C  ". Thus, the B-value ! corresponds to many C-values. Its domain is the set of all real numbers, Ð∞ß ∞Ñ. 804

We can represent this diagram in table form.

17.

C % " !

eÐ&ß "Ñß Ð$ß #Ñß Ð%ß *Ñß Ð(ß 'Ñf The relation is a function since for each B-value, there is only one C-value.

The domain is the set of B-values: e&ß $ß %ß (f. The range is the set of C-values: e"ß #ß *ß 'f.

ÖÐ#ß %Ñß Ð!ß #Ñß Ð#ß &Ñ× The relation is not a function since the B-value # has two different C-values associated with it, % and &. The domain is the set of B-values: e#ß !f. The range is the set of C-values: e%ß #ß &f.

eÐ$ß "Ñß Ð%ß "Ñß Ð#ß (Ñf The relation is a function since for each B-value, there is only one C-value.

The domain is the set of B-values: e$ß %ß #f. The range is the set of C-values: e"ß (f.

ÖÐ"ß "Ñß Ð"ß "Ñß Ð!ß !Ñß Ð#ß %Ñß Ð#ß %Ñ× The relation is not a function since the B-value " has two different C-values associated with it, " and ". (A similar statement can be made for B œ #Þ)

The domain is the set of B-values: e"ß !ß #f. The range is the set of C-values: e"ß "ß !ß %ß %f.

The relation can be described by the set of ordered pairs eÐ#ß "Ñß Ð&ß "Ñß Ð""ß (Ñß Ð"(ß #!Ñß Ð$ß #!ÑfÞ

The relation is a function since for each B-value, there is only one C-value.

Graphs, Linear Equations, and Functions

19.

The domain is the set of B-values: e#ß &ß ""ß "(ß $f. The range is the set of C-values: e"ß (ß #!fÞ

C œ 'B Each value of B corresponds to one C-value. For example, if B œ $, then C œ 'Ð$Ñ œ "). Therefore, C œ 'B defines C as a function of B. Since any B-value, positive, negative, or zero, can be multiplied by ', the domain is Ð∞ß ∞Ñ.

35.

C œ #B  ' For any value of B, there is exactly one value of C, so this equation defines a function. The domain is the set of all real numbers, Ð∞ß ∞Ñ.

37.

C œ B# Each value of B corresponds to one C-value. For example, if B œ $, then C œ $# œ *. Therefore, C œ B# defines C as a function of B. Since any B-value, positive, negative, or zero, can be squared, the domain is Ð∞ß ∞Ñ.

39.

B œ C' The ordered pairs '%ß # and '%ß # both satisfy the equation. Since one value of B, '%, corresponds to two values of C, # and #, the relation does not define a function. Because B is equal to the sixth power of C, the values of B must always be nonnegative. The domain is Ò!ß ∞Ñ.

41.

BC % For a particular B-value, more than one C-value can be selected to satisfy B  C  %. For example, if B œ # and C œ !, then

The relation can be described by the set of ordered pairs eÐ"ß &Ñß Ð"ß #Ñß Ð"ß "Ñß Ð"ß %ÑfÞ

The relation is not a function since the B-value " has four different C-values associated with it, &, #, ", and %.

21.

33.

The domain is the set of B-values: e"f. The range is the set of C-values: e&ß #ß "ß %fÞ

The relation can be described by the set of ordered pairs eÐ%ß $Ñß Ð#ß $Ñß Ð!ß $Ñß Ð#ß $ÑfÞ

The relation is a function since for each B-value, there is only one C-value.

23.

The domain is the set of B-values: e%ß #ß !ß #f. The range is the set of C-values: e$fÞ

The relation can be described by the set of ordered pairs eÐ#ß #Ñß Ð!ß $Ñß Ð$ß #ÑfÞ

The relation is a function since for each B-value, there is only one C-value.

25.

27.

29.

31.

The domain is the set of B-values: e#ß !ß $f. The range is the set of C-values: e#ß $fÞ

#  !  %. True

Using the vertical line test, we find any vertical line will intersect the graph at most once. This indicates that the graph represents a function. This graph extends indefinitely to the left Ð∞Ñ and indefinitely to the right Ð∞Ñ. Therefore, the domain is Ð∞ß ∞Ñ. This graph extends indefinitely downward Ð∞Ñ, and indefinitely upward Ð∞Ñ. Thus, the range is Ð∞ß ∞Ñ. Since a vertical line, such as B œ %, intersects the graph in two points, the relation is not a function. The domain is Ð∞ß !Ó, and the range is Ð∞ß ∞Ñ. Using the vertical line test, we find any vertical line will intersect the graph at most once. This indicates that the graph represents a function. This graph extends indefinitely to the left Ð∞Ñ and indefinitely to the right Ð∞Ñ. Therefore, the domain is Ð∞ß ∞Ñ. This graph extends indefinitely downward Ð∞Ñ, and reaches a high point at C œ %. Therefore, the range is Ð∞ß %Ó. Since a vertical line can intersect the graph of the relation in more than one point, the relation is not a function. The domain, the B-values of the points on the graph, is Ò%ß %Ó. The range, the C-values of the points on the graph, is Ò$ß $Ó.

Now, if B œ # and C œ ", then #  "  %. Also true Therefore, B  C  % does not define C as a function of B. The graph of B  C  % is equivalent to the graph of C  B  %, which consists of the shaded region below the dashed line C œ B  %, which extends indefinitely from left to right. Therefore, the domain is Ð∞ß ∞Ñ. 43.

45.

C œ ÈB For any value of B, there is exactly one corresponding value for C, so this relation defines a function. Since the radicand must be a nonnegative number, B must always be nonnegative. The domain is Ò!ß ∞Ñ.

C œ ÈB  $ is a function because each value of B in the domain corresponds to exactly one value of C. Since the quantity under the radical must be nonnegative, the domain is the set of real numbers that satisfy the condition B$ ! B   $Þ Therefore, the domain is Ò$ß ∞Ñ. 805

Graphs, Linear Equations, and Functions 47.

C œ È%B  # is a function because each value of B in the domain corresponds to exactly one value of C. Since the quantity under the radical must be nonnegative, the domain is the set of real numbers that satisfy the condition

59.

61.

%B  #   ! %B   # B    "# Þ Therefore, the domain is Ò "# ß ∞Ñ. 49.

51.

53.

55.

57.

B% Cœ & Given any value of B, C is found by adding %, then dividing the result by &. This process produces exactly one value of C for each B-value in the domain, so the relation represents a function. The denominator is never !, so the domain is Ð∞ß ∞Ñ. # Cœ B Given any value of B, C is found by dividing that value into #, and negating that result. This process produces exactly one value of C for each B-value in the domain, so the relation represents a function. The domain includes all real numbers except those that make the denominator !, namely !. The domain is Ð∞ß !Ñ ∪ Ð!ß ∞Ñ. # B% Given any value of B, C is found by subtracting %, then dividing the result into #. This process produces exactly one value of C for each B-value in the domain, so the relation represents a function. The domain includes all real numbers except those that make the denominator !, namely %. The domain is Ð∞ß %Ñ ∪ Ð%ß ∞Ñ. Cœ

BC œ " Rewrite BC œ " as C œ B" . Note that B can never equal !, otherwise the denominator would equal !. The domain is Ð∞ß !Ñ ∪ Ð!ß ∞Ñ. Each nonzero B-value gives exactly one C-value. Therefore, BC œ " defines C as a function of B. (a) Each year corresponds to exactly one percentage, so the table defines a function.

63.

6

C œ (B  "# C œ (Ð$Ñ  "# C œ * C œ $B  ) C œ $Ð$Ñ  ) Cœ"

Let x = 3.

Let x = 3.

#B  %C œ ( %C œ #B  ( C œ "# B  (%

Function Notation and Linear Functions

6 Now Try Exercises N1.

0 ÐBÑ œ %B  $ 0 Ð#Ñ œ %Ð#Ñ  $ œ )  $ œ &

Replace x with 2. Multiply. Add.

N2. 0 ÐBÑ œ #B#  %B  " (a) 0 Ð#Ñ œ #Ð#Ñ#  %Ð#Ñ  " œ )  )  " œ "( (b) 0 Ð+Ñ œ #+#  %+  " N3.

1ÐBÑ œ )B  & 1Ð+  #Ñ œ )Ð+  #Ñ  & œ )+  "'  & œ )+  #"

N4. (a) When B œ ", C œ %, so 0 Ð"Ñ œ %. (b)

0 ÐBÑ œ B#  "# 0 Ð"Ñ œ Ð"Ñ#  "# œ "  "# œ ""

N5. (a) When B œ ", C œ !, so 0 Ð"Ñ œ !. (b) 0 ÐBÑ œ # is equivalent to C œ #, and C œ # when B œ ". N6. %B#  C œ & Solving for C gives us C œ %B#  &. 0 ÐBÑ œ %B#  & Replace y with f(x). 0 Ð$Ñ œ %Ð$Ñ#  & œ $'  & œ %" 0 Ð2Ñ œ %2#  & N7. The graph of 1ÐBÑ œ "$ B  # is a line with slope and C-intercept #.

(b) The domain is e#!!%ß #!!&ß #!!'ß #!!(ß #!!)f. The range is e%#Þ$ß %#Þ)ß %$Þ(ß %$Þ)f. (c) Answers will vary. Two possible answers are Ð#!!&ß %#Þ$Ñ and Ð#!!)ß %$Þ)Ñ. The domain and range are Ð∞ß ∞ÑÞ 806

" $

Graphs, Linear Equations, and Functions 6 Section Exercises 1. 3.

5.

7.

9.

11.

13.

15.

0 Ð$Ñ is the value of the dependent variable when the independent variable is $—choice B.

27.

(b) When B œ ", C œ "!, so 0 Ð"Ñ œ "!Þ 29.

0 ÐBÑ œ $B  % 0 Ð!Ñ œ $Ð!Ñ  % œ!% œ%

31.

1ÐBÑ œ B#  %B  " 1Ð#Ñ œ Ð#Ñ#  %Ð#Ñ  " œ Ð%Ñ  )  " œ ""

33.

0 ÐBÑ œ $B  % 0 Ð "$ Ñ œ $Ð "$ Ñ  % œ "  % œ$

35.

1ÐBÑ œ B#  %B  " 1Ð!Þ&Ñ œ Ð!Þ&Ñ#  %Ð!Þ&Ñ  " œ !Þ#&  #  " œ #Þ(&

(a) When B œ #, C œ %, so 0 Ð#Ñ œ %Þ (b) When B œ ", C œ ", so 0 Ð"Ñ œ "Þ (a) The point Ð#ß $Ñ is on the graph of 0 , so 0 Ð#Ñ œ $Þ (b) The point Ð"ß $Ñ is on the graph of 0 , so 0 Ð"Ñ œ $Þ (a) The point Ð#ß $Ñ is on the graph of 0 , so 0 Ð#Ñ œ $Þ (b) The point Ð"ß #Ñ is on the graph of 0 , so 0 Ð"Ñ œ #Þ (a) 0 ÐBÑ œ $: when C œ $, B œ # (b) 0 ÐBÑ œ ": when C œ ", B œ ! (c) 0 ÐBÑ œ $: when C œ $, B œ "

37.

(a) Solve the equation for C. B  $C œ "# $C œ "#  B "#  B Cœ $

0 ÐBÑ œ $B  % 0 Ð:Ñ œ $Ð:Ñ  % œ $:  %

Since C œ 0 ÐBÑ,

0 ÐBÑ œ $B  % 0 ÐBÑ œ $ÐBÑ  % œ $B  % 0 ÐBÑ œ $B  % 0 ÐB  #Ñ œ $ÐB  #Ñ  % œ $B  '  % œ $B  #

(a) When B œ #, C œ "&, so 0 Ð#Ñ œ "&Þ

0 ÐBÑ œ (b) 0 Ð$Ñ œ 39.

"#  B " œ  B  %. $ $

"#  $ * œ œ$ $ $

(a) Solve the equation for C. C  #B# œ $ C œ $  #B#

1ÐBÑ œ B#  %B  " 1Ð1Ñ œ 1#  %1  "

Since C œ 0 ÐBÑ,

19.

0 ÐBÑ œ $B  % 0 ÐB  2Ñ œ $ÐB  2Ñ  % œ $B  $2  %

(b) 0 Ð$Ñ œ $  #Ð$Ñ# œ $  #Ð*Ñ œ "&

21.

0 ÐBÑ œ $B  %, 1ÐBÑ œ B#  %B  "

17.

0 ÐBÑ œ $  #B# .

41.

(a) Solve the equation for C.

0 Ð%Ñ  1Ð%Ñ œ Ò$Ð%Ñ  %Ó  ÒÐ%Ñ#  %Ð%Ñ  "Ó œ Ò)Ó  Ò"Ó œ * 23.

0 œ ÖÐ#ß #Ñß Ð"ß "Ñß Ð#ß "Ñ× (a) When B œ #, C œ ", so 0 Ð#Ñ œ "Þ (b) When B œ ", C œ ", so 0 Ð"Ñ œ "Þ

25.

(a) When B œ #, C œ #, so 0 Ð#Ñ œ #Þ (b) When B œ ", C œ $, so 0 Ð"Ñ œ $Þ

%B  $C œ ) $C œ )  %B )  %B Cœ $ Since C œ 0 ÐBÑ, 0 ÐBÑ œ

)  %B % ) œ B . $ $ $

)  %Ð$Ñ )  "# œ $ $ % % œ œ $ $

(b) 0 Ð$Ñ œ

807

Graphs, Linear Equations, and Functions 43.

The equation #B  C œ % has a straight line as its graph. To find C in Ð$ß CÑ, let B œ $ in the equation.

51.

1ÐBÑ œ  % Using a C-intercept of Ð!ß %Ñ and a slope of 7 œ !, we graph the horizontal line. From the graph we see that the domain is Ð∞ß ∞Ñ. The range is e%fÞ

53.

0 ÐBÑ œ ! Draw the horizontal line through the point Ð!ß !Ñ. On the horizontal line the value of B can be any real number, so the domain is Ð∞ß ∞Ñ. The range is e!fÞ

55.

0 ÐBÑ œ !, or C œ !, is the B-axis.

57.

(a) 0 ÐBÑ œ $Þ(&B 0 Ð$Ñ œ $Þ(&Ð$Ñ œ ""Þ#& (dollars)

#B  C œ % #Ð$Ñ  C œ % 'C œ% C œ # So one point that lies on the graph is Ð$ß # Ñ. To use functional notation for #B  C œ %, solve for C to get C œ #B  %. Replace C with 0 ÐBÑ to get 0 ÐBÑ œ #B  % . 0 Ð$Ñ œ #Ð$Ñ  % œ # Because C œ # when B œ $, the point Ð $ ß # Ñ lies on the graph of the function. 45.

47.

0 ÐBÑ œ #B  & The graph will be a line. The intercepts are !ß & and Ð &# ß !Ñ. The domain is Ð∞ß ∞Ñ. The range is Ð∞ß ∞Ñ.

(b) $ is the value of the independent variable, which represents a package weight of $ pounds; 0 Ð$Ñ is the value of the dependent variable representing the cost to mail a 3-pound package.

2ÐBÑ œ "# B  # The graph will be a line. The intercepts are !ß # and %ß ! . The domain is Ð∞ß ∞Ñ. The range is Ð∞ß ∞Ñ.

(c) $Þ(&Ð&Ñ œ $").(&, the cost to mail a 5-lb package. Using function notation, we have 0 Ð&Ñ œ ")Þ(&Þ 59.

(a) Since the length of a man's femur is given, use the formula 2Ð œ !. Here, 0 Ð!Ñ œ !, which means the pool is empty at time !. (e) 0 Ð#&Ñ œ $!!!; After #& hours, there are $!!! gallons of water in the pool.

809

Graphs, Linear Equations, and Functions 2.

4.

BC œ) For B œ #:

&B  (C œ #) To find the B-intercept, let C œ !. &B  (C œ #) &B  (Ð!Ñ œ #) &B œ #) B œ #) &

#C œ) C œ ' C œ ' Ð#ß 'Ñ For C œ $:

The B-intercept is Ð #) & ß !Ñ. To find the C-intercept, let B œ !.

B  Ð$Ñ œ ) B$œ) B œ & Ð&ß $Ñ

&B  (C œ #) &Ð!Ñ  (C œ #) (C œ #) Cœ%

For B œ $: $C œ) C œ & C œ & Ð$ß &Ñ

The C-intercept is Ð!ß %Ñ. Plot the intercepts and draw the line through them.

For C œ #: B  Ð#Ñ œ ) B#œ) B œ ' Ð'ß #Ñ Plot the ordered pairs, and draw the line through them. 5.

3.

#B  &C œ #! To find the B-intercept, let C œ !. #B  &C œ #! #B  &Ð!Ñ œ #! #B œ #! B œ "!

%B  $C œ "# To find the B-intercept, let C œ !.

The B-intercept is "!ß ! . To find the C-intercept, let B œ !.

%B  $C œ "# %B  $Ð!Ñ œ "# %B œ "# Bœ$

#B  &C œ #! #Ð!Ñ  &C œ #! &C œ #! Cœ%

The B-intercept is Ð$ß !Ñ. To find the C-intercept, let B œ !.

The C-intercept is Ð!ß %Ñ. Plot the intercepts and draw the line through them.

%B  $C œ "# %Ð!Ñ  $C œ "# $C œ "# C œ % The C-intercept is Ð!ß %Ñ. Plot the intercepts and draw the line through them.

6.

B  %C œ ) To find the B-intercept, let C œ !. B  %C œ ) B  %Ð!Ñ œ ) Bœ)

810

Graphs, Linear Equations, and Functions The B-intercept is )ß ! . To find the C-intercept, let B œ !.

15.

!  %C œ ) %C œ ) C œ #

Perpendicular to $B  C œ % Solve for C. C œ $B  % The slope is $; the slope of a line perpendicular to it is  "$ since

The C-intercept is Ð!ß #Ñ. Plot the intercepts and draw the line through them.

$Ð "$ Ñ œ ". 16.

Through "ß & and "ß % 7œ

?C %  & * Undefined œ œ ?B "  Ð"Ñ !

This is a vertical line; it has undefined slope. 17. 7.

By the midpoint formula, the midpoint of the segment with endpoints Ð)ß "#Ñ and Ð)ß "'Ñ is )  ) "#  "' ! % ß Œ  œ Œ ß  œ Ð!ß #ÑÞ # # # #

8.

Through "ß # and %ß & 7œ

10.

7œ 18.

change in C &  # ( ( œ œ œ change in B %  Ð"Ñ & &

Through !ß $ and #ß % C#  C " %$ " " œ œ œ B#  B " #  ! # #

11.

The slope of C œ #B  $ is #, the coefficient of B.

12.

$B  %C œ & Write the equation in slope-intercept form.

The line goes up from left to right, so it has positive slope.

20.

The line goes down from left to right, so it has negative slope.

21.

The line is vertical, so it has undefined slope.

22.

The line is horizontal, so it has ! slope.

23.

To rise " foot, we must move % feet in the horizontal direction. To rise $ feet, we must move $Ð%Ñ œ "# feet in the horizontal direction.

24.

Let B" ß C" œ Ð"*)!ß #",!!!Ñ and B# ß C# œ Ð#!!(ß '",%!!Ñ. Then 7œ

%C œ $B  & C œ $% B  &% The slope is $% . B œ & is a vertical line and has undefined slope.

14.

Parallel to $C œ #B  & Write the equation in slope-intercept form. $C œ #B  & C œ #$ B  &$ The slope of $C œ #B  & is #$ ; all lines parallel to it will also have a slope of #$ .

?C '",%!!  #",!!! %!,%!! œ œ ?B #!!(  "*)! #( ¸ "%*'.

The average rate of change is $"%*' per year (to the nearest dollar). 25.

13.

change in C #! # œ œ œ "Þ change in B !# #

19.

Let B" ß C" œ Ð!ß $Ñ and B# ß C# œ Ð#ß %Ñ. 7œ

?C "  Ð"Ñ # " œ œ œ . ?B $  $ ' $

The B-intercept is Ð#ß !Ñ and the C-intercept is Ð!ß #ÑÞ The slope is 7œ

By the midpoint formula, the midpoint of the segment with endpoints Ð!ß &Ñ and Ð*ß )Ñ is !  Ð*Ñ &  ) * $ * $ ß Œ  œ Œ ß  œ Œ ß Þ # # # # # #

9.

Through $ß " and $ß "

(a) Slope  "$ , C-intercept Ð!ß "Ñ Use the slope-intercept form with 7 œ  "$ and , œ "Þ C œ 7B  , C œ  "$ B  " (b)

C œ  "$ B  " $C œ B  $ B  $C œ $

811

Graphs, Linear Equations, and Functions 26.

(a) Slope !, C-intercept Ð!ß #Ñ Use the slope-intercept form with 7 œ ! and , œ #Þ

31.

7œ

C œ 7B  , C œ Ð!ÑB  # C œ #

C  ' œ (& B  C œ (& B 

C  C" œ 7ÐB  B" Ñ C  ( œ  %$ ÐB  #Ñ C œ  %$ B  (b)

28.

(b)

(a) Vertical, through Ð#ß &Ñ The equation of any vertical line is in the form B œ 5 . Since the line goes through Ð#ß &Ñ, the equation is B œ #. (Slope-intercept form is not possible.) (a) Through #ß & and "ß % Find the slope. 7œ

?C %  Ð&Ñ * œ œ œ * ?B "# "

Use the point-slope form with 7 œ * and B" ß C" œ Ð#ß &Ñ. C  C" œ 7ÐB  B" Ñ C  Ð&Ñ œ *ÐB  #Ñ C  & œ *B  ") C œ *B  "$ (b)

812

C œ *B  "$ *B  C œ "$

and

"% & "' &

C œ (& B  "' & &C œ (B  "' (B  &C œ "' (B  &C œ "'

(a) From Exercise 18, we have 7 œ " and a C-intercept of Ð!ß #Ñ. The slope-intercept form is C œ "B  # or C œ B  #Þ (b)

33.

C œ B  # BC œ#

(a) Parallel to %B  C œ $ and through Ð(ß "Ñ Writing %B  C œ $ in slope-intercept form gives us C œ %B  $, which has slope %. Lines parallel to it will also have slope %. The line with slope % through Ð(ß "Ñ is : C  C" œ 7ÐB  B" Ñ C  Ð"Ñ œ %ÐB  (Ñ C  " œ %B  #) C œ %B  #*

C œ $B  ( $B  C œ ( $B  C œ (

(b) B œ # is already in standard form. 30.

32.

(a) Slope $, through Ð"ß %Ñ Use the point-slope form with 7 œ $ and B" ß C" œ Ð"ß %Ñ. C  C" œ 7ÐB  B" Ñ C  % œ $cB  Ð"Ñd C  % œ $ÐB  "Ñ C  % œ $B  $ C œ $B  (

29.

(b)

) $ #* $

C œ  %$ B  #* $ $C œ %B  #* %B  $C œ #*

( &

C  C" œ 7ÐB  B" Ñ C  ' œ (& ÐB  #Ñ

(a) Slope  %$ , through Ð#ß (Ñ Use the point-slope form with 7 œ  %$ and B" ß C" œ Ð#ß (Ñ.

C  ( œ  %$ B 

?C '  Ð"Ñ ( œ œ ?B #  Ð$Ñ &

Use the point-slope form with 7 œ B" ß C" œ Ð#ß 'Ñ.

(b) C œ # is already in standard form. 27.

(a) Through $ß " and #ß ' Find the slope.

(b)

34.

C œ %B  #* %B  C œ #* %B  C œ #*

(a) Perpendicular to #B  &C œ ( and through Ð%ß $Ñ Write the equation in slope-intercept form. #B  &C œ ( &C œ #B  ( C œ #& B  (& C œ #& B  (& has slope #& and is perpendicular to lines with slope  &# . The line with slope  &# through Ð%ß $Ñ is C  C" œ 7ÐB  B" Ñ C  $ œ  &# ÐB  %Ñ C  $ œ  &# B  "! C œ  &# B  "$

Graphs, Linear Equations, and Functions (b)

35.

C œ  &# B  "$ #C œ &B  #' &B  #C œ #'

38.

&B  C  ' Graph &B  C œ ' as a dashed line through !ß ' and Ð '& ß !Ñ. Use Ð!ß !Ñ as a test point. Since Ð!ß !Ñ does not satisfy the inequality, shade the region on the side of the line that does not contain Ð!ß !Ñ.

39.

#B  C Ÿ " and B   #C Graph #B  C œ " as a solid line through Ð "# ß !Ñ and Ð!ß "Ñ, and shade the region on the side containing Ð!ß !Ñ since it satisfies the inequality. Next, graph B œ #C as a solid line through Ð!ß !Ñ and Ð#ß "Ñ, and shade the region on the side containing Ð#ß !Ñ since #  #Ð!Ñ or #  ! is true. The intersection is the region where the graphs overlap.

40.

B   # or C   #

(a) The fixed cost is $"&*, so that is the value of , . The variable cost is $&(, so C œ 7B  , œ &(B  "&*Þ The cost of a 1-year membership can be found by substituting "# for B. C œ &(Ð"#Ñ  "&* œ ')%  "&* œ )%$ The cost is $)%$Þ (b) As in part (a), C œ %(B  "&*Þ C œ %(Ð"#Ñ  "&* œ &'%  "&* œ (#$ The cost is $(#$Þ

36.

(a) Use Ð$ß ")$*Ñ and Ð(ß #%"%Ñ. 7œ

?C #%"%  ")$* &(& œ œ ¸ "%$Þ(& ?B ($ %

Use the point-slope form of a line. C  C" œ 7ÐB  B" Ñ C  ")$* œ "%$Þ(&ÐB  $Ñ C  ")$* œ "%$Þ(&B  %$"Þ#& C œ "%$Þ(&B  "%!(Þ(&

Graph B œ # as a solid vertical line through Ð#ß !Ñ. Shade the region to the right of B œ #.

The slope, "%$Þ(&, indicates that the revenue from skiing facilities increased by an average of $"%$Þ(& million each year from 2003 to 2007.

Graph C œ # as a solid horizontal line through Ð!ß #Ñ. Shade the region above C œ #. The graph of

(b) The year 2008 corresponds to B œ )Þ

B   # or C   #

C œ "%$Þ(&Ð)Ñ  "%!(Þ(& œ ""&!  "%!(Þ(& œ #&&(Þ(&

includes all the shaded regions.

According to the equation from part (a), we estimate the revenue from skiing facilities to be $#&&) million (to the nearest million). 37.

$B  #C Ÿ "# Graph $B  #C œ "# as a solid line through !ß ' and %ß ! . Use Ð!ß !Ñ as a test point. Since Ð!ß !Ñ satisfies the inequality, shade the region on the side of the line containing Ð!ß !Ñ.

41.

42.

In C  %B  $, the "  " symbol indicates that the graph has a dashed boundary line and that the shading is below the line, so the correct choice is D.

eÐ%ß #Ñß Ð%ß #Ñß Ð"ß &Ñß Ð"ß &Ñf The domain, the set of B-values, is e%ß "f. The range, the set of C-values, is e#ß #ß &ß &f. Since each B-value has more than one C-value, the relation is not a function. 813

Graphs, Linear Equations, and Functions 43.

The relation can be described by the set of ordered pairs

50.

eÐ*ß $#Ñß Ð""ß %(Ñß Ð%ß %(Ñß Ð"(ß '*Ñß Ð#&ß "%Ñf. The relation is a function since for each B-value, there is only one C-value. The domain is the set of B-values: e*ß ""ß %ß "(ß #&f. The range is the set of C-values: e$#ß %(ß '*ß "%fÞ 44.

The domain, the B-values of the points on the graph, is c%ß %d. The range, the C-values of the points on the graph, is c!ß #d. Since a vertical line intersects the graph of the relation in at most one point, the relation is a function.

45.

The B-values are negative or zero, so the domain is Ð∞ß !Ó. The C-values can be any real number, so the range is Ð∞ß ∞Ñ. A vertical line, such as B œ $, will intersect the graph twice, so by the vertical line test, the relation is not a function.

51.

B œ C# The ordered pairs %ß # and %ß # both satisfy the equation. Since one value of B, %, corresponds to two values of C, # and #, the equation does not define a function. Because B is equal to the square of C, the values of B must always be nonnegative. The domain is Ò!ß ∞Ñ. ( B' Given any value of B, C is found by subtracting ', then dividing the result into (. This process produces exactly one value of C for each B-value in the domain, so the equation defines a function. The domain includes all real numbers except those that make the denominator !, namely '. The domain is Ð∞ß 'Ñ ∪ Ð'ß ∞Ñ. Cœ

In Exercises 52–55, use

46.

47.

48.

49.

C œ $B  $ For any value of B, there is exactly one value of C, so the equation defines a function, actually a linear function. The domain is the set of all real numbers, Ð∞ß ∞Ñ. C B# For any value of B, there are many values of C. For example, Ð"ß !Ñ and Ð"ß "Ñ are both solutions of the inequality that have the same B-value but different C-values. The inequality does not define a function. The domain is the set of all real numbers, Ð∞ß ∞Ñ. C œ kBk For any value of B, there is exactly one value of C, so the equation defines a function. The domain is the set of all real numbers, Ð∞ß ∞Ñ.

C œ È%B  ( Given any value of B, C is found by multiplying B by %, adding (, and taking the square root of the result. This process produces exactly one value of C for each B-value in the domain, so the equation defines a function. Since the radicand must be nonnegative, %B  (   ! %B   ( B    (% . The domain is Ò (% ß ∞Ñ.

814

0 ÐBÑ œ #B#  $B  '. 52.

0 Ð!Ñ œ #Ð!Ñ#  $Ð!Ñ  ' œ '

53.

0 Ð#Þ"Ñ œ #Ð#Þ"Ñ#  $Ð#Þ"Ñ  ' œ )Þ)#  'Þ$  ' œ )Þ&#

54.

0 Ð "# Ñ œ #Ð "# Ñ#  $Ð "# Ñ  ' œ  "# 

$ #

 ' œ )

55.

0 Ð5Ñ œ #5 #  $5  '

56.

#B#  C œ ! C œ #B# C œ #B# Since C œ 0 ÐBÑ, 0 ÐBÑ œ #B# , and 0 Ð$Ñ œ #Ð$Ñ# œ #Ð*Ñ œ ").

57.

Solve for C in terms of B. #B  &C œ ( #B  ( œ &C # ( &B  & œ C Thus, choice C is correct.

58.

(a) For each year, there is exactly one life expectancy associated with the year, so the table defines a function. (b) The domain is the set of years, that is, Ö"*'!ß "*(!ß "*)!ß "**!ß #!!!ß #!!*×. The range is the set of life expectancies, that is, Ö'*Þ(ß (!Þ)ß ($Þ(ß (&Þ%ß ('Þ)ß ()Þ"×Þ (c) Answers will vary. Two possible answers are Ð"*'!ß '*Þ(Ñ and Ð#!!*ß ()Þ"ÑÞ (d) 0 Ð"*)!Ñ œ ($Þ(. In 1980, life expectancy at birth was ($Þ( yr. (e) Since 0 Ð#!!!Ñ œ ('Þ), B œ #!!!Þ

Graphs, Linear Equations, and Functions 59.

Because it falls from left to right, the slope is negative.

60.

Use the points "ß & and $ß " . 7œ

61.

62.

?C "  & ' $ œ œ œ ?B $  Ð"Ñ % #

Since 7 œ  $# , the slope of any line parallel to this line is also  $# , whereas the slope of any line perpendicular to this line is #$ since #$ is the negative reciprocal of  $# . #C œ $B  ( To find the B-intercept, let C œ !.

68.

0 ÐBÑ œ ! is equivalent to C œ !, which is the equation we solved in Exercise 62 to find the B-intercept. The solution set is ˜ ($ ™.

69.

The graph is below the B-axis for B  ($ , so the solution set of 0 ÐBÑ  ! is Ð ($ ß ∞Ñ.

70.

The graph is above the B-axis for B  ($ , so the solution set of 0 ÐBÑ  ! is Ð∞ß ($ Ñ.

Test 1.

#Ð!Ñ œ $B  ( $B œ ( B œ ($

#Ð"Ñ  $C œ "# #  $C œ "# $C œ "! ˆ"ß  "! ‰ C œ  "! $ $

The B-intercept is Ð ($ ß !Ñ. 63.

For B œ $:

#C œ $B  ( To find the C-intercept, let B œ !.

#Ð$Ñ  $C œ "# '  $C œ "# $C œ ' C œ #

#C œ $Ð!Ñ  ( #C œ ( C œ (#

#B  $Ð%Ñ œ "# #B  "# œ "# #B œ ! B œ ! !ß %

Solve #C œ $B  ( for C. C œ  $# B  (# Since C œ 0 ÐBÑ, 0 ÐBÑ œ  $# B  (# .

65.

0 ÐBÑ œ  $# B 

( #

0 Ð)Ñ œ  $# Ð)Ñ  66.

œ  #% # 

( #

0 ÐBÑ œ  $# B 

( #

œ  "( # ( #

) œ  "' œ $B  ( #$ œ $B B œ #$ $ 67.

0 ÐBÑ   !  (#   !

 $# B

2.

$B  #C œ #! To find the B-intercept, let C œ !. $B  #Ð!Ñ œ #! $B œ #! B œ #! $

( #

 $# B

$ß #

For C œ %:

The C-intercept is Ð!ß (# Ñ. 64.

#B  $C œ "# For B œ ":

Let f(x) œ 8. Multiply by 2. Subtract 7. Divide by 3.

‰ The B-intercept is ˆ #! $ ß! . To find the C-intercept, let B œ !. $Ð!Ñ  #C œ #! #C œ #! C œ "! The C-intercept is Ð!ß "!Ñ. Draw the line through these two points.

 $# B    (# B Ÿ Ð (# ÑÐ #$ Ñ BŸ

( $

815

Graphs, Linear Equations, and Functions 3.

The graph of C œ & is the horizontal line with slope ! and C-intercept Ð!ß &Ñ. There is no B-intercept.

9.

Use the points Ð"*)!ß ""*,!!!Ñ and Ð#!!)ß *$,!!!Ñ. average rate of change œ

change in C *$,!!!  ""*,!!! œ change in B #!!)  "*)! #',!!! œ ¸ *#* #)

The average rate of change is about *#* farms per year, that is, the number of farms decreased by about *#* each year from 1980 to 2008. 4.

The graph of B œ # is the vertical line with B-intercept at Ð#ß !Ñ. There is no C-intercept.

10.

Through Ð%ß "Ñ; 7 œ & (a) Let 7 œ & and B" ß C" œ Ð%ß "Ñ in the point-slope form. C  C" œ 7ÐB  B" Ñ C  Ð"Ñ œ &ÐB  %Ñ C  " œ &B  #! C œ &B  "* (b)

5.

Through 'ß % and %ß " 7œ

11.

6.

The graph of a line with undefined slope is the graph of a vertical line.

7.

Find the slope of each line.

Through Ð$ß "%Ñ; horizontal

(b) C œ "% is already in standard form. 12.

Through #ß $ and 'ß " (a) First find the slope. 7œ

&B  C œ ) C œ &B  ) C œ &B  )

C  C" œ 7ÐB  B" Ñ C  $ œ  "# cB  Ð#Ñd

&C œ B  $ C œ  "& B  $&

C  $ œ  "# ÐB  #Ñ C  $ œ  "# B  " C œ  "# B  #

The slope is  "& . Since &Ð "& Ñ œ ", the two slopes are negative reciprocals and the lines are perpendicular.

The slope is $# . $C œ #B  & C œ #$ B  &$ The slope is #$ . The lines are neither parallel nor perpendicular. 816

C œ  "# B  #

(b) " #B

Find the slope of each line. #C œ $B  "# C œ $# B  '

?C "  $ % " œ œ œ ?B '  Ð#Ñ ) #

Use 7 œ  "# and B" ß C" œ Ð#ß $Ñ in the pointslope form.

The slope is &.

8.

From part (a) Standard form

(a) A horizontal line has equation C œ 5 . Here 5 œ "%, so the line has equation C œ "%.

?C "  % & " œ œ œ ?B %  ' "! #

The slope of the line is "# .

C œ &B  "* &B  C œ "*

C œ#

#Ð "# B  CÑ œ # # B  #C œ % 13.

From part (a) Variable terms on one side Multiply by 2. Standard form

Through Ð&ß 'Ñ; vertical (a) The equation of any vertical line is in the form B œ 5 . Since the line goes through Ð&ß 'Ñ, the equation is B œ &. Writing B œ & in slopeintercept form is not possible since there is no C-term. (b) From part (a), the standard form is B œ &.

Graphs, Linear Equations, and Functions 14.

(b) C œ #!()Ð&Ñ  &",&&( Let x œ 5. œ $'",*%(, which is more than the actual value.

Through Ð(ß #Ñ and parallel to $B  &C œ ' (a) To find the slope of $B  &C œ ', write the equation in slope-intercept form by solving for C. $B  &C œ ' &C œ $B  ' C œ  $& B  '&

18.

$B  #C  ' Graph the line $B  #C œ ', which has intercepts #ß ! and !ß $ , as a dashed line since the inequality involves  . Test Ð!ß !Ñ, which yields !  ', a false statement. Shade the region that does not include Ð!ß !Ñ.

19.

C  #B  " and B  C  $ First graph C œ #B  " as a dashed line through #ß $ and !ß " . Test Ð!ß !Ñ, which yields !  ", a false statement. Shade the side of the line not containing Ð!ß !Ñ. Next, graph B  C œ $ as a dashed line through $ß ! and !ß $ . Test Ð!ß !Ñ, which yields !  $, a true statement. Shade the side of the line containing Ð!ß !Ñ. The intersection is the region where the graphs overlap.

20.

Choice D is the only graph that passes the vertical line test.

21.

Choice D does not define a function, since its domain (input) element ! is paired with two different range (output) elements, " and #.

22.

The B-values are greater than or equal to zero, so the domain is Ò!ß ∞ÑÞ Since C can be any value, the range is Ð∞ß ∞Ñ.

23.

The domain is the set of B-values: Ö!ß #ß %×. The range is the set of C-values: Ö"ß $ß )×.

24.

0 ÐBÑ œ B#  #B  "

The slope is  $& , so a line parallel to it also has slope  $& . Let 7 œ  $& and B" ß C" œ Ð(ß #Ñ in the point-slope form. C  C" œ 7ÐB  B" Ñ C  # œ  $& cB  Ð(Ñd C  # œ  $& B  ( C  # œ  $& B  C œ  $& B  C œ  $& B 

(b) $ &B

#" & "" &

"" &

From part (a) Variable terms on one side Multiply by 5. Standard form

 C œ  "" &

&Ð $& B  CÑ œ &Ð "" & Ñ $B  &C œ "" 15.

Through Ð(ß #Ñ and perpendicular to C œ #B (a) Since C œ #B is in slope-intercept form Ð, œ !Ñ, the slope, 7, of C œ #B is #. A line perpendicular to it has a slope that is the negative reciprocal of #, that is,  "# . Let 7 œ  "# and B" ß C" œ Ð(ß #Ñ in the point-slope form. C  C" œ 7ÐB  B" Ñ C  # œ  "# ÐB  (Ñ C  # œ  "# B  C œ  "# B  C œ  "# B 

(b) " #B

 C œ  $#

#Ð "# B  CÑ œ #Ð $# Ñ B  #C œ $ 16.

17.

$ #

( # $ #

From part (a) Variable terms on one side Multiply by 2. Standard form

Positive slope means that the line goes up from left to right. The only line that has positive slope and a negative C-coordinate for its C-intercept is choice B. (a) Use the points Ð"ß &$,'$&Ñ and Ð(ß '',"!$Ñ. 7œ

?C '',"!$  &$,'$& "#,%') œ œ ?B (" ' œ #!()

Use the point-slope form with 7 œ #!() and B" ß C" œ Ð"ß &$,'$&Ñ. C  &$,'$& œ #!()ÐB  "Ñ C  &$,'$& œ #!()B  #!() C œ #!()B  &",&&(

(a) 0 Ð"Ñ œ Ð"Ñ#  #Ð"Ñ  " œ "  #  " œ! (b) 0 Ð+Ñ œ +#  #+  " 817

Graphs, Linear Equations, and Functions 25.

818

0 ÐBÑ œ #$ B  " This function represents a line with C-intercept Ð!ß "Ñ and B-intercept Ð $# ß !Ñ. Draw the line through these two points. The domain is Ð∞ß ∞Ñ, and the range is Ð∞ß ∞Ñ.

Check Ð""ß %Ñ: "" œ $  )à %%  "# œ $#

SYSTEMS OF LINEAR EQUATIONS 1

N4.

Systems of Linear Equations in Two Variables

The solution set is eÐ""ß %Ñf. &B  C œ ( $B  #C œ #&

Step 1 To use the substitution method, first solve one of the equations for B or C. Since the coefficient of C in equation Ð"Ñ is ", it is easiest to solve for C in this equation.

1 Now Try Exercises N1. To determine whether Ð#ß &Ñ is a solution of the system B  #C œ ) $B  #C œ !,

&B  C œ ( C œ &B  (

replace B with # and C with & in each equation. B  #C œ )

Ð"Ñ

$B  #C œ !

? #  #Ð&Ñ œ )

? $Ð#Ñ  #Ð&Ñ œ !

? #  "! œ ) )œ)

? '  "! œ ! % œ !

True

$B  #C œ #& $B  #Ð&B  (Ñ œ #&

False

$B  "!B  "% œ #& "$B  "% œ #& "$B œ $* Bœ$

When the equations are graphed, the point of intersection appears to be Ð%ß $Ñ. To check, substitute % for B and $ for C in each equation of the system. ? $Ð%Ñ  #Ð$Ñ œ ' 'œ'

Ð"Ñ

BC œ"

Ð#Ñ

True

? %$œ " "œ"

True

Since Ð%ß $Ñ makes both equations true, the solution set is ÖÐ%ß $Ñ×.

Ð#Ñ

Step 3

Ð"Ñ Ð#Ñ

$B  #C œ '

Ð"Ñ

Step 2 Substitute &B  ( for C in equation Ð#Ñ and solve for B.

Ð#Ñ

The ordered pair Ð#ß &Ñ is not a solution of the system, since it does not make both equations true. N2. $B  #C œ ' BC œ"

Ð"Ñ Ð#Ñ

Step 4 Since C œ &B  ( and B œ $, C œ &Ð$Ñ  ( œ "&  ( œ ). Step 5 Check Ð$ß )Ñ: "&  ) œ (à *  "' œ #& N5.

The solution set is eÐ$ß )Ñf. " "! B

 &$ C œ #& #B  $C œ "

Ð"Ñ Ð#Ñ

Multiply equation Ð"Ñ by the LCD, "!. The new system is B  'C œ % #B  $C œ ".

Ð$Ñ Ð#Ñ

Solve equation Ð$Ñ for B. N3.

B œ $  #C %B  $C œ $#

Ð"Ñ Ð#Ñ

Since equation Ð"Ñ is solved for B, substitute $  #C for B in equation Ð#Ñ. Ð#Ñ %B  $C œ $# %Ð$  #CÑ  $C œ $# "#  )C  $C œ $# "#  &C œ $# &C œ #! Cœ% Since B œ $  #C and C œ %, B œ $  #Ð%Ñ œ "".

B œ 'C  % Substitute 'C  % for B in equation Ð#Ñ and solve for C. #B  $C œ " #Ð'C  %Ñ  $C œ " "#C  )  $C œ " *C  ) œ " *C œ * C œ "

Ð#Ñ

Since B œ 'C  % and C œ ", B œ 'Ð"Ñ  % œ '  % œ #.

From Chapter 4 of Student’s Solutions Manual for Intermediate Algebra, Eleventh Edition. Margaret L. Lial, John Hornsby, Terry McGinnis. Copyright © 2012 by Pearson Education, Inc. Publishing as Addison-Wesley. All rights reserved.

819

Systems of Linear Equations Check Ð#ß "Ñ:  "& 

$ &

œ #& à %  $ œ "

The solution set is eÐ#ß "Ñf.

N6. )B  #C œ & &B  #C œ ")

N9. #B  &C œ ' 'B  "&C œ %

Multiply equation Ð"Ñ by $ and add the result to equation Ð#Ñ.

Ð"Ñ Ð#Ñ

'B  "&C œ ") $ ‚ Ð"Ñ 'B  "&C œ % Ð#Ñ False ! œ ##

Eliminate C by adding equations Ð"Ñ and Ð#Ñ. )B  #C œ & Ð"Ñ &B  #C œ ") Ð#Ñ Add. "$B œ "$ Solve for x. B œ "

The system is inconsistent. The solution set is g.

To find C, replace B with " in either equation Ð"Ñ or Ð#Ñ. )B  #C œ & )Ð"Ñ  #C œ & )  #C œ & #C œ "$ C œ  "$ #

Ð"Ñ

The graphs of the equations are parallel lines. N10. (a) Solve each equation for C. #B  $C œ $ $C œ #B  $ C œ #$ B  "

™ The solution set is ˜Ð"ß  "$ # Ñ .

(b) Solve each equation for C. &C œ B  % C œ  "& B  %&

Ð"Ñ Ð#Ñ

To eliminate C, multiply equation Ð"Ñ by # and add the result to equation Ð#Ñ multiplied by &. %B  "!C œ ) #&B  "!C œ &! #*B œ &) B œ #

# ‚ Ð"Ñ & ‚ Ð#Ñ Add. Solve for x.

To find C, substitute # for B in equation Ð"Ñ or Ð#Ñ. &B  #C œ "! &Ð#Ñ  #C œ "! "!  #C œ "! #C œ ! Cœ!

Ð#Ñ

N8.

B  $C œ ( $B  *C œ #"

Ð"Ñ Ð#Ñ

Multiply equation Ð"Ñ by $, and add the result to equation Ð#Ñ. $B  *C œ #" $ ‚ Ð"Ñ $B  *C œ #" Ð#Ñ True ! œ !

% &

1 Section Exercises 1.

If Ð%ß $Ñ is a solution of a linear system in two variables, then substituting % for B and $ for C leads to true statements in both equations.

3.

If the solution process leads to a false statement such as ! œ $, the solution set is g .

5.

If the two lines forming a system have the same slope and different C-intercepts, the system has ! solutions. (The lines are parallel.)

7.

D; The ordered pair solution must be in quadrant IV, since that is where the graphs of the equations intersect.

9.

(a) B  C œ ! implies that B œ C , so the coordinates of the solution must be equal—this limits our choice to B or C. Since B  C œ ', graph B is correct. (b) As in part (a), we must have B or C. Since B  C œ ', graph C is correct.

The equations are dependent.

(c) B  C œ ! implies that B œ C , so the coordinates of the solution must be opposites— this limits our choice to A or D. Since B  C œ ', graph A is correct.

Equations Ð"Ñ and Ð#Ñ have the same graph.

(d) As in part (c), we must have A or D. Since B  C œ ', graph D is correct.

The solution set is eÐBß CÑ l B  $C œ (f.

820

"!C œ #B  ) C œ  "& B 

The system has infinitely many solutions.

Check Ð#ß !Ñ: %  ! œ %à "!  ! œ "! The solution set is eÐ#ß !Ñf.

%B  'C œ ' 'C œ %B  ' C œ #$ B  "

The system has no solution.

Check Ð"ß  "$ # Ñ: )  "$ œ &à &  "$ œ ") N7. #B  &C œ % &B  #C œ "!

Ð"Ñ Ð#Ñ

Systems of Linear Equations 11.

B  C œ "( B  C œ " To decide if Ð)ß *Ñ is a solution, replace B with ) and C with * in each equation of the system. B  C œ "( ?

)  Ð*Ñ œ "( "( œ "( True

B  C œ " ? )  Ð*Ñ œ " " œ " True

Since Ð)ß *Ñ makes both equations true, Ð)ß *Ñ is a solution of the system. 13.

$B  &C œ "# B C œ "

19.

Replace B with " and C with #.

%B  C œ ' C œ #B

Ð"Ñ Ð#Ñ

Equation Ð#Ñ is already solved for C, so substitute #B for C in equation Ð"Ñ.

$Ð"Ñ  &Ð#Ñ œ "# ? $  "! œ "# "$ œ "# False

15.

To check, substitute # for B and # for C in each equation of the system. Since Ð#ß #Ñ makes both equations true, the solution set of the system is eÐ#ß #Ñf.

%B  C œ ' %B  #B œ ' 'B œ ' Bœ"

Ð"Ñ Let y œ 2x.

It is not necessary to check if Ð"ß #Ñ satisfies the second equation since it does not satisfy the first equation. Therefore, Ð"ß #Ñ is not a solution of the system.

Substitute " for B in Ð#Ñ.

B  C œ & #B  C œ "

The solution "ß # checks. The solution set is e "ß # f.

Graph the line B  C œ & through its intercepts, &ß ! and !ß & , and the line #B  C œ " through its intercepts, Ð "# ß !Ñ and !ß " . The lines appear to intersect at Ð#ß $Ñ.

C œ #Ð"Ñ œ #

21.

B  %C œ "% C œ #B  "

Ð"Ñ Ð#Ñ

Substitute #B  " for C in equation Ð"Ñ and solve for B. B  %C œ "% B  %Ð#B  "Ñ œ "% B  )B  % œ "% *B œ ") Bœ#

Ð"Ñ

Substitute # for B in Ð#ÑÞ Check this ordered pair in the system.

C œ #Ð#Ñ  " œ $

? Ð#Ñ  Ð$Ñ œ & & œ & True ? #Ð#Ñ  Ð$Ñ œ " %  Ð$Ñ œ " True

The solution set is e #ß $ f. 17.

BC œ% #B  C œ # Graph the line B  C œ % through its intercepts, %ß ! and !ß % , and the line #B  C œ # through its intercepts, "ß ! and !ß # . The lines appear to intersect at Ð#ß #Ñ.

23.

The solution #ß $ checks. The solution set is e #ß $ f. $B  %C œ ## $B  C œ !

Ð"Ñ Ð#Ñ

Solve equation Ð#Ñ for C to get C œ $B. Substitute $B for C in equation Ð"Ñ. $B  %C œ ## $B  %Ð$BÑ œ ## $B  "#B œ ## *B œ ## B œ ## * œ

Ð"Ñ Let y = 3x.

## *

821

Systems of Linear Equations Substitute

## *

for B in C œ $B to get C œ $Ð ## * Ñœ

$C œ "! C œ  "! $

## $ .

Since B œ $# C and C œ  "! $ ,

## The solution Ð ## * ß $ Ñ checks. ## ™ The solution set is ˜Ð ## * ß $ Ñ .

25.

&B  %C œ * $  #C œ B

"! B œ $# Ð "! $ Ñ œ  # œ &.

Ð"Ñ Ð#Ñ

The solution Ð&ß  "! $ Ñ checks. ™ The solution set is ˜Ð&ß  "! $ Ñ .

Solve equation Ð#Ñ for B. $  #C œ B #C  $ œ B

31.

Ð#Ñ Ð$Ñ

" ‚ Ð#Ñ

$B  #Ð$BÑ œ ") $B  'B œ ") *B œ ") Bœ# Substitute # for B in C œ $B. C œ $Ð#Ñ œ '

The solution &ß % checks. The solution set is e &ß % f.

The solution #ß ' checks. The solution set is e #ß ' f.

Ð"Ñ Ð#Ñ

33.

Solve equation Ð#Ñ for B. Substitute #C  ( for B in Ð"Ñ.

B œ #Ð$Ñ  ( œ '  ( œ " The solution Ð"ß $Ñ checks. The solution set is ÖÐ"ß $Ñ×. 29.

B œ $C  & B œ $# C

Ð"Ñ Ð#Ñ

Both equations are given in terms of B. Choose equation Ð#Ñ, and substitute $# C for B in equation Ð"Ñ. B œ $C  & œ $C  & $C œ 'C  "! $ #C

822

Ð"Ñ Let x œ 32 y. Multiply by 2.

Ð"Ñ Ð#Ñ

%B  #C œ ! %B  #Ð#BÑ œ ! %B  %B œ ! !œ!

Ð"Ñ Let x = 2y  7.

Substitute $ for C in B œ #C  (.

C œ #B %B  #C œ !

From equation Ð"Ñ, substitute #B for C in equation Ð#Ñ.

B œ #C  (

%B  &C œ "" %Ð#C  (Ñ  &C œ "" )C  #)  &C œ "" "$C  #) œ "" "$C œ $* Cœ$

Ð$Ñ

From Ð#Ñ, we have C œ $B, so substitute $B for C in Ð$Ñ.

Bœ# % $œ&

%B  &C œ "" B  #C œ (

Ð"Ñ Ð#Ñ

$B  #C œ ")

Ð"Ñ Let x = 2y  3.

Substitute % for C in B œ #C  $.

27.

 "$ C œ $ $B  C œ !

Multiply Ð"Ñ by its LCD, ', to eliminate fractions

Substitute #C  $ for B in equation Ð"Ñ. &B  %C œ * &Ð#C  $Ñ  %C œ * "!C  "&  %C œ * 'C  "& œ * 'C œ #% Cœ%

" #B

Ð#Ñ

True

The equations are dependent, and the solution is the set of all points on the line. The solution set is e Bß C l C œ #Bf. 35.

B œ &C &B  #&C œ &

Ð"Ñ Ð#Ñ

From equation Ð"Ñ, substitute &C for B in equation Ð#Ñ. &B  #&C œ & &Ð&CÑ  #&C œ & #&C  #&C œ & !œ&

Ð#Ñ

False

The system is inconsistent. Since the graphs of the equations are parallel lines, there are no ordered pairs that satisfy both equations. The solution set is g.

Systems of Linear Equations 37.

C œ !Þ&B "Þ&B  !Þ&C œ &Þ!

$B  C œ ) Ð#Ñ $Ð$Ñ  C œ ) *C œ) C œ "

Ð"Ñ Ð#Ñ

Multiply both equations by # to eliminate decimals. (We could just substitute !Þ&B for C in the second equation.) #C œ B $B  C œ "!

Ð$Ñ Ð%Ñ

The ordered pair Ð$ß "Ñ satisfies both equations, so it checks. The solution set is e $ß " f.

# ‚ Ð"Ñ # ‚ Ð#Ñ

From Ð$Ñ, we have B œ #C, so substitute #C for B in Ð%Ñ.

45.

"&B  #!C œ $! "&B  *C œ $ ""C œ $$ C œ $

Substitute # for C in B œ #C. B œ #Ð#Ñ œ %

39.

The ordered pair Ð!ß !Ñ clearly satisfies the equations EB  FC œ ! and GB  HC œ !, so if there is a single solution, then it must be Ð!ß !Ñ.

41.

#B  $C œ "' #B  &C œ #%

The ordered pair Ð#ß $Ñ satisfies both equations, so it checks. The solution set is e #ß $ f.

To eliminate B, add the equations. Ð"Ñ Ð#Ñ

œ "' œ #% œ ) œ %

47.

#B  &C œ "" $B  C œ )

Ð"Ñ Ð#Ñ

To eliminate C, multiply equation Ð#Ñ by & and add the result to equation Ð"Ñ. #B  &C œ "" "&B  &C œ %! "(B œ &" B œ $

Ð"Ñ Ð$Ñ

& ‚ Ð#Ñ

To find C, substitute $ for B in equation Ð#Ñ.

Ð"Ñ Ð#Ñ

"%B  %C œ "# "%B  %C œ "# ! œ !

Ð#Ñ

The ordered pair Ð#ß %Ñ satisfies both equations, so it checks. The solution set is e #ß % f.

(B  #C œ ' "%B  %C œ "#

To eliminate C, multiply equation Ð"Ñ by # and add the result to equation Ð#Ñ.

To find B, substitute % for C in equation Ð#Ñ.

43.

& ‚ Ð"Ñ $ ‚ Ð#Ñ

&B  $C œ " Ð#Ñ &B  $Ð$Ñ œ " &B  * œ " &B œ "! Bœ#

Ð"Ñ Ð#Ñ

#B  &C œ #% #B  &Ð%Ñ œ #% #B  #! œ #% #B œ % Bœ#

Ð$Ñ Ð%Ñ

To find B, substitute $ for C in equation Ð#Ñ.

The solution %ß # checks. The solution set is e %ß # f.

$C &C #C C

Ð"Ñ Ð#Ñ

To eliminate B, multiply equation Ð"Ñ by & and equation Ð#Ñ by $. Then add the results.

$Ð#CÑ  C œ "! 'C  C œ "! &C œ "! Cœ#

#B  #B 

$B  %C œ ' &B  $C œ "

Ð$Ñ Ð#Ñ True

# ‚ Ð"Ñ

Multiplying equation Ð"Ñ by # gives equation Ð#Ñ. The equations are dependent, and the solution is the set of all points on the line. The solution set is e Bß C l (B  #C œ 'f.

49.

$B  $C œ ! %B  #C œ $

Ð"Ñ Ð#Ñ

To eliminate C, multiply equation Ð"Ñ by # and equation Ð#Ñ by $. Then add the results. 'B  'C œ ! Ð$Ñ "#B  'C œ * Ð%Ñ 'B œ * $ B œ * ' œ # To find C, substitute

$ #

# ‚ Ð"Ñ $ ‚ Ð#Ñ

for B in equation Ð"Ñ.

823

Systems of Linear Equations $B  $C œ ! $Ð $# Ñ  $C œ ! * #

55.

Ð"Ñ

B  "# C œ B 

 $C œ !

# &C

œ

#

Ð"Ñ

 )&

Ð#Ñ

To eliminate B, add equations Ð"Ñ and Ð#Ñ to get

$C œ  *#

 "# C  #& C œ #  )& Þ

C œ "$ Ð *# Ñ œ  $# The solution Ð $# ß  $# Ñ checks. The solution set is ˜Ð $# ß  $# Ñ™.

Multiply by the LCD, "!. &C  %C œ #!  "' C œ % C œ %

When you get a solution that has non-integer components, it is sometimes more difficult to check the problem than it was to solve it. A graphing calculator can be very helpful in this case. Just store the values for B and C in their respective memory locations, and then type the expressions as shown in the following screen. The results ! and $ (the right sides of the equations) indicate that we have found the correct solution.

Substitute % for C in equation Ð"Ñ to find B. B  "# Ð%Ñ œ # B#œ# Bœ! The solution Ð!ß %Ñ checks. The solution set is eÐ!ß %Ñf. 57.

" #B " #B

 "$ C œ  #C œ

%* ") % $

Ð"Ñ Ð#Ñ

Eliminate the fractions by multiplying equation Ð"Ñ by ") and equation Ð#Ñ by '.

51.

&B  &C œ $ B  C œ "#

*B  'C œ %* $B  "#C œ )

Ð"Ñ Ð#Ñ

Multiply equation Ð$Ñ by # and add the resulting equations to eliminate C.

To eliminate B, multiply equation Ð#Ñ by & and add the result to equation Ð"Ñ. &B  &C œ $ &B  &C œ '! ! œ &(

Ð"Ñ Ð$Ñ False

")B  "#C œ *) $B  "#C œ ) "&B œ *! B œ '

& ‚ Ð#Ñ

Ð"Ñ Ð#Ñ

The solution Ð'ß  &' Ñ checks. The solution set is ˜ˆ'ß  &' ‰™.

To eliminate C, multiply equation Ð"Ñ by # and add the result to equation Ð#Ñ. #B  #C œ #B  #C œ %B œ B œ

! ! ! !

Ð$Ñ Ð#Ñ

# ‚ Ð"Ñ

Substitute ! for B in Ð"Ñ to get C œ !. The solution Ð!ß !Ñ checks. The solution set is eÐ!ß !Ñf.

824

# ‚ Ð$Ñ

$B  "#C œ ) Ð%Ñ $Ð'Ñ  "#C œ ) ")  "#C œ ) "#C œ "! C œ  &'

The solution set is g. B C œ! #B  #C œ !

Ð&Ñ Ð%Ñ

To find C, substitute ' for B in equation Ð%Ñ.

The system is inconsistent. Since the graphs of the equations are parallel lines, there are no ordered pairs that satisfy both equations. 53.

Ð$Ñ ") ‚ Ð"Ñ Ð%Ñ ' ‚ Ð#Ñ

59.

$B  (C œ % 'B  "%C œ $

Ð"Ñ Ð#Ñ

Write each equation in slope-intercept form by solving for C. $B  (C œ % (C œ $B  % C œ  $( B  %(

Ð"Ñ

Systems of Linear Equations 'B  "%C œ $ "%C œ 'B  $ ' $ C œ  "% B  "% C œ  $( B 

$ "%

Since the equations have the same slope,  $( , but $ different C-intercepts, %( and "% , the lines when graphed are parallel. The system is inconsistent and has no solution. 61.

#B œ $C  " 'B œ *C  $

$Ð "$ Ñ  #C œ ! "  #C œ ! #C œ " C œ "#

Ð#Ñ

Check Ð "$ ß "# Ñ: "  " œ !à $  % œ ( The solution set is ˜Ð "$ ß "# Ñ™. 67.

Ð"Ñ Ð#Ñ

C œ $B  ")

C œ  #$ B 

#B  $Ð$B  ")Ñ œ "! #B  *B  &% œ "! ""B  &% œ "! ""B œ %% B œ %

Ð#Ñ

'B œ *C  $ *C œ 'B  $ C œ  '* B  $*

Substitute % for B in Ð$Ñ.

" $

C œ $Ð%Ñ  ") œ '

Since both equations are the same, the solution set is all points on the line C œ  #$ B  "$ . The system has infinitely many solutions.

69.

Ð"Ñ Ð#Ñ

Substitute $ for B in Ð#Ñ. 71.

Ð"Ñ Ð#Ñ

Multiply equation Ð"Ñ by % and add the result to equation Ð#Ñ. "#B  )C œ ! *B  )C œ ( #"B œ ( B œ "$ Substitute

" $

 ") C œ  "% %B  C œ #

Ð"Ñ Ð#Ñ

Ð$Ñ Ð#Ñ

) ‚ Ð"Ñ

The equations are dependent. The solution set is eÐBß CÑ l %B  C œ #f. !Þ$B  !Þ#C œ !Þ% !Þ&B  !Þ%C œ !Þ(

Ð"Ñ Ð#Ñ

To eliminate C, multiply equation Ð"Ñ by #! and equation Ð#Ñ by "!, then add.

The solution Ð$ß #Ñ checks. The solution set is e $ß # f. $B  #C œ ! *B  )C œ (

" #B

%B  C œ # %B  C œ # ! œ !

$B  C œ ( B  C œ & %B œ "# B œ $

$  C œ & C œ # Cœ#

The solution Ð%ß 'Ñ checks. The solution set is e %ß ' f.

Multiply equation Ð"Ñ by ) and add to equation Ð#Ñ.

Add the equations to eliminate C.

65.

Ð$Ñ

Substitute $B  ") for C in Ð"Ñ.

Ð"Ñ

#B œ $C  " $C œ #B  " C œ  #$ B  "$

$B  C œ ( B  C œ &

Ð"Ñ Ð#Ñ

Solve equation Ð#Ñ for C.

Write each equation in slope-intercept form by solving for C.

63.

#B  $C œ "! $B  C œ ")

% ‚ Ð"Ñ Ð#Ñ

for B in equation Ð"Ñ.

'B  %C œ ) &B  %C œ ( B œ " B œ "

Ð$Ñ

#! ‚ Ð"Ñ "! ‚ Ð#Ñ

Substitute " for B in equation Ð$Ñ. &Ð"Ñ  %C œ ( %C œ # C œ "# Check Ð"ß "# Ñ: !Þ$  !Þ" œ !Þ%; !Þ&  !Þ# œ !Þ( The solution set is ˜ˆ"ß "# ‰™.

825

Systems of Linear Equations 73.

The table shows that when X œ $, Y" œ % and Y# œ %. Since no other values of X in the table give the same values for Y" and Y# , and since the functions are linear, the point Ð$ß %Ñ is the only point of intersection for the two graphs.

75.

C" œ $B  & C# œ %B  #

81.

(b) LCD flat panel sales were greater than plasma flat panel sales between 2005 and 2006. (c) The graph for LCD flat panel sales intersects the graph for plasma flat panel sales between 2005 and 2006. The sales value is about $%!!! million.

Both of the graphs in B have negative C-intercepts. But the C-intercept of C# is positive, so the graph in B is not acceptable. The slope of C" is positive, and C" has a negative C-intercept. The slope of C# is negative, and C# has a positive C-intercept. This fits graph A. 77.

(a)

B  C œ "! #B  C œ & $B œ "& B œ &

(d) The ordered pair is Ð#!!&ß %!!!Ñ. (e) Sales of front projection displays were fairly constant. Sales of plasma flat panel displays increased from 2003 to 2006, but then declined. Sales of LCD flat panel displays increased over the whole period, fairly slowly at first, and then very rapidly.

Ð"Ñ Ð#Ñ

Substitute & for B in equation Ð"Ñ. B  C œ "! &  C œ "! Cœ&

83.

The graphs intersect at about "Þ& (years since 2000). So the sales of digital cameras were less than the sales of digital camcorders for 2000, 2001, and the first half of 2002.

85.

""*B  C œ #)$) '#%B  C œ ")#$

Ð"Ñ

The solution set is e &ß & f.

Ð"Ñ Ð#Ñ

Adding the equations gives us (%$B œ "!"&, so B œ "!"& (%$ ¸ "Þ%.

(b) Solve Ð"Ñ and Ð#Ñ for C. B  C œ "! C œ B  "! #B  C œ & #B  & œ C

(a) The graph for plasma flat panel displays is above the graph for front projection displays for years 2004–2008, so those are the years in which the sales for plasma flat panel displays exceeded the sales for front projection displays.

Ð"Ñ Ð$Ñ

Solve equation Ð#Ñ for C. C œ '#%B  ")#$

Ð#Ñ Ð%Ñ

Substitute

"!"& (%$

Ð$Ñ

for B in Ð$Ñ.

C œ '#%Ð "!"& (%$ Ñ  ")#$ ¸ #'(&Þ%

Now graph Ð$Ñ and Ð%Ñ.

Written as an ordered pair, the solution is Ð"Þ%ß #'(&Þ%Ñ. (Values may vary slightly based on the method of solution used.) 87.

$ % &  œ B C # & $ (  œ B C %

Ð"Ñ Ð#Ñ

If : œ B" and ; œ C" , equations Ð"Ñ and Ð#Ñ can be written as 79.

(a) The supply and demand graphs intersect at %, so supply equals demand at a price of $% per halfgallon. (b) At a price of $% per half-gallon, the supply and demand are both about $!! half-gallons. (c) At a price of $# per half-gallon, the supply is #!! half-gallons and the demand is %!! halfgallons.

826

$:  %; œ &:  $; œ

& # ( %.

Ð$Ñ Ð%Ñ

To eliminate ; , multiply equation Ð$Ñ by $ and equation Ð%Ñ by %. Then add the results. *:  "#; œ #!:  "#; œ

"& #

#*:

#* # " #

œ : œ

(

Ð&Ñ Ð'Ñ

$ ‚ Ð$Ñ % ‚ Ð%Ñ

Systems of Linear Equations To find ; , substitute

" #

$:  %; œ

& # & # & #

$Ð "# Ñ  %; œ $ #

Since : œ Since ; œ

Substitute

for : in equation Ð$Ñ. Let p œ 12 .

The solution ˆ +- ß !‰ checks. The solution set is ˜ˆ +- ß !‰™.

and : œ #" , #" œ B" and B œ #. and ; œ %" , %" œ C" and C œ %.

93.

The solution Ð#ß %Ñ checks. The solution set is e #ß % f. 89.

# $ ""  œ B C # " #   œ " B C

Let : œ

" B

Ð#Ñ

Bœ

'; ); "%; ;

œ "" œ % œ ( œ "#

C œ &+ˆ +" ‰ œ &

Ð$Ñ Ð%Ñ

Ð&Ñ Ð'Ñ

The solution ˆ +" ß &‰ checks. The solution set is ˜ˆ +" ß &‰™.

# ‚ Ð$Ñ % ‚ Ð%Ñ

Since : œ

" B

and : œ #,

# œ B" , so #B œ " and B œ #" . Since ; œ

" C

and ; œ #" , " #

œ

" C

and C œ #.

The solution Ð "# ß #Ñ checks. The solution set is ˜Ð "# ß #Ñ™. 91.

95.

%Ð#B  $C  DÑ œ %Ð&Ñ )B  "#C  %D œ #!

97.

B  #C  $D "  #Ð#Ñ  $D "  %  $D $D D

for ; in Ð%Ñ. :  #Ð "# Ñ œ " :  " œ " : œ # :œ#

+B  ,C œ +B  #,C œ -

99.

2

œ* œ* œ* œ "# œ%

Let x = 1, y = 2.

Multiplying the first equation by $ will give us $B, which when added to $B in the second equation, gives us !.

Systems of Linear Equations in Three Variables

2 Now Try Exercises N1.

B  C  #D œ " $B  #C  (D œ ) $B  %C  *D œ "!

Ð"Ñ Ð#Ñ Ð$Ñ

Step 1 Eliminate B by adding equations Ð#Ñ and Ð$Ñ.

Ð"Ñ Ð#Ñ

To eliminate C, multiply Ð"Ñ by # and add the result to equation Ð#Ñ. Ð$Ñ #+B  #,C œ #Ð#Ñ +B  #,C œ $+B œ $$B œ œ $+ +

$ " œ $+ +

Substitute  +" for B in Ð#Ñ.

To eliminate :, multiply Ð$Ñ by # and Ð%Ñ by %. Then add the results.

" #

Ð"Ñ Ð#Ñ

#+B  &+B œ $ $+B œ $

Ð"Ñ

#:  $; œ "" # :  #; œ ".

Substitute

#+B  C œ $ C œ &+B

Substitute &+B for C in Ð"Ñ.

and ; œ C" . Rewrite the system as

%:  %: 

for B in Ð"Ñ. +Ð +- Ñ  ,C œ -  ,C œ ,C œ ! Cœ!

Ð$Ñ

 %; œ %; œ " ; œ "%

" B " C

+

# ‚ Ð"Ñ

$B  #C  (D œ ) $B  %C  *D œ "!  #C  "'D œ #

Ð#Ñ Ð$Ñ Ð%Ñ

Step 2 To eliminate B again, multiply equation Ð"Ñ by $ and add the result to equation Ð$Ñ. $B  $C  'D œ $ $B  %C  *D œ "!  (C  "&D œ (

$ ‚ Ð"Ñ Ð$Ñ Ð&Ñ 827

Systems of Linear Equations Step 3 Use equations Ð%Ñ and Ð&Ñ to eliminate C. Multiply equation Ð%Ñ by ( and add the result to # times equation Ð&Ñ. "%C  ""#D œ "% "%C  $!D œ "%  )#D œ ! D œ !

D œ % Dœ% Substitute % for D in equation Ð#Ñ to find C.

Step 4 Substitute ! for D in equation Ð&Ñ to find C.

Equation Ð"Ñ: '  % œ "! Equation Ð#Ñ: %  #! œ #% Equation Ð$Ñ: #  ' œ )

Step 5 Substitute ! for D and " for C in equation Ð"Ñ to find B. B  C  #D œ " B"!œ" Bœ#

Check Ð#ß "ß %Ñ

Ð&Ñ Let z = 0.

(C  "&D œ ( (C  "&Ð!Ñ œ ( (C œ ( Cœ"

N3.

Ð"Ñ Let y = 1, z = 0.

The solution set is eÐ#ß "ß !Ñf.

True True True

# ‚ Ð"Ñ

N4.

$ ‚ Ð#Ñ # ‚ Ð$Ñ Ð%Ñ

Use equation Ð%Ñ together with equation Ð"Ñ to eliminate D . Multiply equation Ð"Ñ by "& and add it to equation Ð%Ñ. %&B  "&D œ "&! #B  "&D œ &' %(B œ *% B œ #

"& ‚ Ð"Ñ Ð%Ñ

Substitute # for B in equation Ð"Ñ to find D . Ð"Ñ Let x = 2.

Ð$Ñ False

The solution set is g.

Ð"Ñ Ð#Ñ Ð$Ñ

"#C  "&D œ (# #B  "#C œ "' #B  "&D œ &'

828

Ð"Ñ Ð#Ñ Ð$Ñ

Since a false statement results, the system is inconsistent.

B  $C  #D œ "! #B  'C  %D œ #! " $ #B  #C  D œ &

Ð"Ñ Ð#Ñ Ð$Ñ

Since equation Ð#Ñ is # times equation Ð"Ñ and equation Ð$Ñ is "# times equation Ð"Ñ, the three equations are dependent. All three have the same graph.

Since equation Ð"Ñ is missing C, eliminate C again from equations Ð#Ñ and Ð$Ñ. Multiply equation Ð#Ñ by $ and add the result to # times equation Ð$Ñ.

$B  D œ "! $Ð#Ñ  D œ "! '  D œ "!

B  &C  #D œ % $B  C  D œ ' #B  "!C  %D œ (

#B  "!C  %D œ ) #B  "!C  %D œ ( ! œ "&

#"!œ" Equation Ð"Ñ: '#!œ) Equation Ð#Ñ: Equation Ð$Ñ: '  %  ! œ "!

 D œ "! %C  &D œ #% B  'C œ )

The solution set is eÐ#ß "ß %Ñf.

True True True

Multiply equation Ð"Ñ by # and add the result to equation Ð$Ñ.

Step 6 Check Ð#ß "ß !Ñ

N2. $B

Ð#Ñ Let z = 4.

%C  &D œ #% %C  &Ð%Ñ œ #% %C  #! œ #% %C œ % Cœ"

( ‚ Ð%Ñ # ‚ Ð&Ñ

N5.

The solution set is eÐBß Cß DÑ l B  $C  #D œ "!f. B  $C  #D œ %  C  #$ D œ (

Ð"Ñ Ð#Ñ

 $# C  D œ #

Ð$Ñ

" $B " #B

Eliminate the fractions in equations Ð#Ñ and Ð$Ñ. B  $C  #D œ % B  $C  #D œ #% B  $C  #D œ %

Ð"Ñ Ð%Ñ Ð&Ñ

$ ‚ Ð#Ñ # ‚ Ð$Ñ

Equations Ð"Ñ and Ð&Ñ are dependent (they have the same graph). Equations Ð"Ñ and Ð%Ñ are not equivalent. Since they have the same coefficients, but a different constant term, their graphs have no points in common (the planes are parallel). Thus, the system is inconsistent and the solution set is g.

Systems of Linear Equations 5.

2 Section Exercises 1.

Substitute " for B, # for C, and $ for D in $B  #C  D, which is the left side of each equation.

$B  #C  D œ ) B  %C  D œ #! %B  'C œ #)

Ð"Ñ Ð#Ñ Ð$Ñ

#B  )C  #D œ %! #B  $C  #D œ "' %B  &C œ #%

Eliminate B by adding equation Ð"Ñ to # times equation Ð#Ñ. #B  &C  $D œ " Ð"Ñ #B  )C  %D œ ")  "$C  (D œ "* Ð%Ñ

# ‚ Ð#Ñ

%B  'C %B  &C  C C

Ð#Ñ " ‚ Ð$Ñ

%B  &C œ #% %B  &Ð%Ñ œ #% %B  #! œ #% %B œ % Bœ"

Use equations Ð%Ñ and Ð&Ñ to eliminate D . Multiply equation Ð%Ñ by # and add the result to ( times equation Ð&Ñ. # ‚ Ð%Ñ ( ‚ Ð&Ñ

œ "% œ "% œ "% œ# œ"

B  %C  D "  %Ð%Ñ  D "  "'  D "(  D D D

Ð&Ñ

Substitute # for C and " for D in equation Ð$Ñ to find B. B  #C  %D œ & B  #Ð#Ñ  %Ð"Ñ œ & B  %  % œ & B  ) œ & Bœ$

Ð$Ñ

The solution $ß #ß " checks in all three of the original equations. The solution set is e $ß #ß " f.

" ‚ Ð%Ñ Ð&Ñ

Ð&Ñ

Substitute " for B and % for C in equation Ð$Ñ to find D .

Substitute # for C in equation Ð&Ñ to find D . 'C  #D 'Ð#Ñ  #D "#  #D #D D

œ #) œ #% œ % œ %

Substitute % for C in equation Ð&Ñ to find B.

Ð&Ñ

#'C  "%D œ $) %#C  "%D œ *) ')C œ "$' C œ #

# ‚ Ð$Ñ Ð#Ñ Ð&Ñ

Use equations Ð%Ñ and Ð&Ñ to eliminate B. Multiply equation Ð%Ñ by " and add the result to equation Ð&Ñ.

Now eliminate B by adding equation Ð#Ñ to " times equation Ð$Ñ. B  %C  #D œ * B  #C  %D œ & 'C  #D œ "%

Ð"Ñ Ð$Ñ Ð%Ñ

To get another equation without D , multiply equation Ð$Ñ by # and add the result to equation Ð#Ñ.

Choice B is correct since its right side is %. #B  &C  $D œ " B  %C  #D œ * B  #C  %D œ &

Ð"Ñ Ð#Ñ Ð$Ñ

Eliminate D by adding equations Ð"Ñ and Ð$Ñ.

$Ð"Ñ  #Ð#Ñ  Ð$Ñ œ $  %  $ œ($ œ% 3.

$B  #C  D œ ) #B  $C  #D œ "' B  %C  D œ #!

7.

œ #! œ #! œ #! œ #! œ$ œ $

Ð$Ñ

The solution "ß %ß $ checks in all three of the original equations. The solution set is e "ß %ß $ f. #B  &C  #D œ ! %B  (C  $D œ " $B  )C  #D œ '

Ð"Ñ Ð#Ñ Ð$Ñ

Add equations Ð"Ñ and Ð$Ñ to eliminate D . #B  &C  #D œ ! $B  )C  #D œ ' &B  $C œ '

Ð"Ñ Ð$Ñ Ð%Ñ

Multiply equation Ð"Ñ by $ and equation Ð#Ñ by #. Then add the results to eliminate D again.

829

Systems of Linear Equations Ð"Ñ  #Ð!Ñ  D œ % Dœ$

$ ‚ Ð"Ñ # ‚ Ð#Ñ

'B  "&C  'D œ ! )B  "%C  'D œ # "%B  C œ #

Ð&Ñ

Solve the system

11.

&B  $C œ ' "%B  C œ #.

Ð%Ñ Ð&Ñ

$ ‚ Ð&Ñ

Ð%Ñ

9.

Ð"Ñ

Substitute " for B and

$ &

Ð"Ñ Ð$Ñ Ð&Ñ

Ð'Ñ $ ‚ Ð&Ñ Ð%Ñ

Substitute " for B in equation Ð&Ñ and solve for C. #Ð"Ñ  C œ # Cœ! Substitute " for B and ! for C in Ð"Ñ.

Ð"Ñ

 'D œ $ Dœ

Ð"Ñ Ð#Ñ Ð%Ñ

Multiply equation Ð&Ñ by $.

830

for C in equation Ð"Ñ.

'D œ

Add Ð"Ñ and Ð$Ñ to eliminate D .

'B  $C œ ' $B  $C œ $ $B œ $ B œ "

$ "!

Ð%Ñ

B  #C  'D œ # $ "  #Ð "! Ñ  'D œ #

Add Ð"Ñ and Ð#Ñ to eliminate D .

B  #C  D œ % B  C  D œ # #B  C œ #

Ð%Ñ

&B  "!C œ ) &Ð"Ñ  "!C œ ) "!C œ $ $ C œ "!

Ð"Ñ Ð#Ñ Ð$Ñ

B  #C  D œ % #B  C  D œ " $B  $C œ $

# ‚ Ð$Ñ

Substitute " for B in equation Ð%Ñ.

The solution set is e !ß #ß & f. B  #C  D œ % #B  C  D œ " B  C  D œ #

Ð#Ñ

$B  #C  'D œ ' #B  )C  'D œ # &B  "!C œ )

To find D , substitute B œ ! and C œ # in equation Ð"Ñ. œ! œ! œ! œ "! œ &

" ‚ Ð#Ñ

Eliminate D by adding equation Ð#Ñ to # times equation Ð$Ñ.

To find C, substitute B œ ! into equation Ð%Ñ.

#B  &C  #D #Ð!Ñ  &Ð#Ñ  #D "!  #D #D D

Ð"Ñ

B  #C  'D œ # $B  #C  'D œ ' %B œ % B œ "

Ð%Ñ

&B  $C œ ' &Ð!Ñ  $C œ ' Cœ#

Ð"Ñ Ð#Ñ Ð$Ñ

B  #C  'D œ # $B  #C  'D œ ' B  %C  $D œ "

Eliminate C and D by adding equation Ð"Ñ to " times equation Ð#Ñ.

Multiply equation Ð&Ñ by $ then add this result to Ð%Ñ. &B  $C œ ' %#B  $C œ ' %(B œ ! B œ !

The solution set is e "ß !ß $ f.

"# & # &

$ # ‰™ The solution set is ˜ˆ"ß "! ß& .

13.

Ð"Ñ Ð#Ñ Ð$Ñ

B  C  D œ # #B  C  D œ & B  #C  $D œ %

Eliminate C and D by adding equations Ð"Ñ and Ð#Ñ. B  C  D œ #B  C  D œ $B œ B œ

# & (  ($

Ð"Ñ Ð#Ñ

To get another equation without C, multiply equation Ð#Ñ by # and add the result to equation Ð$Ñ. %B  #C  #D œ "! B  #C  $D œ % $B  D œ "%

# ‚ Ð#Ñ Ð$Ñ Ð%Ñ

Substitute  ($ for B in equation Ð%Ñ to find D .

Systems of Linear Equations $B  D $Ð ($ Ñ  D (  D D D

œ "% œ "% œ "% œ ( œ(

Ð%Ñ

"!B  "*!D œ "#! "!B  #(D œ "#! #"(D œ ! D œ !

Substituting ! for D in Ð(Ñ gives us B œ "#. Substitute "# for B and ! for D in Ð%Ñ.

Substitute  ($ for B and ( for D in equation Ð"Ñ to find C. B  C  D œ #  ($  C  ( œ # (  $C  #" œ ' $C  #) œ ' $C œ ## C œ ## $

#Ð"#Ñ  C  %Ð!Ñ œ ' #%  C œ ' C œ ")

Ð"Ñ Multiply by 3. 17.

‰™ The solution set is ˜ˆ ($ ß ## $ ß( . A calculator check reduces the probability of making any arithmetic errors and is highly recommended. The following screen shows the substitution of the solution for B, C, and D along with the left sides of the three original equations. The evaluation of the three expressions, #, &, and % (the right sides of the three equations), indicates that we have found the correct solution.

Ð)Ñ

The solution set is e "#ß ")ß ! f.

Ð"Ñ Ð#Ñ Ð$Ñ

&Þ&B  #Þ&C  "Þ'D œ ""Þ)$ #Þ#B  &Þ!C  !Þ"D œ &Þ*( $Þ$B  (Þ&C  $Þ#D œ #"Þ#&

Multiply the equations by "! to eliminate the decimals for the coefficients of the variables. &&B  #&C  "'D œ "").$ ##B  &!C  D œ &*Þ( $$B  (&C  $#D œ #"#Þ&

Ð%Ñ Ð&Ñ Ð'Ñ

Multiply equation Ð%Ñ by $ and add the result to equation Ð'Ñ to eliminate C. "'&B  (&C  %)D œ $&%Þ* $$B  (&C  $#D œ #"#Þ& "$#B  "'D œ "%#Þ%

$ ‚ Ð%Ñ Ð'Ñ Ð(Ñ

Multiply equation Ð%Ñ by # and add the result to equation Ð&Ñ to eliminate C. # ‚ Ð%Ñ

""!B  &!C  $#D œ #$'Þ' ##B  &!C  D œ &*Þ( "$#B  $"D œ "('Þ* 15.

" $B  $% B " #B

  

" 'C " $C $ #C

  

# $D " %D $ %D

œ "

Ð"Ñ

œ

$

Ð#Ñ

œ #"

Ð$Ñ

Now add equations Ð(Ñ and Ð)Ñ to eliminate B. "$#B  "'D "$#B  $"D "&D D

Eliminate all fractions. #B  C  %D œ ' *B  %C  $D œ $' #B  'C  $D œ )%

Ð%Ñ ' ‚ Ð"Ñ Ð&Ñ "# ‚ Ð#Ñ Ð'Ñ % ‚ Ð$Ñ

Multiply Ð%Ñ by %. )B  %C  "'D œ #% *B  %C  $D œ $' B  "*D œ "#

Ð&Ñ Ð(Ñ

Multiply Ð%Ñ by '. "#B  'C  #%D œ $' #B  'C  $D œ )% "!B  #(D œ "#! Multiply Ð(Ñ by "!.

Ð&Ñ Ð)Ñ

Ð'Ñ Ð)Ñ

œ "%#Þ% œ "('Þ* œ $%Þ& œ #Þ$

Ð(Ñ Ð)Ñ

Substitute #Þ$ for D in equation Ð)Ñ and solve for B. "$#B  $"Ð#Þ$Ñ œ "('Þ* "$#B  ("Þ$ œ "('Þ* "$#B œ "!&Þ' B œ !Þ) Use equation Ð&Ñ to solve for C. ##B  &!C  D œ &*Þ( ##Ð!Þ)Ñ  &!C  #Þ$ œ &*Þ( &!C œ (& C œ "Þ&

The solution set is e !Þ)ß "Þ&ß #Þ$ f.

831

Systems of Linear Equations 19.

#B  $C  #D œ " B  #C  D œ "( #C  D œ (

B  %C  D œ % #  %C  # œ % %C  % œ % %C œ ) Cœ#

Ð"Ñ Ð#Ñ Ð$Ñ

Multiply equation Ð#Ñ by #, and add the result to equation Ð"Ñ. #B  $C  #D œ " Ð"Ñ #B  %C  #D œ $% # ‚ Ð#Ñ  (C œ $& C œ &

23.

œ( œ( œ( œ $ œ$

Ð$Ñ

'B  "!D œ "% # ‚ Ð$Ñ 'B  #D œ ## Ð%Ñ  "#D œ $' D œ $

Ð"Ñ

To find B, substitute $ for D into equation Ð$Ñ. $B  &D œ ( $B  &Ð$Ñ œ ( $B œ ) B œ )$

Ð"Ñ Ð#Ñ Ð$Ñ

Equation Ð$Ñ is missing C. Eliminate C again by multiplying equation Ð"Ñ by # and adding the result to equation Ð#Ñ.

Use equations Ð$Ñ and Ð%Ñ to eliminate B. Multiply equation Ð$Ñ by * and add the result to equation Ð%Ñ. *B  ")D œ ") * ‚ Ð$Ñ *B  &D œ ) Ð%Ñ "$D œ #' D œ # Substitute # for D in equation Ð$Ñ to find B. Ð$Ñ

Substitute # for B and # for D in equation Ð#Ñ to find C.

Ð$Ñ

To find C, substitute $ for D into equation Ð#Ñ. $C  #D œ % $C  #Ð$Ñ œ % $C œ # C œ #$

)B  %C  'D œ "# # ‚ Ð"Ñ B  %C  D œ % Ð#Ñ *B  &D œ ) Ð%Ñ

832

Ð"Ñ Ð#Ñ Ð$Ñ

Since equation Ð$Ñ does not have a C-term, we can multiply equation Ð$Ñ by # and add the result to equation Ð%Ñ to eliminate B and solve for D .

The solution set is e %ß &ß $ f.

B  #D œ # B  #Ð#Ñ œ # B  % œ # B œ # Bœ#

C œ ' $C  #D œ %  &D œ (

'B  $C œ ") $ ‚ Ð"Ñ $C  #D œ % Ð#Ñ 'B  #D œ ## Ð%Ñ

#B  $C  #D œ " #B  $Ð&Ñ  #Ð$Ñ œ " #B  * œ " #B œ ) Bœ%

%B  #C  $D œ ' B  %C  D œ % B  #D œ #

#B 

To eliminate C, multiply equation Ð"Ñ by $ and add the result to equation Ð#Ñ.

To find B, substitute C œ & and D œ $ into equation Ð"Ñ.

21.

The solution set is e #ß #ß # f.

$B

To find D , substitute & for C in equation Ð$Ñ. #C  D #Ð&Ñ  D "!  D D D

Ð#Ñ

Ð#Ñ

The solution set is ˜ˆ )$ ß #$ ß $‰™. 25.

&B  #C  D œ & $B  #C  D œ $ B  'C œ "

Ð"Ñ Ð#Ñ Ð$Ñ

Add Ð"Ñ and Ð#Ñ to eliminate C and DÞ )B œ ) B œ " Substitute " for B in equation Ð$Ñ. B  'C œ " Ð"Ñ  'C œ " 'C œ ! Cœ!

Ð$Ñ

Substitute " for B and ! for C in Ð"Ñ.

Systems of Linear Equations &B  #C  D &Ð"Ñ  #Ð!Ñ  D &D D

27.

œ& œ& œ& œ!

Substitute ' for D in Ð%Ñ.

Ð"Ñ

The solution set is eÐ"ß !ß !Ñf. (B

 $D œ $% #C  %D œ #! $ " œ # %B  'C

Ð"Ñ Ð#Ñ Ð$Ñ

31.

Eliminate fractions from equation Ð$Ñ by multiplying by "#. *B  #C œ #%

*B  %D œ %%Þ Ð&Ñ Eliminate D by adding % times Ð"Ñ to $ times Ð&Ñ. 33.

35.

#C  %Ð#Ñ œ #! #C  ) œ #! #C œ "# Cœ'

$ &C " $B

  

D œ ' œ !

Ð"Ñ Ð#Ñ

œ &

Ð$Ñ

" #D # $D

'C  &D œ !  #D œ "&

Ð%Ñ Ð&Ñ

"! ‚ Ð#Ñ $ ‚ Ð$Ñ

"# # ‚ Ð"Ñ "& Ð&Ñ #( $

Substitute $ for B in Ð&Ñ. B  #D $  #D #D D

œ "& œ "& œ "# œ '

Ð"Ñ Ð#Ñ Ð$Ñ

# ‚ Ð"Ñ Ð#Ñ Ð%Ñ

Eliminate B by adding Ð$Ñ to $ times Ð"Ñ.

Eliminate D by adding Ð&Ñ to # times Ð"Ñ. )B  #D œ B  #D œ *B œ B œ

B  &C  #D œ " #B  )C  D œ % $B  C  &D œ "*

#B  "!C  %D œ # #B  )C  D œ % ")C  $D œ '

Eliminate fractions first. B

The solution set is e Bß Cß D l B  C  %D œ )f.

Eliminate B by adding Ð#Ñ to # times Ð"Ñ.

The solution set is eÐ%ß 'ß #Ñf. %B

Ð"Ñ Ð#Ñ Ð$Ñ

&B  &C  #!D œ %! B  C  %D œ ) $B  $C  "#D œ #%

Dividing equation Ð"Ñ by & gives equation Ð#Ñ. Dividing equation Ð$Ñ by $ also gives equation Ð#Ñ. The resulting equations are the same, so the three equations are dependent.

œ %% œ %% œ) œ#

Substitute # for D in Ð#Ñ.

29.

Ð"Ñ Ð#Ñ Ð$Ñ

The solution set is g; inconsistent system.

Substitute % for B in Ð&Ñ. *Ð%Ñ  %D $'  %D %D D

#B  #C  'D œ & $B  C  D œ # B  C  $D œ %

The solution set is e $ß &ß ' f.

#B  #C  'D œ & Ð"Ñ #B  #C  'D œ ) # ‚ Ð$Ñ ! œ "$ False

Adding equations Ð#Ñ and Ð%Ñ gives us

% ‚ Ð"Ñ $ ‚ Ð&Ñ

Ð%Ñ

Multiply equation Ð$Ñ by # and add the result to equation Ð"Ñ.

Ð%Ñ "# ‚ Ð$Ñ

#)B  "#D œ "$' #(B  "#D œ "$# B œ %

'C  &D œ ! 'C  &Ð'Ñ œ ! 'C œ $! Cœ&

Ð&Ñ

$B  "&C  'D œ $ $B  C  &D œ "*  "'C  ""D œ ##

$ ‚ Ð"Ñ Ð$Ñ Ð&Ñ

To eliminate D from equations Ð%Ñ and Ð&Ñ, we could first divide equation Ð%Ñ by $ and then multiply the resulting equation by "", or we could simply multiply equation Ð%Ñ by "" $ and add it to equation Ð&Ñ. ''C  ""D œ ## "'C  ""D œ ## &!C œ ! C œ !

"" $

‚ Ð%Ñ

Ð&Ñ

Substitute ! for C in equation Ð&Ñ.

833

Systems of Linear Equations "'C  ""D "'Ð!Ñ  ""D ""D D

B  C  #D !  !  #D #D D

Ð&Ñ

œ ## œ ## œ ## œ#

Substitute ! for C and # for D in equation Ð"Ñ. B  &C  #D œ " B  &Ð!Ñ  #Ð#Ñ œ " B  % œ " Bœ$

37.

Ð"Ñ

41.

Ð"Ñ Ð#Ñ Ð$Ñ

Ð"Ñ Ð#Ñ Ð$Ñ

Eliminate D by adding equations Ð#Ñ and Ð$Ñ. $B  C  D œ ! %B  #C  D œ ! (B  C œ !

Ð#Ñ Ð$Ñ Ð%Ñ

To get another equation without D , multiply equation Ð#Ñ by # and add the result to equation Ð"Ñ. 'B  #C  #D œ ! # ‚ Ð#Ñ B  C  #D œ ! Ð"Ñ (B  C œ ! Ð&Ñ Add equations Ð%Ñ and Ð&Ñ to find B. (B  C œ (B  C œ "%B œ B œ

! Ð%Ñ ! Ð&Ñ ! !

Substitute ! for B in equation Ð%Ñ to find C. (B  C œ ! (Ð!Ñ  C œ ! !C œ! Cœ!

B  #C  "$ D œ % $B  'C  D œ "# 'B  "#C  #D œ $

Ð"Ñ Ð#Ñ Ð$Ñ

Ð%Ñ

Substitute ! for B and ! for C in equation Ð"Ñ to find D .

Ð%Ñ Ð&Ñ Ð$Ñ

' ‚ Ð"Ñ # ‚ Ð#Ñ

We can now easily see that Ð%Ñ and Ð&Ñ are dependent equations (they have the same graph— in fact, their graph is the same plane). Equation Ð$Ñ has the same coefficients, but a different constant term, so its graph is a plane parallel to the other plane—that is, there are no points in common. Thus, the system is inconsistent and the solution set is g.

The solution set is e Bß Cß D l #B  C  D œ 'f. B  C  #D œ ! $B  C  D œ ! %B  #C  D œ !

The solution set is e !ß !ß ! f.

'B  "#C  #D œ #% 'B  "#C  #D œ #% 'B  "#C  #D œ $

Multiplying equation Ð"Ñ by # gives equation Ð#Ñ. Multiplying equation Ð$Ñ by % also gives equation Ð#Ñ. The resulting equations are the same, so the three equations are dependent. 39.

Ð"Ñ

The coefficients of D are "$ , ", and #. We'll multiply the equations by values that will eliminate fractions and make the coefficients of D easy to compare.

The solution set is e $ß !ß # f. #B  C  D œ ' %B  #C  #D œ "# B  "# C  "# D œ $

œ! œ! œ! œ!

43.

B #B B B

 C  D  A  C  D  A  #C  $D  A  C  D  #A

œ & œ $ œ ") œ )

Ð"Ñ Ð#Ñ Ð$Ñ Ð%Ñ

Eliminate A. Add equations Ð"Ñ and Ð#Ñ. B  C  D  A œ & #B  C  D  A œ $ $B  #C œ )

Ð"Ñ Ð#Ñ Ð&Ñ

Eliminate A again. Add equations Ð"Ñ and Ð$Ñ. B  C  D  A œ & B  #C  $D  A œ ") #B  C  %D œ #$

Ð"Ñ Ð$Ñ Ð'Ñ

Eliminate A again. Multiply equation Ð#Ñ by #. Add the result to equation Ð%Ñ. %B  #C  #D  #A œ ' # ‚ Ð#Ñ B  C  D  #A œ ) Ð%Ñ &B  $C  $D œ # Ð(Ñ Equations Ð&Ñ, Ð'Ñ, and Ð(Ñ do not contain a A-term. Since Ð&Ñ does not have a D -term, we will find another equation without a D -term. Eliminate D . Multiply equation Ð'Ñ by $ and equation Ð(Ñ by %. Then add the results. 'B  $C  "#D œ '* $ ‚ Ð'Ñ #!B  "#C  "#D œ ) % ‚ Ð(Ñ #'B  *C œ '" Ð)Ñ

834

Systems of Linear Equations Eliminate C. Multiply equation Ð&Ñ by * and equation Ð)Ñ by #. Then add the results.

#"C  "#D œ "# $ ‚ Ð'Ñ $'B  %)C  "#D œ )% % ‚ Ð(Ñ $'B  #(C œ (# Ð)Ñ

* ‚ Ð&Ñ # ‚ Ð)Ñ

#(B  ")C œ (# &#B  ")C œ "## #&B œ &! B œ #

Eliminate C using Ð&Ñ and Ð)Ñ. $B  $C œ ' %B  $C œ ) B œ # B œ #

To find C, substitute B œ # into equation Ð&Ñ. $B  #C œ ) $Ð#Ñ  #C œ ) #C œ # Cœ"

Ð&Ñ

œ #$ œ #$ œ #! œ&

Substitute ! for C in Ð'Ñ.

Ð'Ñ

(Ð!Ñ  %D œ % %D œ % Dœ"

BCDAœ& #"&Aœ& A œ $ Aœ$

 C  D  %C  D  $C  &D  %C  &D

 A  A  A  #A

Substitute # for B, ! for C, and " for D in Ð"Ñ. $Ð#Ñ  Ð!Ñ  Ð"Ñ  A œ $ '  "  A œ $ Aœ%

Ð"Ñ

The solution set is e #ß !ß "ß % f.

The solution set is e #ß "ß &ß $ f. $B #B #B &B

Ð)Ñ ƒ *

&Ð#Ñ  &C œ "! &C œ ! Cœ!

To find A, substitute B œ #, C œ ", and D œ & into equation Ð"Ñ.

45.

‚ Ð&Ñ

Substitute # for B in Ð&Ñ.

To find D , substitute B œ # and C œ " into equation Ð'Ñ. #B  C  %D #Ð#Ñ  "  %D %D D

$ &

œ $ œ ( œ $ œ (

Ð"Ñ Ð#Ñ Ð$Ñ Ð%Ñ

47. Let B œ the length of the longest side. Then &' B œ the length of the shortest side, and B  "( œ the length of the medium side. The perimeter is $#$, so B  &' B  B  "( œ $#$. 'B  &B  ' B  "( œ 'Ð$#$Ñ 'B  &B  'B  "!# œ "*$) "(B œ #!%! B œ "#!

Eliminate A. Add equations Ð"Ñ and Ð#Ñ. $B  C  D  A œ $ Ð"Ñ #B  %C  D  A œ ( Ð#Ñ &B  &C œ "! Ð&Ñ

Since B œ "#!, &' B œ "!!, and B  "( œ "!$. The lengths of the sides are "!! inches, "!$ inches, and "#! inches.

Eliminate A again. Add equations Ð#Ñ and Ð$Ñ. #B  %C  D  A œ ( Ð#Ñ #B  $C  &D  A œ $ Ð$Ñ (C  %D œ % Ð'Ñ Eliminate A again. Multiply equation Ð#Ñ by #. Add the result to equation Ð%Ñ. %B  )C  #D  #A œ "% # ‚ Ð#Ñ &B  %C  &D  #A œ ( Ð%Ñ *B  "#C  $D œ #" Ð(Ñ Equations Ð&Ñ, Ð'Ñ, and Ð(Ñ do not contain a A-term. Since Ð&Ñ does not have a D -term, we will find another equation without a D -term.

Multiply by 6.

49.

Let B œ the least number. Then $B œ the greatest number, and $B  % œ the middle number. The sum is "', so B  Ð$BÑ  Ð$B  %Ñ œ "'. &B  % œ "' &B œ #! B œ % Since B œ %, $B œ "#, and $B  % œ ). The three numbers are %, ), and "#.

Eliminate D . Multiply equation Ð'Ñ by $ and equation Ð(Ñ by %. Then add the results. 835

Systems of Linear Equations

3

Applications of Systems of Linear Equations

3 Now Try Exercises

Step 5 The cost of a general admission ticket is $$*Þ** and the cost of a ticket for children under %) inches tall is $$"Þ!!.

N1. Step 2 Let [ œ the width of the field and P œ the length of the field. Step 3 Since the length is "! feet more than twice its width, P œ #[  "!.

Ð"Ñ

The perimeter of a rectangle is given by #[  #P œ T . With perimeter T œ '#! feet, #[  #P œ '#!.

Ð#Ñ

Step 4 Substitute #[  "! for P in equation Ð#Ñ. #[  #Ð#[  "!Ñ œ '#! #[  %[  #! œ '#! '[ œ '!! [ œ "!! Substitute [ œ "!! into equation Ð"Ñ. P œ #[  "! œ #!!  "! œ #"! Step 5 The length of the field is #"! feet and the width is "!! feet. Step 6 Check #"! is "! more than # times "!! and #Ð#"!Ñ  #Ð"!!Ñ œ '#!, as required. N2. Step 2 Let B œ the cost of a general admission ticket, and C œ the cost of a ticket for children under %) inches tall. Step 3 From the given information, #B  $C œ "(#Þ*) B  %C œ "'$Þ**.

œ "(#Þ*) œ $#(Þ*) œ "&&Þ!! œ $"Þ!!

Let C œ $"Þ!! in equation Ð#Ñ.

836

Step 6 Check

Ð"Ñ Ð#Ñ

Ð"Ñ # ‚ Ð#Ñ

#Ð$*Þ**Ñ  $Ð$"Þ!!Ñ œ "(#Þ*) and $*Þ**  %Ð$"Þ!!Ñ œ "'$Þ**.

N3. Step 2 Let B œ the amount of "&% acid solution and C œ the amount of #&% acid solution. Liters of Solution B C $!

Percent (as a decimal) "&% œ !Þ"& #&% œ !Þ#& ")% œ !Þ")

Liters of Pure Acid !Þ"&B !Þ#&C !Þ")Ð$!Ñ œ &Þ%

Step 3 Write the system from the columns. B C œ $! !Þ"&B  !Þ#&C œ &Þ%

Ð"Ñ Ð#Ñ

Step 4 Multiply Ð#Ñ by "!! to clear the equation of decimals, and multiply Ð"Ñ by "&. "&B  "&C "&B  #&C "!C C

œ %&! œ &%! œ *! œ *

"& ‚ Ð"Ñ "!! ‚ Ð#Ñ

Substitute * for C in Ð"Ñ. B  * œ $! B œ $!  * œ #" Step 5 #" L of "&% and * L of #&% should be mixed. Step 6 Check

Step 4 Multiply equation Ð#Ñ by # and add to equation Ð"Ñ. #B  $C #B  )C  &C C

B  %Ð$"Þ!!Ñ œ "'$Þ** B  "#%Þ!! œ "'$Þ** B œ $*Þ**

#"  * œ $! and !Þ"&Ð#"Ñ  !Þ#&Ð*Ñ œ &Þ%.

N4. Step 2 Let B œ Vann's rate and C œ Ivy's rate. Make a table. Use > œ .< . Vann Ivy

Distance

Rate

Time

&!

B

&!

C

%!

%!

Step 3 Since the times are the same,

B C

Systems of Linear Equations &! %! œ B C &!C œ %!B %!B  &!C œ !.

Multiply by xy. Ð"Ñ

Since Vann's rate, B, is # mph faster than Ivy's rate, C, B œ C  #.

Ð#Ñ

Step 4 Substitute C  # for B in equation Ð"Ñ to find C. %!B  &!C œ ! %!ÐC  #Ñ  &!C œ ! %!C  )!  &!C œ ! "!C œ )! Cœ)

Ð"Ñ

The time allocated for piecework is (% hours, so )B  #C  "!D œ (%.

The time allocated for machine quilting is %# hours, so %B  #C  &D œ %#.

Ð#Ñ

The time allocated for finishing is #% hours, so B  C  D œ 1#.

Ð$Ñ

Solve the system of equations Ð"Ñ, Ð#Ñ, and Ð$Ñ. To eliminate D , multiply equation Ð$Ñ by "! and add the result to equation Ð"Ñ.

Step 5 Vann's rate is "! mph, and Ivy's rate is ) mph. Step 6 Check "! mph is # mph faster than ) mph. It would take Vann & hours to travel &! miles at "! mph, which is the same amount of time it would take Ivy to travel %! miles at ) mph. N5. Let B œ the number of bottles of $#Þ!! Gatorade sold, C œ the number of $"Þ&! pretzels sold, and D œ the number of $"Þ!! candy items sold. The number of pretzels was four times the number of candy items, so C œ %D . Ð"Ñ The number of Gatorades was $"! more than the number of pretzels, so B œ $"!  C. Ð#Ñ Sales for these items totaled $"&#!, so

"!B  "!C  "!D œ "#! )B  #C  "!D œ (% #B  )C œ %'

"! ‚ Ð$Ñ Ð"Ñ

Divide the last equation by # to get B  %C œ #$Þ Ð%Ñ To eliminate D again, multiply equation Ð$Ñ by & and add the result to equation Ð#Ñ. & ‚ Ð$Ñ

&B  &C  &D œ '! %B  #C  &D œ %# B  $C œ ")

Ð#Ñ Ð&Ñ

Add equations Ð%Ñ and Ð&Ñ to eliminate C. B  %C œ #$ B  $C œ ") C œ &

Ð%Ñ Ð&Ñ

Substitute & for C in equation Ð%Ñ.

#B  "Þ&C  "D œ "&#!. Ð$Ñ Substitute $"!  C for B in Ð$Ñ. #Ð$"!  CÑ  "Þ&C  "D œ "&#!

Ð"Ñ

#B  #C  #D œ #%, or equivalently,

Since C œ ) and B œ C  #, B œ "!.

Ð%Ñ

Now substitute %D for C in Ð%Ñ. #Ð$"!  %DÑ  "Þ&Ð%DÑ  "D '#!  )D  'D  "D "&D D

N6. Let B œ the number of lone star quilts, C œ the number of bandana quilts, and D œ the number of log cabin quilts.

œ "&#! œ "&#! œ *!! œ '!

From Ð"Ñ, C œ %Ð'!Ñ œ #%!. From Ð#Ñ, B œ $"!  #%! œ &&!. Workers sold '! candy items, #%! pretzels, and &&! bottles of Gatorade.

B  %C œ #$ B  %Ð&Ñ œ #$ B  #! œ #$ Bœ$

Ð%Ñ

Substitute $ for B and & for C in equation Ð$Ñ. BCD $&D )D D

œ 1# œ "# œ "# œ%

Ð$Ñ

Each month, the quilting shop should make $ lone star quilts, & bandana quilts, and % log cabin quilts.

837

Systems of Linear Equations 5.

3 Section Exercises 1.

Step 2 Let B œ the number of games that the Dodgers won and let C œ the number of games that they lost.

The total revenue was $##"Þ% billion. B  C œ ##"Þ%

Step 3 They played "'# games, so Ð"Ñ

C œ B  #'Þ'

They won #) more games than they lost, so Ð#Ñ

B  ÐB  #'Þ'Ñ œ ##"Þ% #B  #'Þ' œ ##"Þ% #B œ "*%Þ) B œ *(Þ%

Ð#)  CÑ  C œ "'# #)  #C œ "'# #C œ "$% C œ '(

Substitute *(Þ% for B in equation Ð#Ñ. C œ B  #'Þ' œ *(Þ%  #'Þ' œ "#%Þ! Verizon's revenue was $*(Þ% billion and AT&T's revenue was $"#%Þ! billion.

Substitute '( for C in Ð#Ñ. B œ #)  '( œ *&

7.

Step 5 The Dodgers win-loss record was *& wins and '( losses.

B  C œ *!.

C œ *!  B.

Ð$Ñ

Substitute *!  B for C in equation Ð"Ñ. Ð$B  "!Ñ  Ð*!  BÑ œ ")! #B  "!! œ ")! #B œ )! B œ %!

Ð"Ñ

The perimeter of a rectangle is given by #[  #P œ T .

Substitute B œ %! into equation Ð$Ñ to get

With perimeter T œ ##) ft,

C œ *!  B œ *!  %! œ &!.

Ð#Ñ

The angles measure %!° and &!°.

Substitute [  %# for P in equation Ð#Ñ. #[  #Ð[  %#Ñ œ ##) #[  #[  )% œ ##) %[ œ "%% [ œ $' Substitute [ œ $' into equation Ð"Ñ. Ð"Ñ

The length is () ft and the width is $' ft.

838

Ð#Ñ

Solve equation Ð#Ñ for C to get

Since the length is %# ft more than the width,

P œ [  %# P œ $'  %# œ ()

Ð"Ñ

Also, the angles B and C are complementary, so

Let [ œ the width of the tennis court and P œ the length of the court.

#[  #P œ ##).

From the figure in the text, the angles marked C and $B  "! are supplementary, so Ð$B  "!Ñ  C œ ")!.

Step 6 *& is #) more than '( and the sum of *& and '( is "'#.

P œ [  %#.

Ð#Ñ

Substitute B  #'Þ' for C in equation Ð"Ñ.

Step 4 Substitute #)  C for B in Ð"Ñ.

3.

Ð"Ñ

AT&T's revenue was $#'Þ' billion more than that of Verizon.

B  C œ "'#.

B œ #)  C.

Let B œ the revenue for Verizon and C œ the revenue for AT&T (both in billions of dollars).

9.

Let B œ the hockey FCI and C œ the basketball FCI. The sum is $&)!Þ"', so B  C œ &)!Þ"'.

Ð"Ñ

The hockey FCI was $$Þ(! less than the basketball FCI, so B œ C  $Þ(!.

Ð#Ñ

From Ð#Ñ, substitute C  $Þ(! for B in Ð"Ñ.

Systems of Linear Equations Gallons of Solution B C #!

ÐC  $Þ(!Ñ  C œ &)!Þ"' #C  $Þ(! œ &)!Þ"' #C œ &)$Þ)' C œ #*"Þ*$ From Ð#Ñ, The hockey FCI was $#))Þ#$ and the basketball FCI was $#*"Þ*$.

B C œ #! !Þ#&B  !Þ$&C œ 'Þ%

Let B œ the cost of a single Junior Roast Beef sandwich, and C œ the cost of a single Big Montana sandwich.

#&B  #&C #&B  $&C "!C C

Ð"Ñ

$! Junior Roast Beef sandwiches and & Big Montana sandwiches cost $)%Þ'&. $!B  &C œ )%Þ'&

Ð#Ñ

B  C œ #! B  "% œ #! Bœ'

"&B  "!C œ (&Þ#& "&B  "!Ð%Þ$*Ñ œ (&Þ#& "&B  %$Þ*! œ (&Þ#& "&B œ $"Þ$& B œ #Þ!* A single Junior Roast Beef sandwich costs $#Þ!* and a single Big Montana sandwich costs $4Þ$*. 13.

Use the formula (rate of percent) • (base amount) œ amount (percentage) of pure acid to compute parts (a) – (d). (a) !Þ"!Ð'!Ñ œ ' oz (b) !Þ#&Ð'!Ñ œ "& oz (c) !Þ%!Ð'!Ñ œ #% oz (d) !Þ&!Ð'!Ñ œ $! oz

15.

The cost is the price per pound, $#Þ#*, times the number of pounds, B, or $#Þ#*B.

17.

Let B œ the amount of #&% alcohol solution, and C œ the amount of $&% alcohol solution. Make a table. The percent times the amount of solution gives the amount of pure alcohol in the third column.

Ð"Ñ

Mix ' gal of #&% solution and "% gal of $&% solution.

# ‚ Ð"Ñ Ð#Ñ

Substitute %Þ$* for C in equation Ð"Ñ.

#& ‚ Ð"Ñ "!! ‚ Ð#Ñ

œ &!! œ '%! œ "%! œ "%

Substitute C œ "% into equation Ð"Ñ.

Multiply equation Ð"Ñ by # and add to equation Ð#Ñ. $!B  #!C œ "&!Þ&! $!B  &C œ )%Þ'&  "&C œ '&Þ)& C œ %Þ$*

Ð"Ñ Ð#Ñ

Solve the system. Multiply equation Ð"Ñ by #& and equation Ð#Ñ by "!!. Then add the results.

"& Junior Roast Beef sandwiches and "! Big Montana sandwiches cost $(&Þ#&Þ "&B  "!C œ (&Þ#&

Gallons of Pure Alcohol !Þ#&B !Þ$&C !Þ$#Ð#!Ñ œ 'Þ%

The third row gives the total amounts of solution and pure alcohol. From the columns in the table, write a system of equations.

B œ C  $Þ(! œ #*"Þ*$  $Þ(! œ #))Þ#$.

11.

Percent (as a decimal) #&% œ !Þ#& $&% œ !Þ$& $#% œ !Þ$#

19.

Let B œ the amount of pure acid and C œ the amount of "!% acid. Make a table. Liters of Solution B C &%

Percent (as a decimal) "!!% œ " "!% œ !Þ"! #!% œ !Þ#!

Liters of Pure Acid "Þ!!B œ B !Þ"!C !Þ#!Ð&%Ñ œ "!Þ)

Solve the following system. B C œ &% B  !Þ"!C œ "!Þ)

Ð"Ñ Ð#Ñ

Multiply equation Ð#Ñ by "! to clear the decimals. "!B  C œ "!)

Ð$Ñ

To eliminate C, multiply equation Ð"Ñ by " and add the result to equation Ð$Ñ. B  C œ &% "!B  C œ "!) *B œ &% B œ '

" ‚ Ð"Ñ Ð$Ñ

Since B œ ', B  C œ &% '  C œ &% C œ %).

Ð"Ñ

Use ' L of pure acid and %) L of "!% acid. 839

Systems of Linear Equations 21.

The times are equal, so "&! %!! œ . B C

Complete the table.

Nuts Cereal Mixture

Number of Kilograms B C $!

Price per Kilogram #Þ&! "Þ!! "Þ(!

Value

The rate of the plane is #! km per hour less than $ times the rate of the train, so

#Þ&!B "Þ!!C "Þ(!Ð$!Ñ œ &"

$B  #! œ CÞ Ð#Ñ

From the "Number of Kilograms" column,

Multiply Ð"Ñ by BC (or use cross products). "&!C œ %!!B

B  C œ $!Þ Ð"Ñ

"&!Ð$B  #!Ñ œ %!!B %&!B  $!!! œ %!!B &!B œ $!!! B œ '!

#Þ&!B  "Þ!!C œ &"Þ Ð#Ñ Solve the system. "! ‚ Ð"Ñ "! ‚ Ð#Ñ

From Ð"Ñ, "%  C œ $!, so C œ "'. The party mix should be made from "% kg of nuts and "' kg of cereal. 23.

From the "Principal" column in the text, B  C œ $!!!. Ð"Ñ From the "Interest" column in the text, !Þ!#B  !Þ!%C œ "!!. Ð#Ñ Multiply equation Ð#Ñ by "!! to clear the decimals. #B  %C œ "!,!!!

Ð$Ñ

To eliminate B, multiply equation Ð"Ñ by # and add the result to equation Ð$Ñ. #B  #C #B  %C #C C

œ '!!! # ‚ Ð"Ñ œ "!,!!! Ð$Ñ œ %!!! œ #!!!

From Ð"Ñ, B  #!!! œ $!!!, so B œ "!!!. $"!!! is invested at #%, and $#!!! is invested at %%. 25.

(a) The rate of the boat going upstream is decreased by the rate of the current, so it is Ð"!  BÑ mph. (b) The rate of the boat going downstream is increased by the rate of the current, so it is Ð"!  BÑ mph.

27.

840

Ð$Ñ

From Ð#Ñ, substitute $B  #! for C in Ð$Ñ.

From the "Value" column,

"!B  "!C œ $!! #&B  "!C œ &"! "&B œ #"! B œ "%

Ð"Ñ

Let B œ the rate of the train and C œ the rate of the plane.


.

Train

B

"&!

Plane

C

"&! B %!! C

%!!

From Ð#Ñ, C œ $Ð'!Ñ  #! œ "'!. The rate of the train is '! km/hr, and the rate of the plane is "'! km/hr. 29.

Let B œ the rate of the boat in still water and C œ the rate of the current. Furthermore, rate upstream œ B  C rate downstream œ B  C.

and

Use these rates and the information in the problem to make a table. Upstream Downstream

< BC BC

> # "Þ&

. $' $'

From the table, use the formula . œ to write a system of equations. $' œ #ÐB  CÑ $' œ "Þ&ÐB  CÑ Remove the parentheses and move the variables to the left side. #B  #C œ $' "Þ&B  "Þ&C œ $'

Ð"Ñ Ð#Ñ

Solve the system. Multiply equation Ð"Ñ by $ and equation Ð#Ñ by %. Then add the results. 'B  'B 

'C 'C "#C C

œ "!) œ "%% œ $' œ $

$ ‚ Ð"Ñ % ‚ Ð#Ñ

Substitute C œ $ into equation Ð"Ñ. #B  #C œ $' #B  #Ð$Ñ œ $' #B œ %# B œ #"

Ð"Ñ

The rate of the boat is #" mph, and the rate of the current is $ mph.

Systems of Linear Equations 31.

Let B œ the number of pounds of the $!Þ(&-per-lb candy and C œ the number of pounds of the $"Þ#&per-lb candy.

35.

"* citrons and ( fragrant wood apples is "!(" gives us

Make a table. Price Number per Pound of Pounds Less Expensive Candy More Expensive Candy Mixture

$!Þ(&

*B  (C œ "!(Þ Ð"Ñ

Value

"( citrons and * fragrant wood apples is "!"" gives us

$!Þ(&B

B

Let B œ the price for a citron and let C œ the price for a wood apple.

(B  *C œ "!"Þ Ð#Ñ $"Þ#&

C

$"Þ#&C

$!Þ*'

*

$!Þ*'Ð*Ñ œ $)Þ'%

Multiply equation Ð"Ñ by ( and equation Ð#Ñ by *. Then add. '$B  %*C '$B  )"C $#C C

From the "Number of Pounds" column, B  C œ *.

Ð"Ñ

From the "Value" column,

Substitute & for C in equation Ð"Ñ.

!Þ(&B  "Þ#&C œ )Þ'%. Ð#Ñ

*B  (Ð&Ñ œ "!( *B  $& œ "!( *B œ (# Bœ)

Solve the system. (&B  (&C (&B  "#&C &!C C

(& ‚ Ð"Ñ "!! ‚ Ð#Ñ

œ '(& œ )'% œ ")* œ $Þ()

From Ð"Ñ, B  $Þ() œ *, so B œ &Þ##. Mix &Þ## pounds of the $!Þ(&-per-lb candy with $Þ() pounds of the $"Þ#&-per-lb candy to obtain * pounds of a mixture that sells for $!Þ*' per pound. 33.

Let B œ the number of general admission tickets and C œ the number of student tickets. Make a table. Ticket General Student Totals

Number B C ")%

Value of Tickets & • B œ &B % • C œ %C )"#

Solve the system. B  C œ ")% &B  %C œ )"#

Ð"Ñ Ð#Ñ

To eliminate C, multiply equation Ð"Ñ by % and add the result to equation Ð#Ñ. %B  %C œ ($' &B  %C œ )"# B œ ('

( ‚ Ð"Ñ * ‚ Ð#Ñ

œ (%* œ *!* œ "'! œ &

% ‚ Ð"Ñ Ð#Ñ

From Ð"Ñ, ('  C œ ")%, so C œ "!). (' general admission tickets and "!) student tickets were sold.

The prices are ) for a citron and & for a wood apple. 37.

Let B œ the measure of one angle, C œ the measure of another angle, and D œ the measure of the last angle. Two equations are given, so or and

D œ B  "! B  D œ "! B  C œ "!!.

Ð"Ñ Ð#Ñ

Since the sum of the measures of the angles of a triangle is ")!°, the third equation of the system is B  C  D œ ")!. Ð$Ñ Equation Ð"Ñ is missing C. To eliminate C again, multiply equation Ð#Ñ by " and add the result to equation Ð$Ñ. B  C œ "!! B  C  D œ ")! D œ )!

" ‚ Ð#Ñ Ð$Ñ

Since D œ )!, B  D œ "! B  )! œ "! B œ (! B œ (!.

Ð"Ñ

From Ð#Ñ, (!  C œ "!!, so C œ $!. The measures of the angles are (!°, $!°, and )!°.

841

Systems of Linear Equations 39.

Let B œ the measure of the first angle, C œ the measure of the second angle, and D œ the measure of the third angle.

B  C  D œ B  C  D œ #B œ B œ

The sum of the angles in a triangle equals ")!°, so B  C  D œ ")!.

Substitute $$ for B in Ð$Ñ.

Ð"Ñ

B  #D $$  #D #D D

The measure of the second angle is "!° greater than $ times that of the first angle, so C œ $B  "!. Ð#Ñ The third angle is equal to the sum of the other two, so

43.

Substitute *! for D and $B  "! for C in equation Ð$Ñ.

B œ D  &Þ Ð"Ñ

Ð$Ñ

The number of silver medals earned was $& less than twice the number of bronze medals, so C œ #D  $&Þ Ð#Ñ The total number of medals earned was (#, so

Ð$Ñ

B  C  D œ (#Þ Ð$Ñ Substitute D  & for B and #D  $& for C in Ð$Ñ. ÐD  &Ñ  Ð#D  $&Ñ  D œ (# %D  %! œ (# %D œ ""# D œ #)

The three angles have measures of #!°, (!°, and *!°. 41.

Let B œ the length of the longest side, C œ the length of the middle side, and D œ the length of the shortest side.

From Ð"Ñ, B œ #)  & œ #$Þ From Ð#Ñ, C œ #Ð#)Ñ  $& œ &'  $& œ #".

Perimeter is the sum of the measures of the sides, so B  C  D œ (!. Ð"Ñ The longest side is % cm less than the sum of the other sides, so BœCD% B  C  D œ %.

or

Ð#Ñ

Twice the shortest side is * cm less than the longest side, so or

#D œ B  * B  #D œ *

Ð$Ñ

Add equations Ð"Ñ and Ð#Ñ to eliminate C and D . 842

Let B œ the number of gold medals, C œ the number of silver medals, and D œ the number of bronze medals. Russia earned & fewer gold medals than bronze, so

Substitute B œ #! and D œ *! into equation Ð$Ñ. D œBC *! œ #!  C (! œ C

Ð"Ñ

The shortest side is "# cm long, the middle side is #& cm long, and the longest side is $$ cm long.

Ð"Ñ

D œBC *! œ B  Ð$B  "!Ñ )! œ %B #! œ B

Ð$Ñ

B  C  D œ (! $$  C  "# œ (! C  %& œ (! C œ #&

Solve the system. Substitute D for B  C in equation Ð"Ñ. œ ")! œ ")! œ ")! œ *!

œ * œ * œ #% œ "#

Substitute $$ for B and "# for D in Ð"Ñ.

D œ B  C. Ð$Ñ

ÐB  CÑ  D DD #D D

Ð"Ñ Ð#Ñ

(! % '' $$

Russia earned #$ gold, #" silver, and #) bronze medals. 45.

Let B œ the number of $"' tickets, C œ the number of $#$ tickets, and D œ the number of VIP $%! tickets. Nine times as many $"' tickets have been sold as VIP tickets, so B œ *DÞ Ð"Ñ The number of $"' tickets was && more than the sum of the number of $#$ tickets and the number of VIP tickets, so B œ &&  C  DÞ Ð#Ñ

Systems of Linear Equations Sales of these tickets totaled $%',&(&. "'B  #$C  %!D œ %',&(&

49.

Ð$Ñ

Since B is in terms of D in Ð"Ñ, we'll substitute *D for B in Ð#Ñ and then get C in terms of D . *D œ &&  C  D )D  && œ C

Let B, C, and D denote the number of kilograms of the first, second, and third chemicals, respectively. The mix must include '!% of the first and second chemicals, so B  C œ !Þ'!Ð(&!Ñ œ %&!Þ Ð"Ñ The second and third chemicals must be in a ratio of % to $ by weight, so

Ð%Ñ

C œ %$ DÞ Ð#Ñ

Substitute *D for B and )D  && for C in Ð$Ñ. "'Ð*DÑ  #$Ð)D  &&Ñ  %!D "%%D  ")%D  "#'&  %!D $')D D

The total is (&!, so

œ %',&(& œ %',&(& œ %(,)%! œ "$!

B  C  D œ (&!Þ Ð$Ñ From Ð"Ñ, B œ %&!  CÞ Ð%Ñ From Ð#Ñ, D œ $% CÞ Ð&Ñ

From Ð"Ñ, B œ *Ð"$!Ñ œ ""(!Þ From Ð%Ñ, C œ )Ð"$!Ñ  && œ *)&Þ

Substitute %&!  C for B and $% C for D in Ð$Ñ.

There were ""(! $"' tickets, *)& $#$ tickets, and "$! $%! tickets sold. 47.

Ð%&!  CÑ  C  $% C œ (&! $ %C

œ $!! C œ %!!

Let B œ the number of T-shirts shipped to bookstore A, C œ the number of T-shirts shipped to bookstore B, and D œ the number of T-shirts shipped to bookstore C. Twice as many T-shirts were shipped to bookstore B as to bookstore A, so C œ #B. Ð"Ñ The number shipped to bookstore C was %! less than the sum of the numbers shipped to the other two bookstores, so

From Ð%Ñ, B œ %&!  %!! œ &!Þ From Ð&Ñ, D œ $% Ð%!!Ñ œ $!!Þ Use &! kg of the first chemical, %!! kg of the second chemical, and $!! kg of the third chemical to make the plant food. 51.

Step 2 Let B œ the number of wins, C œ the number of losses in regulation play, and D œ the number of overtime losses. Step 3 They played )# games, so

D œ B  C  %!. Ð#Ñ The total number of T-shirts shipped was )!!, so B  C  D œ )!!. Ð$Ñ Substitute #B for C [from Ð"Ñ] into equation Ð#Ñ to get D in terms of B. D œ B  C  %! D œ B  Ð#BÑ  %! D œ $B  %!

B  C  D œ )#.

The points from their wins and overtime losses totaled ""', so #B  D œ ""'.

Substitute #B for C and $B  %! for D in Ð$Ñ. B  Ð#BÑ  Ð$B  %!Ñ œ )!! 'B  %! œ )!! 'B œ )%! B œ "%! From Ð"Ñ, C œ #Ð"%!Ñ œ #)!. From Ð%Ñ, D œ $Ð"%!Ñ  %! œ $)!. The number of T-shirts shipped to bookstores A, B, and C was "%!, #)!, and $)!, respectively.

Ð#Ñ

They had * more losses in regulation play than overtime losses, so

Ð#Ñ Ð%Ñ

Ð"Ñ

C œ D  *.

Ð$Ñ

Step 4 Solving equation Ð#Ñ for B gives us B œ &)  "# D . Now substitute for B and C in Ð"Ñ. Ð&) 

" # DÑ

B  C  D œ )#  ÐD  *Ñ  D œ )# "Þ&D  '( œ )# "Þ&D œ "& D œ "!

Ð"Ñ

Since C œ D  *, C œ "!  * œ "*Þ Since B œ &)  "# D , B œ &)  "# Ð"!Ñ œ &$Þ

843

Systems of Linear Equations Step 5 The Bruins won &$ games, lost "* games, and had "! overtime losses. Step 6 Adding &$, "*, and "! gives )# total games. The points from their wins and overtime losses totaled #Ð&$Ñ  "! œ "!'  "! œ ""', and there were * more losses in regulation play than overtime losses. The solution is correct. 53.

(a) The additive inverse of ' is Ð'Ñ œ '. (b) The multiplicative inverse (reciprocal) of ' " is ' œ  '" .

55.

(a) The additive inverse of

( )

is  () .

(b) The multiplicative inverse (reciprocal) of " ) (Î) œ ( .

4

( )

is

Solving Systems of Linear Equations by Matrix Methods

" # & × Ô " # " " " Õ # " $ "% Ø " # & × Ô" ! $ " ' Õ # " $ "% Ø Ô" ! Õ!

" $ $

Ô" ! Õ! Ô" ! Õ! Ô" ! Õ!

" ”! " ”!

$ "

$ "# • $ ") •

" ' R$

B  C  #D œ & C  "$ D œ # D œ $.

C  "$ Ð$Ñ œ # C œ "

$ #•

B  Ð"Ñ  #Ð$Ñ œ & B  ( œ & Bœ#

#R"  R#  "* R#

B  $C œ $ C œ #. Substitute # for C in the first equation. B  $Ð#Ñ œ $ B'œ$ B œ $ The solution set is eÐ$ß #Ñf.

844

$R#  R$

Substitute " for C and $ for D in the first equation.

This matrix gives the system

N2.

" $ R#

This matrix gives the system

Write the augmented matrix for this system. $ $ $ *

#R"  R$

Substitute $ for D in the second equation.

B  $C œ $ #B  $C œ "#

" ”#

& × ' #% Ø

" # & × "  "$ # ( #% Ø $ " # & × "  "$ # ! ' ") Ø " # & × "  "$ # ! " $Ø

4 Now Try Exercises N1.

# " (

R"  R#

The solution set is eÐ#ß "ß $Ñf. N3. (a)

$B  C œ ) 'B  #C œ %

Write the augmented matrix for this system. $ ” ' $ ”!

" ) # %• " ) ! #! •

#R"  R#

This matrix gives the system $B  C œ ) ! œ #!.

False

B  C  #D œ & B  #C  D œ " #B  C  $D œ "%

The false statement indicates that the system is inconsistent and has no solution.

Write the augmented matrix for this system.

The solution set is g.

Systems of Linear Equations (b)

7.

B  #C œ ( B  #C œ (

BC œ& BC œ$

Write the augmented matrix for this system.

Write the augmented matrix for this system.

" ” "

" ”"

" ”!

# ( # ( • # ( ! !•

" ”!

R"  R#

" ”!

This matrix gives the system B  #C œ ( ! œ !.

" & " $ • " & # # • " & " "•

BC œ& C œ ".

The solution set is eÐBß CÑ l B  #C œ (f.

Substitute C œ " in the first equation. BC œ& B"œ& Bœ%

4 Section Exercises Ô # ! Õ "

$ & %

"× $ )Ø

The solution set is e %ß " f. 9.

(a) The elements of the second row are !, &, and $. (b) The elements of the third column are ", $, and ). (c) The matrix is square since the number of rows (three) is the same as the number of columns. (d) The matrix obtained by interchanging the first and third rows is Ô " ! Õ #

% & $

)× $ Þ "Ø

(e) The matrix obtained by multiplying the first row by  "# is

Ô #Ð # Ñ ! Õ " "

$Ð "# Ñ & %

 "# R#

This matrix gives the system

True

The true statement indicates that the system has dependent equations.

1.

"R"  R#

"Ð "# Ñ × Ô "  $#  "# × œ ! $ & $ . ) Ø Õ" % )Ø

(f) The matrix obtained by multiplying the third row by $ and adding to the first row is Ô #  $Ð"Ñ $  $Ð%Ñ "  $Ð)Ñ × Ô " "& #& × ! & $ œ ! & $ . Õ Ø Õ" % )Ø " % ) 3.

There are $ rows and # columns, so the dimensions are written as $ ‚ #.

5.

There are % rows and # columns, so the dimensions are written as % ‚ #.

#B  %C œ ' $B  C œ # Write the augmented matrix. # ”$

% "

' #•

The easiest way to get a " in the first row, first column position is to multiply the elements in the first row by "# . " ”$

# "

$ #•

" # R"

To get a ! in row two, column ", we need to subtract $ from the $ that is in that position. To do this we will multiply row " by $ and add the result to row #. " ”! " ”!

# ( # "

$ ( •

$ "•

$R"  R#  "( R#

This matrix gives the system B  #C œ $ C œ ". Substitute C œ " in the first equation. B  #C œ $ B  #Ð"Ñ œ $ Bœ"

The solution set is e "ß " f.

845

Systems of Linear Equations 11.

17.

$B  %C œ "$ #B  $C œ "%

$B  #C œ ! B C œ!

Write the augmented matrix.

Write the augmented matrix.

$ ”#

$ ” " " ” $

" ”# " ”! " ”!

"$ "% • #( "% • ( #( "( ') • % $ ( $

( "

#( %•

"R#  R"

! !• ! !•

# " " #

#R"  R#

We use Ç to represent the interchanging of two rows.

"  "( R#

" ”! " ”!

This matrix gives the system B  (C œ #( C œ %.

" " " "

! !• ! !•

%B  "#C œ $' B  $C œ *

" % R"

R"  R#

B  $C œ * ! œ ")

False

which is inconsistent and has no solution. The solution set is g. #B  C œ % %B  #C œ ) Write the augmented matrix. # ”% ”

" % " ” !

" # " #

# " #

!

% )•

# • ) # • !

The solution set is e !ß ! f. B  C  $D œ " #B  C  D œ * $B  C  %D œ )

Write the augmented matrix.

The corresponding system is

Ô" # Õ$ Ô" ! Õ! Ô" ! Õ! Ô" Ö! Õ!

Ô" ! Õ!

" $ " × " " * " % ) Ø " $ " × $ ( ( # & &Ø " $ "× "  ($  ($ # & &Ø " $ "× ( "  $  ($ Ù " " Ø ! $ $ " " !

$  ($ "

"×  ($ "Ø

#R"  R# $R"  R$  "$ R#

#R#  R$

$R$

This matrix gives the system " # R"

%R"  R#

Row #, ! œ !, indicates that the system has dependent equations. The solution set is eÐBß CÑ l #B  C œ %f.

846

B!œ! Bœ!

19.

Write the augmented matrix.

15.

Substitute C œ 0 in the first equation.

The solution set is e "ß % f.

"# $' $ *• $ * $ * • $ * ! ") •

R# BC œ! C œ !.

B  (C œ #( B  (Ð%Ñ œ #( B  #) œ #( B œ "

% ” " " ” " " ” !

$R"  R#

This matrix gives the system

Substitute C œ % in the first equation.

13.

R" Ç R#

B  C  $D œ " C  ($ D œ  ($ D œ ". Substitute D œ " in the second equation. C  ($ D œ  ($ C  ($ Ð"Ñ œ  ($ Cœ!

Systems of Linear Equations Ô" ! Õ!

Substitute C œ ! and D œ " in the first equation. B  C  $D œ " B  !  $Ð"Ñ œ " B$œ" Bœ%

21.

" " !

B  C œ " C  D œ ' D œ %.

B  C  D œ ' #B  C  D œ * B  #C  $D œ "

Substitute D œ % in the second equation. CD œ' C  Ð%Ñ œ ' Cœ#

Write the augmented matrix.

Ô" ! Õ! Ô" ! Õ!

" " # " $ $

" " $ " $ %

" " $

" " %

" " !

" " "

'× * "Ø '× #" & Ø '× ( & Ø

'× ( "' Ø

Substitute C œ # in the first equation. BC œ" B#œ" Bœ$

#R"  R# "R"  R$

The solution set is e $ß #ß % f.

 "$ R# 25. $R#  R$

Ô " $ Õ #

BCD œ' CD œ( D œ "'.

Ô " " Õ "

Substitute D œ "' in the second equation. C  "' œ ( C œ #$

Ô" ! Õ!

Substitute C œ #$ and D œ "' in the first equation. B  #$  "' œ ' B(œ' B œ "

The solution set is e "ß #$ß "' f. B  C œ " C  D œ ' B  D œ " Write the augmented matrix. Ô" ! Õ" Ô" ! Õ! Ô" ! Õ!

" " ! " " " " " !

! " " ! " " ! " #

"× ' " Ø "× ' # Ø "× ' ) Ø

"R"  R$

"R#  R$

B  #C  D œ % $B  'C  $D œ "# #B  %C  #D œ ) Write the augmented matrix.

This matrix gives the system

23.

" # R$

This matrix gives the system

The solution set is e %ß !ß " f.

Ô" # Õ" Ô" ! Õ!

"× ' % Ø

! " "

# ' %

" $ #

# # #

" " "

# ! !

" ! !

%× ! !Ø

%× "# ) Ø %× % % Ø

" $ R# " # R$

"R"  R# R"  R$

This augmented matrix represents a system of dependent equations. The solution set is e Bß Cß D l B  #C  D œ %f. 27.

B  #C  $D œ # #B  %C  'D œ & B  C  #D œ ' Write the augmented matrix. Ô" # Õ" Ô" ! Õ!

# % " # ! $

# × & 'Ø $ # × ! " " )Ø $ ' #

#R"  R# R"  R$

From the second row, ! œ ", we see that the system is inconsistent. The solution set is g. 847

Systems of Linear Equations 29.

33.

%B  C œ & #B  C œ $ Enter the augmented matrix as [A]. % ”#

" "

& $•

B  D œ $ C  D œ $ B  C œ ) Enter the augmented matrix as [E].

Ô" ! Õ"

The TI-83/4 screen for A follows. (Use MATRX EDIT.)

Now use the reduced row echelon form (rref) command to simplify the system. (Use MATRX MATH ALPHA B for rref and MATRX 1 for [A].)

! " "

" " !

$ × $ )Ø

The solution set is e "ß (ß % f. 35.

#' œ # • # • # • # • # • # œ '%

37.

Ð&Ñ% œ Ð&ÑÐ&ÑÐ&ÑÐ&Ñ œ '#&

39.

Ð $% Ñ% œ œ

$ $ $ $ % • % • % • %

$•$•$•$ )" œ %•%•%•% #&'

Review Exercises 1.

This matrix gives the system "B  !C œ " !B  "C œ ",

31.

or simply, B œ " and C œ ". The solution set is e "ß " f.

&B  C  $D œ ' #B  $C  D œ & $B  #C  %D œ $

(a) The graphs meet between the years of 1980 and 1985. Therefore, the number of degrees for men equal the number of degrees for women between 1980 and 1985. (b) The number was just less than &!!,!!!Þ

2.

B  $C œ ) #B  C œ #

Ð"Ñ Ð#Ñ

Graph the two lines. They appear to intersect at the point Ð#ß #Ñ. Check Ð#ß #Ñ in both equations.

Enter the augmented matrix as [C].

Ô & # Õ $

" $ #

$ " %

' × & $Ø

B  $C œ )

Ð"Ñ

?

#  $Ð#Ñ œ ) )œ) #B  C œ #

True Ð#Ñ

?

The solution set is e "ß #ß " f. 848

#Ð#Ñ  # œ # #œ#

True

The solution set is e #ß # f.

Systems of Linear Equations 3.

Checking the ordered pairs in choices A, B, and C yields true statements.

6.

For choice D, we have: ? $Ð$Ñ  #Ð#Ñ œ '

*B  C œ % *B  ÐB  %Ñ œ % *B  B  % œ % )B œ ! Bœ!

?

False

So D is the answer. 4.

Ð"Ñ Ð#Ñ

Substitute B  % for C in equation Ð"Ñ and solve for B.

$B  #C œ ' *%œ' &œ'

*B  C œ % C œB%

(a) The graphs of these two linear equations will intersect once. 7.

Ð"Ñ

Since B œ !, C œ B  % œ !  % œ %. The solution set is e !ß % f. &B  #C œ # B  'C œ #'

Ð"Ñ Ð#Ñ

Solve equation Ð#Ñ for B. B œ #'  'C Substitute #'  'C for B in equation Ð"Ñ. &B  #C œ # &Ð#'  'CÑ  #C œ # "$!  $!C  #C œ # "$!  $#C œ # $#C œ "#) Cœ%

(b) The graphs of these two linear equations will not intersect. They are parallel lines.

Ð"Ñ

Since B œ #'  'C and C œ %, B œ #'  'Ð%Ñ œ #'  #% œ #.

(c) The graphs of these two linear equations will be the same line.

8.

The solution set is e #ß % f.

&B  C œ "# Ð"Ñ #B  #C œ ! Ð#Ñ To eliminate C, multiply equation Ð"Ñ by # and add the result to equation Ð#Ñ. "!B  #C œ #% #B  #C œ ! "#B œ #% B œ #

5.

$B  C œ % B œ #$ C Substitute

# $C

Ð"Ñ Ð#Ñ

Ð#Ñ

To find C, substitute # for B in equation Ð#Ñ. #B  #C œ ! Ð#Ñ #Ð#Ñ  #C œ ! % œ #C #œC

for B in equation Ð"Ñ and solve for C .

$B  C œ % $Ð #$ CÑ  C œ % #C  C œ % $C œ % C œ  %$

# ‚ Ð"Ñ

Ð"Ñ

Since B œ #$ C and C œ  %$ , B œ #$ Ð %$ Ñ œ  )* .

The solution set is ˜Ð )* ß  %$ Ñ™.

9.

The ordered pair Ð#ß #Ñ satisfies both equations, so it checks. The solution set is e #ß # f. B  %C œ % $B  C œ "

Ð"Ñ Ð#Ñ

To eliminate C, multiply equation Ð#Ñ by % and add the result to equation Ð"Ñ.

849

Systems of Linear Equations B  %C œ % "#B  %C œ % "$B œ ! B œ !

13.

Ð"Ñ % ‚ Ð#Ñ

B  %C œ % !  %C œ % Cœ"

"!B  )C œ % "!B  )C œ ( ! œ ""

Ð"Ñ

The solution set is e !ß " f. 'B  &C œ % %B  #C œ )

Ð"Ñ Ð#Ñ

14.

"#B  "!C œ ) "#B  'C œ #% "'C œ $# C œ #

15. Ð#Ñ

 "' C œ  "# B  C œ *

To eliminate D again, multiply equation Ð"Ñ by # and add the result to equation Ð#Ñ.

Multiply equation Ð"Ñ by ' to clear the fractions. Add the result to equation Ð#Ñ to eliminate C. ' ‚ Ð"Ñ

12.

Ð#Ñ

The solution set is e 'ß $ f. $B  C œ ' C œ '  $B

Ð"Ñ Ð#Ñ

Since equation Ð#Ñ can be rewritten as $B  C œ ', the two equations are the same, and hence, dependent. The solution set is e Bß C l $B  C œ 'f.

850

%B  'C  #D œ $# B  #C  #D œ $ &B  )C œ $&

# ‚ Ð"Ñ Ð#Ñ Ð&Ñ

Use equations Ð%Ñ and Ð&Ñ to eliminate B. Multiply equation Ð%Ñ by & and add the result to equation Ð&Ñ.

Ð#Ñ

Since B œ ', B  C œ * '  C œ * C œ $ C œ $.

Ð"Ñ Ð$Ñ Ð%Ñ

#B  $C  D œ "' $B  C  D œ & B  %C œ #"

Ð"Ñ Ð#Ñ

B  C œ $ B  C œ * #B œ "# B œ '

Ð"Ñ Ð#Ñ Ð$Ñ

#B  $C  D œ "' B  #C  #D œ $ $B  C  D œ &

To eliminate D , add equations Ð"Ñ and Ð$Ñ.

The solution set is e "ß # f. " 'B

C œ $B  # C œ $B  % Since both equations are in slope-intercept form, their slopes and C-intercepts can be easily determined. The two lines have the same slope, $, but the C-intercepts, Ð!ß #Ñ and Ð!ß %Ñ, are different. Therefore, the lines are parallel, do not intersect, and have no common solution.

# ‚ Ð"Ñ $ ‚ Ð#Ñ

Since C œ #, %B  #C œ ) %B  #Ð#Ñ œ ) %B  % œ ) %B œ % B œ ".

# ‚ Ð"Ñ Ð#Ñ False

Since a false statement results, the system is inconsistent. The solution set is g.

To eliminate B, multiply equation Ð"Ñ by # and equation Ð#Ñ by $. Then add the results.

11.

Ð"Ñ Ð#Ñ

Multiply equation Ð"Ñ by # and add the result to equation Ð#Ñ.

Substitute ! for B in equation Ð"Ñ to find C.

10.

&B  %C œ # "!B  )C œ (

&B  #!C œ "!& &B  )C œ $& #)C œ "%! C œ &

& ‚ Ð%Ñ Ð&Ñ

Substitute & for C in equation Ð%Ñ to find B. B  %C œ #" B  %Ð&Ñ œ #" B  #! œ #" B œ " Bœ"

Ð%Ñ

Substitute " for B and & for C in equation Ð#Ñ to find D .

Systems of Linear Equations B  #C  #D "  #Ð&Ñ  #D "  "!  #D #D D

16.

œ $ œ $ œ $ œ' œ$

C œ #B  $!

Ð#Ñ

The perimeter is &(! ft. #B  #C œ &(!

B

C œ # $C  D œ *  #D œ (

To eliminate C, multiply equation Ð"Ñ by $ and add the result to equation Ð#Ñ.

From Ð"Ñ, C œ #Ð)&Ñ  $! œ #!!Þ The width is )& ft and the length is #!! ft.

$ ‚ Ð"Ñ Ð#Ñ Ð%Ñ

19.

To eliminate D , multiply equation Ð%Ñ by # and add the result to equation Ð$Ñ. #%B  #D œ $! B  #D œ ( #$B œ #$ B œ "

# ‚ Ð%Ñ

#B  $C œ #*'Þ'' $B  #C œ $"*Þ$*

Ð$Ñ

œ( œ( œ' œ$

17.

%B  'C œ &*$Þ$# *B  'C œ *&)Þ"( &B œ $'%Þ)& B œ (#Þ*(

Ð$Ñ

Ð"Ñ

The average price for a Yankees ticket was $(#Þ*( and the average price for a Red Sox ticket was $&!Þ#%.

Ð"Ñ Ð#Ñ Ð$Ñ

20.

# ‚ Ð"Ñ Ð$Ñ False

Since a false statement results, equations Ð"Ñ and Ð$Ñ have no common solution. The system is inconsistent. The solution set is g. 18.

#B  $C œ #*'Þ'' #Ð(#Þ*(Ñ  $C œ #*'Þ'' "%&Þ*%  $C œ #*'Þ'' $C œ "&!Þ(# C œ &!Þ#%

Ð"Ñ

To eliminate C, multiply equation Ð"Ñ by # and add the result to equation Ð$Ñ. 'B  #C  #D œ "' 'B  #C  #D œ "! ! œ '

# ‚ Ð"Ñ $ ‚ Ð#Ñ

Substitute (#Þ*( for B in equation Ð"ÑÞ

The solution set is e "ß #ß $ f. $B  C  D œ ) %B  #C  $D œ "& 'B  #C  #D œ "!

Ð"Ñ Ð#Ñ

To eliminate C, multiply equation Ð"Ñ by # and add the result to $ times equation Ð#Ñ.

Substitute " for B in equation Ð"Ñ to find C. %B  C œ # %Ð"Ñ  C œ # %C œ# C œ # Cœ#

Let B œ the average price for a Yankees ticket and C œ the average price for a Red Sox ticket. From the given information, we get the following system of equations:

Substitute " for B in equation Ð$Ñ to find D . B  #D "  #D #D D

Ð#Ñ

#B  #C œ &(! #B  #Ð#B  $!Ñ œ &(! #B  %B  '! œ &(! 'B œ &"! B œ )&

Ð"Ñ Ð#Ñ Ð$Ñ

"#B  $C œ ' $C  D œ * "#B  D œ "&

Ð#Ñ

Substitute #B  $! for C in equation Ð#ÑÞ

The solution set is e "ß &ß $ f. %B 

Ð"Ñ

Let B œ the width of the rink, and C œ the length of the rink. The length is $! ft longer than twice the width.

Let B œ the speed of the plane and C œ the speed of the wind. Complete the chart. With Wind Against Wind

< BC BC

> "Þ(& #

. "Þ(&ÐB  CÑ #ÐB  CÑ

The distance each way is &'! miles. From the chart, "Þ(&ÐB  CÑ œ &'!. Divide by "Þ(&. B  C œ $#!

Ð"Ñ

From the chart,

851

Systems of Linear Equations #ÐB  CÑ œ &'! B  C œ #)!.

BD # #C œ B  D B  #C  D œ !. Cœ

Ð#Ñ

Solve the system by adding equations Ð"Ñ and Ð#Ñ to eliminate C. B  C œ B  C œ #B œ B œ

Solve the system.

Ð"Ñ Ð#Ñ

$#! #)! '!! $!!

Add equations Ð"Ñ and Ð#Ñ to find B. B  C  D œ B  C  D œ #B œ B œ

The speed of the plane was $!! mph, and the speed of the wind was #! mphÞ Let B œ the amount of $#-per-pound nuts and C œ the amount of $"-per-pound chocolate candy.

Substitute )& for B and '! for C in equation Ð"Ñ to find D .

Solve the system formed from the second and fourth columns.

BCD )&  '!  D "%&  D D

Ð"Ñ Ð#Ñ 23.

Ð$Ñ

Substitute "!!  B for C in equation Ð#Ñ. #B  Ð"!!  BÑ œ "$! B œ $!

Let B œ the measure of the largest angle, C œ the measure of the middle-sized angle, and D œ the measure of the smallest angle. Since the sum of the measures of the angles of a triangle is ")!°,

Ð"Ñ

Since her commissions on the sales totaled $"(,!!!, !Þ"!B  !Þ!'C  !Þ!&D œ "(,!!!. Multiply by "!! to clear the decimals, so "!B  'C  &D œ ",(!!,!!!. Ð#Ñ Since the &% sale amounted to the sum of the other two sales, D œ B  C. Ð$Ñ

B  C  D œ ")!. Ð"Ñ Since the largest angle measures "!° less than the sum of the other two, B œ C  D  "! B  C  D œ "!.

Let B œ the value of sales at "!%, C œ the value of sales at '%, and D œ the value of sales at &%. B  C  D œ #)!,!!!

She should use $! lb of $#-per-pound nuts and (! lb of $"-per-pound chocolate candy.

Ð#Ñ

Since the measure of the middle-sized angle is the average of the other two,

852

Ð"Ñ

Since her total sales were $#)!,!!!,

From Ð$Ñ, C œ "!!  $! œ (!.

or

œ ")! œ ")! œ ")! œ $&

The three angle measures are )&°, '!°, and $&°.

Solve equation Ð"Ñ for C.

22.

Ð"Ñ Ð$Ñ

B  C  D œ ")! B  #C  D œ ! $C œ ")! C œ '!

Number of Price per Value Pounds Pound Nuts B # #B Chocolate C " "C œ C Mixture "!! "Þ$! "Þ$!Ð"!!Ñ œ "$!

C œ "!!  B

Ð"Ñ Ð#Ñ

")! "! "(! )&

Add equations Ð"Ñ and Ð$Ñ, to find C.

Complete the chart.

B  C œ "!! #B  C œ "$!

Ð"Ñ Ð#Ñ Ð$Ñ

B  C  D œ ")! B  C  D œ "! B  #C  D œ !

From Ð"Ñ, $!!  C œ $#!, so C œ #!.

21.

Ð$Ñ

Solve the system. B  C  D œ #)!,!!! "!B  'C  &D œ ",(!!,!!! D œ BC

Ð"Ñ Ð#Ñ Ð$Ñ

Since equation Ð$Ñ is given in terms of D , substitute B  C for D in equations Ð"Ñ and Ð#Ñ.

Systems of Linear Equations B  C  D œ #)!,!!! B  C  ÐB  CÑ œ #)!,!!! #B  #C œ #)!,!!! B  C œ "%!,!!!

Ð"Ñ

"!C  #!D "!C  #)D )D D

Ð%Ñ

"!B  'C  &D œ ",(!!,!!! "!B  'C  &ÐB  CÑ œ ",(!!,!!! "!B  'C  &B  &C œ ",(!!,!!! "&B  ""C œ ",(!!,!!!

Ð#Ñ

""B  ""C œ ",&%!,!!! "&B  ""C œ ",(!!,!!! %B œ "'!,!!! B œ %!,!!!

25.

"" ‚ Ð%Ñ Ð&Ñ

Mantle hit ( fewer than Maris, so B œ C  (. Ð#Ñ Maris hit $* more than Berra, so C œ D  $*

B  C  D œ "$( ÐC  (Ñ  C  ÐC  $*Ñ œ "$( $C  %' œ "$( $C œ ")$ C œ '"

B  C  D œ ). Ð"Ñ Since the final solution will be "#Þ&% hydrogen peroxide, Multiply by "!! to clear the decimals. 26.

Since the amount of )% solution used must be # L more than the amount of #!% solution, B œ D  #. Ð$Ñ

B  C  D œ ) )B  "!C  #!D œ "!! B œ D#

Ð"Ñ Ð#Ñ Ð$Ñ

Since equation Ð$Ñ is given in terms of B, substitute D  # for B in equations Ð"Ñ and Ð#Ñ.

)B  "!C  #!D )ÐD  #Ñ  "!C  #!D )D  "'  "!C  #!D "!C  #)D

œ "!! œ "!! œ "!! œ )%

Ð"Ñ

#B  &C œ % %B  C œ "% Write the augmented matrix. # ”%

Solve the system.

Ð$Ñ

From Ð#Ñ, B œ C  ( œ '"  ( œ &%. From Ð$Ñ, D œ C  $* œ '"  $* œ ##. Mantle hit &% home runs, Maris hit '" home runs, and Berra hit ## home runs.

!Þ!)B  !Þ"!C  !Þ#!D œ !Þ"#&Ð)Ñ.

Ð#Ñ

or D œ C  $*.

Substitute C  ( for B and C  $* for D in Ð"Ñ.

Since the amount of the mixture will be ) L,

BCD œ) ÐD  #Ñ  C  D œ ) C  #D œ '

Let B œ the number of home runs hit by Mantle, C œ the number of home runs hit by Maris, and D œ the number of home runs hit by Berra. B  C  D œ "$(Þ Ð"Ñ

Let B œ the number of liters of )% solution, C œ the number of liters of "!% solution, and D œ the number of liters of #!% solution.

)B  "!C  #!D œ "!!

Ð&Ñ

They combined for "$( home runs, so

From Ð%Ñ, C œ "!!,!!!. From Ð$Ñ, D œ %!,!!!  "!!,!!! œ "%!,!!!. He sold $%!,!!! at "!%, $"!!,!!! at '%, and $"%!,!!! at &%. 24.

"! ‚ Ð%Ñ

From Ð$Ñ, B œ D  # œ $  # œ &Þ From Ð%Ñ, C œ '  #D œ '  #Ð$Ñ œ !. Mix & L of )% solution, none of "!% solution, and $ L of #!% solution.

Ð&Ñ

To eliminate B, multiply equation Ð%Ñ by "" and add the result to equation Ð&Ñ.

œ '! œ )% œ #% œ $

# ”! # ”!

% "% • & % "" ## • & % " # • & "

#R"  R# "  "" R#

This matrix gives the system #B  &C œ % C œ #.

Ð"Ñ Ð%Ñ

Substitute C œ # in the first equation. Ð#Ñ

Ð&Ñ

To eliminate C, multiply equation Ð%Ñ by "! and add the result to equation Ð&Ñ.

#B  &C œ % #B  &Ð#Ñ œ % #B  "! œ % #B œ ' Bœ$

The solution set is e $ß # f.

853

Systems of Linear Equations 27.

Substitute C œ ! and D œ " in the first equation.

'B  $C œ * (B  #C œ "(

B  #C  D œ " B  #Ð!Ñ  Ð"Ñ œ " B"œ" Bœ!

Write the augmented matrix. ' ” (

$ * # "( • " & #' ” ( # "( • " & #' ” ! $$ "'& • " ”!

& "

#' &•

(R"  R# "  $$ R#

This matrix gives the system B  &C œ #' C œ &. Substitute C œ & in the first equation. B  &C œ #' B  &Ð&Ñ œ #' B  #& œ #' B œ "

28.

The solution set is e "ß & f.

# " "× Ô " $ % # # Õ # " " " Ø # " "× Ô" ! # & & Õ! $ " "Ø "× Ô " # " ! " % % Õ ! $ " "Ø " "× Ô" # ! " % % Õ ! ! "$ "$ Ø # " !

" % "

"× % " Ø

B  $C œ ( $B  D œ # C  #D œ % Ô" $ Õ! Ô" ! Õ! Ô" ! Õ!

Ô" ! Õ! Ô" ! Õ!

$R"  R# #R"  R$ R$  R#

$R#  R$

" R$  "$

This matrix gives the system B  #C  D œ " C  %D œ % D œ ". Substitute D œ " in the second equation. C  %D œ % C  %Ð"Ñ œ % Cœ!

854

29.

$ ! " $ * " $ " *

(× # %Ø

! " # ! " # ! # "

(× "* %Ø (× % "* Ø

$R"  R#

R# Ç R$

We use Ç to represent the interchanging of two rows.

B  #C  D œ " $B  %C  #D œ # #B  C  D œ "

Ô" ! Õ!

The solution set is e !ß !ß " f.

R"  R#

$ " !

! # "(

$ " !

! # "

(× % "( Ø

(× % " Ø

*R#  R$

"  "( R$

This matrix gives the system B  $C œ ( C  #D œ % D œ ". Substitute D œ " in the second equation. C  #D œ % C  #Ð"Ñ œ % C#œ% Cœ# Substitute C œ # in the first equation. B  $C œ ( B  $Ð#Ñ œ ( B'œ( Bœ" The solution set is e "ß #ß " f.

Systems of Linear Equations 30.

[1] System B would be easier to solve using the substitution method because the second equation is already solved for C.

31.

[1]

# $B " $B

 "' C œ

"* #

 #* C œ #

34.

[1]

Ð#Ñ

Ð$Ñ Ð%Ñ

$B  "#C œ $B  #C œ "%C œ C œ

' ‚ Ð"Ñ * ‚ Ð#Ñ

# ‚ Ð$Ñ 35.

Substitute "# for B in equation Ð$Ñ to find C.

[2]

Ð$Ñ

#B  &C  D œ "# B  C  %D œ "! )B  #!C  %D œ $"

Since a false statement results, the system is inconsistent. The solution set is g. B œ (C  "! #B  $C œ $

Ð"Ñ Ð#Ñ

Since equation Ð"Ñ is given in terms of B, substitute (C  "! for B in equation Ð#Ñ and solve for C. #Ð(C  "!Ñ  $C œ $ "%C  #!  $C œ $ "(C œ "( C œ " From Ð"Ñ, B œ (Ð"Ñ  "! œ $. The solution set is e $ß " f.

[1]

(B  $C œ "# &B  #C œ )

Ð"Ñ Ð#Ñ

Substitute ! for B in equation Ð"Ñ to find C.

Ð"Ñ Ð#Ñ Ð$Ñ

)B  #!C  %D œ %) % ‚ Ð"Ñ )B  #!C  %D œ $" Ð$Ñ ! œ (* False

[1]

The solution set is e &ß $ f.

"%B  'C œ #% # ‚ Ð"Ñ "&B  'C œ #% $ ‚ Ð#Ñ #*B œ ! B œ !

(B  $C œ "# (Ð!Ñ  $C œ "# $C œ "# Cœ%

Multiply equation Ð"Ñ by % and add the result to equation Ð$Ñ.

33.

Ð"Ñ

To eliminate C, multiply equation Ð"Ñ by # and equation Ð#Ñ by $. Then add the results.

The solution set is e "#ß * f. 32.

Ð#Ñ

B  %C œ "( B  %Ð$Ñ œ "( B  "# œ "( Bœ&

Ð%Ñ

%B  C œ &( %Ð"#Ñ  C œ &( %)  C œ &( Cœ*

$ ‚ Ð"Ñ

&" * %# $

Substitute $ for C in equation Ð"Ñ to find B.

To eliminate C, multiply equation Ð$Ñ by # and add the result to equation Ð%Ñ. )B  #C œ ""% $B  #C œ ") ""B œ "$# B œ "#

Ð"Ñ Ð#Ñ

To eliminate B, multiply equation Ð"Ñ by $ and add the result to equation Ð#Ñ.

Ð"Ñ

Multiply equation Ð"Ñ by ' and equation Ð#Ñ by * to clear the fractions. %B  C œ &( $B  #C œ ")

B  %C œ "( $B  #C œ *

36.

Ð"Ñ

The solution set is e !ß % f.

[1]

#B  &C œ ) $B  %C œ "!

Ð"Ñ Ð#Ñ

To eliminate C, multiply equation Ð"Ñ by % and equation Ð#Ñ by & and add the results. )B  #!C œ $# "&B  #!C œ &! #$B œ )# B œ )# #$

% ‚ Ð"Ñ & ‚ Ð#Ñ

Instead of substituting to find C, we'll choose different multipliers and eliminate B from the original system. 'B  'B 

"&C )C #$C C

œ #% œ #! œ % % œ  #$

$ ‚ Ð"Ñ # ‚ Ð#Ñ

% ™ The solution set is ˜Ð )# #$ ß  #$ Ñ .

855

Systems of Linear Equations 37.

[3] Let B œ the number of liters of &% solution and C œ the number of liters of "!% solution. Liters of Solution B "! C

Percent (as a decimal) !.!& !Þ#! !Þ"!

Test 1.

Liters of Pure Acid !Þ!&B !Þ#!Ð"!Ñ œ # !Þ"!C

(b) Philadelphia's graph indicates that it will experience population decline. (c) In the year 2000, the city populations from least to greatest are Dallas, Phoenix, Philadelphia, and Houston.

Solve the system formed from the first and third columns. B  "! œ C !Þ!&B  # œ !Þ"!C

Ð"Ñ Ð#Ñ

2.

Multiply equation Ð#Ñ by "!! to clear the decimals. &B  #!! œ "!C

&B  #!! œ "!C &B  #!! œ "!ÐB  "!Ñ &B  #!! œ "!B  "!! "!! œ &B #! œ B

3.

When each equation of the system BC œ( BC œ&

Ð$Ñ

is graphed, the point of intersection appears to be Ð'ß "Ñ. To check, substitute ' for B and " for C in each of the equations. Since Ð'ß "Ñ makes both equations true, the solution set of the system is e 'ß " f.

He should use #! L of &% solution. 38.

(a) The graphs for Dallas and Philadelphia intersect in the year 2010. The population for each city will be about "Þ%& million. (b) The graphs for Houston and Phoenix appear to intersect in the year 2025. The population for each city will be about #Þ) million. This can be represented by the ordered pair #!#&ß #Þ) .

Ð$Ñ

Substitute B  "! for C in equation Ð$Ñ and solve for B.

(a) Rising graphs indicate population growth, so Houston, Phoenix, and Dallas will experience population growth.

[3] Let B, C, and D denote the number of medals won by Germany, the United States, and Canada, respectively. The total number of medals won was *$, so B  C  D œ *$Þ Ð"Ñ Germany won seven fewer medals than the United States, so

4.

Canada won eleven fewer medals than the United States, so

#B  $C œ #% #B  $Ð #$ BÑ œ #% #B  #B œ #% %B œ #% Bœ'

D œ C  1"Þ Ð$Ñ Substitute C  ( for B and C  "" for D in Ð"Ñ.

From Ð#Ñ, B œ $(  ( œ $!Þ From Ð$Ñ, D œ $(  "" œ #'. Germany won $! medals, the United States won $( medals, and Canada won #' medals.

856

Ð"Ñ Ð#Ñ

Since equation Ð#Ñ is solved for C, substitute  #$ B for C in equation Ð"Ñ and solve for B.

B œ C  (Þ Ð#Ñ

ÐC  (Ñ  C  ÐC  ""Ñ œ *$ $C  ") œ *$ $C œ """ C œ $(

#B  $C œ #% C œ  #$ B

Ð"Ñ

From Ð#Ñ, C œ  #$ Ð'Ñ œ %. 5.

The solution set is e 'ß % f. $B  C œ ) #B  'C œ $

Ð"Ñ Ð#Ñ

To eliminate C, multiply equation Ð"Ñ by '. Then add that equation and equation Ð#Ñ.

Systems of Linear Equations Substitute # for C in equation Ð"Ñ to find B.

' ‚ Ð"Ñ

")B  'C œ %) #B  'C œ $ Ð#Ñ #!B œ %& * B œ  %& #! œ  %

&B  #C œ % &B  #Ð#Ñ œ % &B  % œ % &B œ ! Bœ!

To find C, substitute  *% for B in equation Ð#Ñ. (We could also use elimination by adding # ‚ Ð"Ñ and $ ‚ Ð#Ñ.) #B  'C œ $ #Ð *% Ñ  'C œ $

Ð#Ñ

The solution set is e !ß # f. 9.

 *#  'C œ $ 'C œ Cœ

or

"& # "& 1#

œ

The solution set is ˜ *% ß &% ™. 6.

"#B  &C œ ) $B œ &% C  # or

B œ

& "# C

Substitute for C.

& "# C

# $





Ð#Ñ 10.

"#B  &C œ )  $# Ñ  &C œ ) &C  )  &C œ ) )œ)

Ð"Ñ

True

To eliminate B, multiply equation Ð"Ñ by " and add the result to equation Ð$Ñ.

Equations Ð"Ñ and Ð#Ñ are dependent. The solution set is e Bß C l "#B  &C œ )f. Ð"Ñ Ð#Ñ

$B  C œ "# #B  C œ $ &B œ "& B œ $

Ð"Ñ Ð#Ñ

Substitute $ for B in equation Ð"Ñ to find C. $B  C œ "# $Ð$Ñ  C œ "# *  C œ "# Cœ$

Ð"Ñ

The solution set is e $ß $ f. &B  #C œ % 'B  $C œ '

$B  &C  $D œ # $B  "!C  #D œ ' &C  &D œ %

'B  "!C  'D œ % 'B  &C  D œ !  &C  &D œ %

#% $! &% #

# ‚ Ð"Ñ Ð#Ñ Ð&Ñ

To eliminate C, add equations Ð%Ñ and Ð&Ñ. &C  &D &C  &D  "!D D

œ % œ % œ ! œ !

Ð%Ñ Ð&Ñ

Substitute ! for D in equation Ð%Ñ to find C.

Ð"Ñ Ð#Ñ

œ œ œ œ

" ‚ Ð"Ñ Ð$Ñ Ð%Ñ

To eliminate B again, multiply equation Ð"Ñ by # and add the result to equation Ð#Ñ.

&C  &D œ % &C  &Ð!Ñ œ % &C  ! œ % &C œ % C œ %&

To eliminate B, multiply equation Ð"Ñ by ' and equation Ð#Ñ by &. Then add the results. $!B  "#C $!B  "&C #(C C

Ð"Ñ Ð#Ñ Ð$Ñ

$B  &C  $D œ # 'B  &C  D œ ! $B  "!C  #D œ '

To eliminate C, add equations Ð"Ñ and Ð#Ñ.

8.

Since a false statement results, the system is inconsistent. The solution set is g.

for B in equation Ð"Ñ and solve

$B  C œ "# #B  C œ $

Ð#Ñ

'B  )C œ "' # ‚ Ð"Ñ 'B  )C œ ( Ð#Ñ ! œ * False

& "#Ð "# C

7.

Ð"Ñ

$B  %C œ ) )C œ (  'B 'B  )C œ (

Multiply equation Ð"Ñ by # and add the result to equation Ð#Ñ.

& %

Ð"Ñ

# $

Ð"Ñ

' ‚ Ð"Ñ & ‚ Ð#Ñ

Substitute find B.

% &

Ð%Ñ

for C and ! for D in equation Ð"Ñ to

857

Systems of Linear Equations $B  &C  $D œ # $B  &Ð %& Ñ  $Ð!Ñ œ # $B  %  ! œ # $B œ # B œ  #$

Ð"Ñ

13.

Make a table. Faster Car Slower Car

The solution set is ˜Ð #$ ß %& ß !Ñ™. 11.

%B  C  D œ "" B  C  D œ % C  #D œ !

$Þ&B  $Þ&C œ %#!. Multiply by "! to clear the decimals.

Ð"Ñ

$&B  $&C œ %#!!

% ‚ Ð#Ñ

$&B  $&C œ $&B  $&C œ (!B œ B œ

Ð%Ñ ƒ Ð&Ñ Ð$Ñ

"!&! %#!! &#&! (&

$& ‚ Ð"Ñ Ð#Ñ

Substitute (& for B in equation Ð"Ñ to find C.

From Ð$Ñ, C  #Ð"Ñ œ !, so C œ #Þ From Ð#Ñ, B  Ð#Ñ  " œ %, so B œ $.

B  C œ $! (&  C œ $! C œ %& C œ %&

The solution set is e $ß #ß " f.

Let B œ the gross (in millions of dollars) for Star Wars Episode IV: A New Hope, and C œ the gross (in millions of dollars) for Indiana Jones and the Kingdom of the Crystal Skull. Together the movies grossed $(()Þ! million, so B  C œ (()Þ!Þ Ð"Ñ Indiana Jones and the Kingdom of the Crystal Skull grossed $"%%Þ! million less than Star Wars Episode IV: A New Hope, so C œ B  "%%Þ!Þ Ð#Ñ

Ð"Ñ

The faster car is traveling at (& mph, and the slower car is traveling at %& mph. 14.

Let B œ the number of liters of #!% solution and C œ the number of liters of &!% solution. Make a table. Liters of Solution B C "#

Percent (as a decimal) !Þ#! !Þ&! !Þ%!

Liters of Pure Alcohol !Þ#!B !Þ&!C !Þ%!Ð"#Ñ œ %Þ)

Since "# L of the mixture are needed,

Substitute B  "%%Þ! for C in equation Ð"Ñ. Ð"Ñ

From Ð#Ñ, C œ %'"Þ!  "%%Þ! œ $"(Þ!Þ Star Wars Episode IV: A New Hope grossed $%'"Þ! million and Indiana Jones and the Kingdom of the Crystal Skull grossed $$"(Þ! million.

858

Ð#Ñ

To eliminate C, multiply equation Ð"Ñ by $& and add the result to equation Ð#Ñ.

Ð%Ñ

B  C œ (()Þ! B  ÐB  "%%Þ!Ñ œ (()Þ! #B œ *##Þ! B œ %'"Þ!

. $Þ&B $Þ&C

Since the cars travel a total of %#! miles,

To eliminate C, divide equation Ð%Ñ by & and add the result to equation Ð$Ñ.

12.

> $Þ& $Þ&

B  C œ $!. Ð"Ñ

To eliminate B, multiply equation Ð#Ñ by % and add the result to equation Ð"Ñ.

C  D œ " C  #D œ ! D œ "

< B C

Since the slow car travels $! mph slower than the fast car,

Ð"Ñ Ð#Ñ Ð$Ñ

%B  C  D œ "" %B  %C  %D œ "' &C  &D œ &

Let B œ the rate of the faster car and C œ the rate of the slower car.

B  C œ "#. Ð"Ñ Since the amount of pure alcohol in the #!% solution plus the amount of pure alcohol in the &!% solution must equal the amount of alcohol in the mixture, !Þ#B  !Þ&C œ %Þ). Multiply by "! to clear the decimals. #B  &C œ %)

Ð#Ñ

Multiply equation Ð"Ñ by # and add the result to equation Ð#Ñ.

Systems of Linear Equations #B  #C #B  &C $C C

œ #% œ %) œ #% œ )

To eliminate D , multiply equation Ð"Ñ by *& and add the result to equation Ð#Ñ.

# ‚ Ð"Ñ Ð#Ñ

From Ð"Ñ, B  ) œ "#, so B œ %. % L of #!% solution and ) L of &!% solution are needed. 15.

Divide equation Ð%Ñ by &. $B  #C œ #%!

Let B œ the price of an AC adaptor and C œ the price of a rechargeable flashlight.

$B  #C œ #%! $Ð#CÑ  #C œ #%! )C œ #%! C œ $!

(B  #C œ )'. Ð"Ñ Since $ AC adaptors and % rechargeable flashlights cost $)%, Solve the system.

B  C  D œ "!! '!  $!  D œ "!! D œ "!

Ð"Ñ Ð#Ñ

# ‚ Ð"Ñ Ð#Ñ

17.

$B  #C œ % &B  &C œ * Write the augmented matrix.

Substitute ) for B in equation Ð"Ñ to find C.

$ ”&

# &

% *•

We could divide row " by $, but to avoid working with fractions, we'll multiply row " by # and then multiply row # by " and add to row " to obtain a """ in the first row.

Ð"Ñ

An AC adaptor costs $), and a rechargeable flashlight costs $"&. 16.

Let B œ the amount of Orange Pekoe, C œ the amount of Irish Breakfast, and D œ the amount of Earl Grey. The owner wants "!! oz of tea, so B  C  D œ "!!Þ Ð"Ñ An equation which relates the prices of the tea is

' ”& " ”& " ”! " ”!

% ) & *• " " & *• " " "! "% • " " ( • " &

Multiply by "!! to clear the decimals. )!B  )&C  *&D œ )$!!

Ð#Ñ

The mixture must contain twice as much Orange Pekoe as Irish Breakfast, so

#R" R#  R"

–&R" + R# " "! R#

This matrix gives the system B  C œ " C œ (& .

!Þ)!B  !Þ)&C  !Þ*&D œ !Þ)$Ð"!!ÑÞ

B œ #CÞ Ð$Ñ

Ð"Ñ

He should use '! oz of Orange Pekoe, $! oz of Irish Breakfast, and "! oz of Earl Grey.

To eliminate C, multiply equation Ð"Ñ by # and add the result to equation Ð#Ñ.

(B  #C œ )' (Ð)Ñ  #C œ )' &'  #C œ )' #C œ $! C œ "&

Ð&Ñ

From Ð$Ñ, B œ #Ð$!Ñ œ '!Þ Substitute '! for B and $! for C in equation Ð"Ñ to find D .

$B  %C œ )%. Ð#Ñ

"%B  %C œ "(# $B  %C œ )% ""B œ )) B œ )

Ð&Ñ

Substitute #C for B in equation Ð&Ñ to find C.

Since ( AC adaptors and # rechargeable flashlights cost $)',

(B  #C œ )' $B  %C œ )%

*& ‚ Ð"Ñ Ð#Ñ Ð%Ñ

*&B  *&C  *&D œ *&!! )!B  )&C  *&D œ )$!! "&B  "!C œ "#!!

Substitute C œ

( &

B

in the first equation. ( &

œ "

B œ "  œ  && 

( & ( &

The solution set is ˜Ð #& ß (& Ñ™.

œ

# &

859

Systems of Linear Equations 18.

B  $C  #D œ "" $B  (C  %D œ #$ &B  $C  &D œ "%

This matrix gives the system B  $C  #D œ "" CD œ& D œ $.

Write the augmented matrix. Ô" $ Õ& Ô" ! Õ! Ô" ! Õ! Ô" ! Õ!

Ô" ! Õ!

860

# "" × % #$ & "% Ø $ # "" × # # "! "# "& '* Ø $ # "" × " " & "# "& '* Ø $ ( $

$ " !

# " $

$ " !

# " "

"" × & * Ø

"" × & $Ø

Substitute D œ $ in the second equation.

$R"  R# &R"  R$  "# R#

"#R#  R$

 "$ R$

CD œ& C$œ& Cœ# Substitute C œ # and D œ $ in the first equation. B  $C  #D œ "" B  $Ð#Ñ  #Ð$Ñ œ "" B  '  ' œ "" B œ "

The solution set is e "ß #ß $ f.

7% , 8 Á !, cannot be simplified because the 8$ bases 7 and 8 are different. The quotient rule does not apply.

EXPONENTS, POLYNOMIALS, AND POLYNOMIAL FUNCTIONS 1

(c)

N6. (a) Ð#7$ Ñ% œ Ð#Ñ% Ð7$ Ñ% œ "'7$ • % œ "'7"#

Integer Exponents and Scientific Notation

1 Now Try Exercises (a) )& • )% œ )&  % œ )* % (

$

%

( $

(b) Ð&B C ÑÐ(BC Ñ œ &Ð(ÑB BC C œ $&B%  " C(  $ œ $&B& C"! (c) : • ; cannot be simplified further because the bases : and ; are not the same. The product rule does not apply. #

N2. (a) &! œ " Any real number (except !) raised to the power zero is equal to ". (b) Since B Á !, &B Á !, and Ð&BÑ! œ ". (c) &! œ Ð&! Ñ œ " (d) "!!  *! œ "  " œ ! N3. (a) *% œ

" *%

(b) Ð$CÑ' œ

" ,CÁ! Ð$CÑ'

(c) %5 $ œ %Œ (d) %"  '" œ

$B# $$ ÐB# Ñ$ #(B# • $ #(B' œ œ œ ,  C$ ÐC$ Ñ$ C$ • $ C* $

CÁ! & N7. Œ  $

$

N1. Apply the product rule for exponents, if possible.

#

(b) Œ

" %  œ  $, 5 Á ! 5$ 5

" " $ # &  œ  œ % ' "# "# "#

" " " &$ œ œ "ƒ $ œ "• œ &$ œ "#& $ " & & " &$ " # "!# " " " #& "! œ œ ƒ & œ (b) • & # " # "! # "!# " #& " $# $# )•% ) œ œ œ œ • "!! " "!! #& • % #&

N4. (a)

N5. Apply the quotient rule for exponents, if possible.

$ $$ #( œŒ  œ $ œ & & "#& $

" N8. (a) B) • B • B% œ B)  "  % œ B$ œ $ B " (b) Ð&$ Ñ# œ &$Ð#Ñ œ &' œ ' & (c)

:# ; % :# ; % œ # • " # " : ; : ; #  Ð#Ñ %  Ð"Ñ œ: ; :% œ :% ; $ œ $ ;

(d) Œ

#B# &B# Œ  # C C # % $ ' # B Ð&Ñ B œ ' • C# C # % $ ' # B Ð&Ñ B œ C' C# )B"! )B"! œ œ #& C % Ð&Ñ# C% $

#

Combination of rules

N9. (a) (,&'!,!!!,!!! œ (^&'!,!!!,!!!. Place a caret after the first nonzero digit, (. Count * places from the decimal point (understood to be after the last !) to the caret. Use a positive exponent on "! since (,&'!,!!!,!!!  (Þ&'. (,&'!,!!!,!!! œ (Þ&' ‚ "!* (b) !Þ!!! !!! #%& œ !Þ!!! !!! #^%& Count ( places. Use a negative exponent on "! since !Þ!!! !!! #%&  #Þ%&. !Þ!!! !!! #%& œ #Þ%& ‚ "!( N10. (a) %Þ%& ‚ "!"! œ %Þ%,&!!,!!!,!!!Þ œ %%,&!!,!!!,!!! Move the decimal "! places to the right.

(a)

>) œ >)  # œ > ' , > Á ! >#

(b) &Þ* ‚ "!& œ !Þ!!!!&Þ* œ !Þ!!! !&*

(b)

%& œ %&  Ð#Ñ œ %( %#

Move the decimal & places to the left.

From Chapter 5 of Student’s Solutions Manual for Intermediate Algebra, Eleventh Edition. Margaret L. Lial, John Hornsby, Terry McGinnis. Copyright © 2012 by Pearson Education, Inc. Publishing as Addison-Wesley. All rights reserved.

861

Exponents, Polynomials, and Polynomial Functions N11.

!Þ!!!'$ ‚ %!!,!!0 'Þ$ ‚ "!% ‚ % ‚ "!& œ 1400 ‚ !Þ!!!!!$ "Þ% ‚ "!$ ‚ $ ‚ "!' 'Þ$ ‚ % ‚ "!% ‚ "!& œ "Þ% ‚ $ ‚ "!$ ‚ "!' 'Þ$ ‚ % ‚ "!" œ "Þ% ‚ $ ‚ "!$ 'Þ$ ‚ % œ ‚ "!% "Þ% ‚ $ œ ' ‚ "!% œ '!,!!!

N12. . œ , so . >œ ÒNote that $!,!!!,!!!,!!! œ $ ‚ "!"! .Ó < "Þ# ‚ "!"& œ $ ‚ "!"! "Þ# œ ‚ "!"&  "! $ œ !Þ% ‚ "!& œ %Þ! ‚ "!% , or %!,!!! sec.

23.

"&! œ ", since +! œ " for any nonzero base +.

25.

)! œ Ð)! Ñ œ Ð"Ñ œ "

27.

Ð#&Ñ! œ " since #& is in parentheses.

29.

$!  Ð$Ñ! œ "  " œ #

31.

$!  $! œ "  " œ !

33.

%!  7! œ "  " œ #

35.

(a) &# œ

" " œ # & #&

(b) &# œ 

" " œ &# #&

(c) Ð&Ñ# œ

" " œ Ð&Ñ# #&

(d) Ð&Ñ# œ 

It would take %Þ! ‚ "!% seconds.

1 Section Exercises 1.

Ð+,Ñ# œ +# , # by a power rule. Since +# , # Á +, # , the expression Ð+,Ñ# œ +, # has been simplified incorrectly. The exponent should apply to both + and , . $ $% Œ  œ %, + Á ! + + $% $% Since % Á , the expression + + %

3.

$ $% Œ  œ + + %

has been simplified incorrectly. 5.

B$ • B% œ B$  % œ B( is correct.

7.

Your friend multiplied the bases, which is incorrect. Instead, keep the same base and add the exponents.

9.

"$ *

• "$

)

œ "$ *

%)

"

œ "$

œ )*  " œ )"!

11.

)

13.

B$ • B& • B* œ B$  &  * œ B"(

15.

Ð$A& ÑÐ*A$ Ñ œ Ð$ÑÐ*ÑA&  $ œ #(A)

17.

Ð#B# C& ÑÐ*BC$ Ñ œ Ð#ÑÐ*ÑB#  " C&  $ œ ")B$ C)

19.

, the coefficient is the degree is ".

173. *B  &B  B  )B  "#B œ Ð*  &  "  )  "#ÑB œ *B

9.

In ) œ )B! , the coefficient is ) and the degree is !.

11.

In B$ œ "B$ , the coefficient is " and the degree is $.

" '

and since

> '

œ "' >" ,

Exponents, Polynomials, and Polynomial Functions 13.

#B$  B  $B#  % The polynomial is written in descending powers of the variable if the exponents on the terms of the polynomial decrease from left to right. #B$  $B#  B  %

47.

$+, #  (+# ,  &+, #  "$+# , œ Ð$  &Ñ+, #  Ð(  "$Ñ+# , œ #+, #  #!+# ,

49.

%  Ð#  $7Ñ  '7  * œ %  #  $7  '7  * œ Ð$  'Ñ7  Ð%  #  *Ñ œ $7  ""

51.

'  $:  Ð#:  "Ñ  Ð#:  *Ñ œ '  $:  #:  "  #:  * œ Ð$  #  #Ñ:  Ð'  "  *Ñ œ :  %

53.

Ð&B#  (B  %Ñ  Ð$B#  'B  #Ñ œ &B#  $B#  (B  'B  %  # œ )B#  B  #

55.

Ð'>#  %>%  >Ñ  Ð$>%  %>#  &Ñ œ %>%  $>%  '>#  %>#  >  & œ Ð%  $Ñ>%  Ð'  %Ñ>#  >  & œ >%  #>#  >  &

57.

ÐC$  $C  #Ñ  Ð%C$  $C#  #C  "Ñ œ C$  %C$  $C#  $C  #C  #  " œ Ð"  %ÑC$  $C#  Ð$  #ÑC  Ð#  "Ñ œ &C$  $C#  &C  "

59.

Ð$<  )Ñ  Ð#<  &Ñ

The leading term is #B$ . The leading coefficient is #. 15.

%:$  ):&  :( œ :(  ):&  %:$ The leading term is :( . The leading coefficient is ".

17.

"!  7$  $7% œ $7%  7$  "! The leading term is $7% . The leading coefficient is $.

19.

#& is one term, so it's a monomial. #& is a nonzero constant, so it has degree zero.

21.

(7  ## has two terms, so it's a binomial. The exponent on 7 is ", so (7  ## has degree ".

23.

(C'  ""C) is a binomial of degree ).

25.

78& is one term, so it's a monomial. Since 7 œ 7" and "  & œ ', the degree is '.

27.

&7$  '7  *7# has three terms, so it's a trinomial. The greatest exponent is $, so the degree is $.

29.

':% ;  $:$ ; #  #:; $  ; % has four terms, so it is classified as none of these. The greatest sum of exponents on any term is &, so the polynomial has degree &.

31.

Only choice A is a trinomial (it has three terms) with descending powers and having degree '.

33.

&D %  $D % œ Ð&  $ÑD % œ )D %

35.

7$  #7$  '7$ œ Ð"  #  'Ñ7$ œ (7$

37.

BBBBB œ Ð"  "  "  "  "ÑB œ &B

39.

7%  $7#  7 is already simplified since there are no like terms to be combined.

41.

&>  %=  '>  *= œ Ð&>  '>Ñ  Ð%=  *=Ñ œ >  "$=

43.

#5  $5 #  &5 #  ( œ Ð$5 #  &5 # Ñ  #5  ( œ Ð$  &Ñ5 #  #5  ( œ )5 #  #5  (

45.

8%  #8$  8#  $8%  8$ œ 8%  $8%  #8$  8$  8# œ Ð"  $Ñ8%  Ð#  "Ñ8$  8# œ #8%  8$  8#

Change all signs in the second polynomial and add. œ Ð$<  )Ñ  Ð#<  &Ñ œ $<  )  #<  & œ $<  #<  )  & œ <  "$ 61.

Ð#+#  $+  "Ñ  Ð%+#  &+  'Ñ œ Ð#+#  $+  "Ñ  Ð%+#  &+  'Ñ œ #+#  %+#  $+  &+  "  ' œ #+#  #+  (

63.

ÐD &  $D #  #DÑ  Ð%D &  #D #  &DÑ œ D &  $D #  #D  %D &  #D #  &D œ D &  %D &  $D #  #D #  #D  &D œ $D &  D #  (D

65.

#":  ) *:  % "#:  % Add vertically.

67.

"#:#  %:  " $:#  (:  ) *:#  "":  * Add vertically.

867

Exponents, Polynomials, and Polynomial Functions 69.

Now perform the subtraction.

Subtract.

Ð%7#  $8#  &8Ñ  Ð8#  #8Ñ œ %7#  $8#  &8  8#  #8 œ %7#  $8#  8#  &8  #8 œ %7#  %8#  (8

"#+  "& (+  $ Change all the signs in the second polynomial, and add. "#+  "& (+  $ &+  ") 71.

83.

ÒÐC%  C#  "Ñ  ÐC%  #C#  "ÑÓ  Ð$C%  $C#  #Ñ œ C%  C#  "  C%  #C#  "  $C%  $C#  # œ C%  C%  $C%  C#  #C#  $C# ""# œ C%  %C#  %

85.

Ò$D #  &D  Ð#D #  'DÑÓ  ÒÐ)D #  Ò&D  D # ÓÑ  #D # Ó œ Ð$D #  &D  #D #  'DÑ  Ð)D #  &D  D #  #D # Ñ œ Ð$D #  #D #  &D  'DÑ  Ð)D #  D #  #D #  &D Ñ œ ÐD #  ""DÑ  Ð""D #  &DÑ œ D #  ""D  ""D #  &D œ D #  ""D #  ""D  &D œ "!D #  "'D

87.

Using the vertical line test, we find any vertical line will intersect the graph at most once. This indicates that the graph represents a function. The domain is the set of B-values: Ð∞ß ∞Ñ. The range is the set of C-values: Ð∞ß ∞Ñ.

89.

Since a vertical line can intersect the graph of the relation in more than one point, the relation is not a function. The domain is the set of B-values: Ò!ß ∞Ñ. The range is the set of C-values: Ð∞ß ∞Ñ.

91.

0 ÐBÑ œ B#  #

Subtract. '7#  ""7  & )7#  #7  " Change all the signs in the second polynomial, and add. '7#  ""7  & )7#  #7  " "%7#  "$7  '

73.

Add column by column to obtain the result on the bottom line. "#D #  ""D  ) &D #  "'D  # %D #  &D  * "$D #  "!D  $

75.

'C$  *C# ) $ # %C  #C  &C "!C$  (C#  &C  )

77.

Subtract. &+%

 )+#  * $ '+  +#  #

Change all the signs in the second polynomial, and add. &+%

 )+#  * $  '+  +#  #

(a) 0 Ð"Ñ œ Ð"Ñ#  # œ"#œ$

&+%  '+$  *+#  "" 79.

81.

3

Simplify the expression in brackets first. #

#

#

Ð$7  &8  #8Ñ  Ð$7 Ñ  %8 œ $7#  &8#  #8  $7#  %8# œ $7#  $7#  &8#  %8#  #8 œ 8#  #8

868

(b) 0 Ð#Ñ œ ##  # œ%#œ'

(C#  'C  &  Ð%C#  #C  $Ñ œ (C#  'C  &  %C#  #C  $ œ (C#  %C#  'C  #C  &  $ œ $C#  %C  # #

Polynomial Functions, Graphs, and Composition

3 Now Try Exercises N1. 0 ÐBÑ œ B$  #B#  ( 0 Ð$Ñ œ Ð$Ñ$  #Ð$Ñ#  ( œ #(  #Ð*Ñ  ( œ $)

Exponents, Polynomials, and Polynomial Functions N2. T ÐBÑ œ !Þ!"((%B#  !Þ()("B  %"Þ#'

N7.

The year #!!# corresponds to #!!#  "*!! œ "#Þ T Ð"#Ñ œ !Þ!"((%Ð"#Ñ#  !Þ()("Ð"#Ñ  %"Þ#' œ %)Þ"&!'% Let x = 12Þ Thus, in 2002 about %)Þ# million students were enrolled in public schools in the United States.

B # " ! " #

0 ÐBÑ œ B#  % ! $ % $ !

N3. 0 ÐBÑ œ B$  $B#  %, 1ÐBÑ œ #B$  B#  "# (a) Ð0  1ÑÐBÑ œ 0 ÐBÑ  1ÐBÑ œ ÐB$  $B#  %Ñ  Ð#B$  B#  "#Ñ œ B$  $B#  %  #B$  B#  "# œ B$  #B#  ) (b) Ð0  1ÑÐBÑ œ 0 ÐBÑ  1ÐBÑ œ ÐB$  $B#  %Ñ  Ð#B$  B#  "#Ñ œ B$  $B#  %  #B$  B#  "# œ $B$  %B#  "'

Any value of B can be used, so the domain is Ð∞ß ∞Ñ. The minimum C-value is % and there is no maximum C-value, so the range is Ò%ß ∞Ñ.

3 Section Exercises 1.

(a) 0 Ð"Ñ œ 'Ð"Ñ  % œ '  % œ "!

N4. 0 ÐBÑ œ B#  %, 1ÐBÑ œ 'B# (a) Ð0  1ÑÐBÑ œ 0 ÐBÑ  1ÐBÑ œ ÐB#  %Ñ  Ð'B# Ñ œ &B#  %

(b) 0 Ð#Ñ œ 'Ð#Ñ  % œ "#  % œ ) 3.

(b) 0 Ð#Ñ œ Ð#Ñ#  $Ð#Ñ  % œ%'%œ# 5.

0 ÐBÑ œ &B%  $B#  ' (a) 0 Ð"Ñ œ &Ð"Ñ%  $Ð"Ñ#  ' œ &•"  $•"  ' œ&$'œ)

N5. 0 ÐBÑ œ $B  ( and 1ÐBÑ œ B  # Definition 1(x) = x  2 Subtract. f(x) = 3x + 7

(b) 0 Ð#Ñ œ &Ð#Ñ%  $Ð#Ñ#  ' œ & • "'  $ • %  ' œ )!  "#  ' œ (% 7.

1 ‰ 0 Ð"Ñ œ 1Ð0 Ð"ÑÑ œ 1Ð"  &Ñ œ 0 Ð'Ñ œ Ð'Ñ#  # œ 

%$Definition f(x) = x  5 Subtract. 1(x) = x# + 2

(b) Ð0 ‰ 1ÑÐBÑ œ 0 Ð1ÐBÑÑ œ 0 ÐB#  #Ñ œ ÐB#  #Ñ  & œ B#  $

Definition 1(x) = x# + 2 f(x) = x  5

0 ÐBÑ œ B#  #B$  ) (a) 0 Ð"Ñ œ Ð"Ñ#  #Ð"Ñ$  ) œ Ð"Ñ  #Ð"Ñ  ) œ "  #  ) œ ""

N6. 0 ÐBÑ œ B  & and 1ÐBÑ œ B#  # (a)

0 ÐBÑ œ B#  $B  % (a) 0 Ð"Ñ œ Ð"Ñ#  $Ð"Ñ  % œ"$%œ)

(b) Ð0  1ÑÐBÑ œ 0 ÐBÑ  1ÐBÑ œ ÐB#  %Ñ  Ð'B# Ñ œ ( B#  % Ð0  1ÑÐ%Ñ œ (Ð%Ñ#  % œ (Ð"'Ñ  % œ "!)

Ð0 ‰ 1ÑÐ(Ñ œ 0 Ð1Ð(ÑÑ œ 0 Ð(  #Ñ œ 0 Ð&Ñ œ $Ð&Ñ  ( œ ##

0 ÐBÑ œ 'B  %

(b) 0 Ð#Ñ œ Ð#Ñ#  #Ð#Ñ$  ) œ Ð%Ñ  # • )  ) œ %  "'  ) œ % 9.

T ÐBÑ œ "''(B#  ##Þ()B  %$!! (a) B œ #!!!  #!!! œ ! T Ð!Ñ œ "''(Ð!Ñ#  ##Þ()Ð!Ñ  %$!! œ %$!! thousand pounds 869

Exponents, Polynomials, and Polynomial Functions (b) B œ #!!$  #!!! œ $

21.

Ð0  2ÑÐBÑ œ 0 ÐBÑ  2ÐBÑ œ ÐB#  *Ñ  ÐB  $Ñ œ B#  *  B  $ œ B#  B  '

23.

Ð0  2ÑÐ$Ñ œ 0 Ð$Ñ  2Ð$Ñ œ ÒÐ$Ñ#  *Ó  cÐ$Ñ  $d œ Ð*  *Ñ  Ð'Ñ œ!'œ'

25.

Ð1  2ÑÐ"!Ñ œ 1Ð"!Ñ  2Ð"!Ñ œ #Ð"!Ñ  cÐ"!Ñ  $d œ #!  Ð"$Ñ œ $$

27.

Ð1  2ÑÐ$Ñ œ 1Ð$Ñ  2Ð$Ñ œ #Ð$Ñ  cÐ$Ñ  $d œ '  Ð'Ñ œ '  ' œ !

T Ð$Ñ œ "''(Ð$Ñ#  ##Þ()Ð$Ñ  %$!! œ "*,$("Þ$% ¸ "*,$(" thousand pounds (c) B œ #!!'  #!!! œ ' T Ð'Ñ œ "''(Ð'Ñ#  ##Þ()Ð'Ñ  %$!! œ '%,%%)Þ') ¸ '%,%%* thousand pounds 11.

T ÐBÑ œ !Þ!!")*B$  !Þ""*$B#  #Þ!#(B  #)Þ"* (a) B œ "*)&  "*)& œ ! T Ð!Ñ œ !Þ!!")*Ð!Ñ$  !Þ""*$Ð!Ñ#  #Þ!#(Ð!Ñ  #)Þ"* œ #)Þ"* Ð$#)Þ# billion) (b) B œ #!!!  "*)& œ "& T Ð"&Ñ œ !Þ!!")*Ð"&Ñ$  !Þ""*$Ð"&Ñ#  #Þ!#(Ð"&Ñ  #)Þ"* œ (*Þ!&)(& Ð$(*Þ" billion)

29.

(c) B œ #!!'  "*)& œ #" T Ð#"Ñ œ !Þ!!")*Ð#"Ñ$  !Þ""*$Ð#"Ñ#  #Þ!#(Ð#"Ñ  #)Þ"* œ "!&Þ)'&!" Ð$"!&Þ* billion) 13.

15.

œ

31.

Ð0  1ÑÐBÑ œ 0 ÐBÑ  1ÐBÑ œ ÐB#  *Ñ  Ð#BÑ œ B#  #B  *

19.

Ð0  1ÑÐ$Ñ œ 0 Ð$Ñ  1Ð$Ñ œ Ð$#  *Ñ  #Ð$Ñ œ!'œ' Alternatively, we could evaluate the polynomial in Exercise 17, B#  #B  *, using B œ $.

870

Ð1  2ш "# ‰ œ 1ˆ "# ‰  2ˆ "# ‰ œ #ˆ "# ‰  ˆ "#  $‰ œ  *#

33.

Answers will vary. Let 0 ÐBÑ œ B$ and 1ÐBÑ œ B% . Ð0  1ÑÐBÑ œ 0 ÐBÑ  1ÐBÑ œ B$  B % Ð1  0 ÑÐBÑ œ 1ÐBÑ  0 ÐBÑ œ B%  B $ Because the two differences are not equal, subtraction of polynomial functions is not commutative.

For Exercises 35–50, 0 ÐBÑ œ B#  %, 1ÐBÑ œ #B  $, and 2ÐBÑ œ B  &. 35.

Ð2 ‰ 1ÑÐ%Ñ œ 2Ð1Ð%ÑÑ œ 2Ð# • %  $Ñ œ 2Ð""Ñ œ ""  & œ '

37.

Ð1 ‰ 0 ÑÐ'Ñ œ 1Ð0 Ð'ÑÑ œ 1Ð'#  %Ñ œ 1Ð%!Ñ œ # • %!  $ œ )$

39.

Ð0 ‰ 2ÑÐ#Ñ œ 0 Ð2Ð#ÑÑ œ 0 Ð#  &Ñ œ 0 Ð(Ñ œ Ð(Ñ#  % œ &$

For Exercises 17–32, let 0 ÐBÑ œ B#  *, 1ÐBÑ œ #B, and 2ÐBÑ œ B  $. 17.

‰  ˆ "" %

œ "  ˆ (# ‰

(a) Ð0  1ÑÐBÑ œ 0 ÐBÑ  1ÐBÑ œ Ð%B#  )B  $Ñ  Ð&B#  %B  *Ñ œ B#  "#B  "# (b) Ð0  1ÑÐBÑ œ 0 ÐBÑ  1ÐBÑ œ Ð%B#  )B  $Ñ  Ð&B#  %B  *Ñ œ Ð%B#  )B  $Ñ  Ð&B#  %B  *Ñ œ *B#  %B  '

" #

œ  *%

(a) Ð0  1ÑÐBÑ œ 0 ÐBÑ  1ÐBÑ œ Ð&B  "!Ñ  Ð$B  (Ñ œ )B  $ (b) Ð0  1ÑÐBÑ œ 0 ÐBÑ  1ÐBÑ œ Ð&B  "!Ñ  Ð$B  (Ñ œ Ð&B  "!Ñ  Ð$B  (Ñ œ #B  "(

Ð1  2ш "% ‰ œ 1ˆ "% ‰  2ˆ "% ‰ œ #ˆ "% ‰  ˆ "%  $‰

Exponents, Polynomials, and Polynomial Functions 41.

Ð0 ‰ 1ÑÐ!Ñ œ 0 Ð1Ð!ÑÑ œ 0 #Ð!Ñ  $ œ 0 Ð$Ñ œ $#  % œ "$

43.

Ð1 ‰ 0 ÑÐBÑ œ 1Ð0 ÐBÑÑ œ 1ÐB#  %Ñ œ #ÐB#  %Ñ  $ œ #B#  )  $ œ #B#  ""

45.

47.

Any B-value can be used, so the domain is Ð∞ß ∞Ñ. From the graph, we see that any C-value can be obtained from the function, so the range is Ð∞ß ∞Ñ. 57.

Ð2 ‰ 1ÑÐBÑ œ 2Ð1ÐBÑÑ œ 2Ð#B  $ œ #B  $  & œ #B  # Ð0 ‰ 2ш "# ‰ œ 0 ˆ2ˆ "# ‰‰

œ 0 Ð "#  &Ñ

B # " ! " #

0 ÐBÑ œ $B# $Ð#Ñ# œ "# $Ð"Ñ# œ $ $Ð!Ñ# œ ! $Ð"Ñ# œ $ $Ð#Ñ# œ "#

Since the greatest exponent is #, the graph of 0 is a parabola.

œ 0 Ð *# Ñ œ Ð *# Ñ#  % œ

49.

)" %



"' %

œ

*( %

0 ‰ 1 ˆ "# ‰ œ 0 ˆ1ˆ "# ‰‰

œ 0 Ò#Ð "# Ñ  $Ó œ 0 Ð#Ñ œ ##  % œ%%œ)

51.

Any B-value can be used, so the domain is Ð∞ß ∞Ñ. From the graph, we see that the C-values are at most !, so the range is Ð∞ß !Ó.

0 ÐBÑ œ "#B, 1ÐBÑ œ &#)!B Ð0 ‰ 1ÑÐBÑ œ 0 1ÐBÑ œ 0 Ð&#)!BÑ œ "#Ð&#)!BÑ œ '$,$'!B

59.

Ð0 ‰ 1ÑÐBÑ computes the number of inches in B mi. 53.

Ñ œ #>, TÐÐ"&=$ >% Ñ œ $Ð"&Ñ=# • =$ • >" • >% œ %&=#  $ >"  % œ %&=& >& N2. (a) $5 $ Ð#5 &  $5 #  %Ñ œ $5 $ Ð#5 & Ñ  $5 $ Ð$5 # Ñ  $5 $ Ð%Ñ œ '5 )  *5 &  "#5 $

N8. 0 ÐBÑ œ $B#  ", 1ÐBÑ œ )B  ( Ð0 1ÑÐBÑ œ 0 ÐBÑ • 1ÐBÑ œ Ð$B#  "Ñ )B  ( œ #%B$  #"B#  )B  (

(b) &BÐ#B  "ÑÐB  %Ñ œ &BÒÐ#B  "ÑÐBÑ  Ð#B  "ÑÐ%ÑÓ œ &BÒ#B#  B  )B  %Ó œ &BÒ#B#  (B  %Ó œ "!B$  $&B#  #!B N3.

$>#  &> > #  *>  "&> $>$  &>#  %> $>$  "%>#  "*>

Using the last result, Ð0 1ÑÐ#Ñ œ #%Ð#Ñ$  #"Ð#Ñ#  )Ð#Ñ  ( œ "*#  )%  "'  ( œ **.

 %  $  "#

4 Section Exercises

 "#

N4. Ð$:  5ÑÐ&:  %5Ñ F O I L œ "&:#  "#5:  &5:  %5 # œ "&:#  (5:  %5 # N5. Use ÐB  CÑÐB  CÑ œ B#  C# .

1.

Ð#B  &ÑÐ$B  %Ñ F O I L œ #BÐ$BÑ  #BÐ%Ñ  Ð&ÑÐ$BÑ  Ð&ÑÐ%Ñ œ 'B#  )B  "&B  #! œ 'B#  (B  #! (Choice C)

3.

Ð#B  &ÑÐ$B  %Ñ F O I L œ #BÐ$BÑ  #BÐ%Ñ  Ð&ÑÐ$BÑ  Ð&ÑÐ%Ñ œ 'B#  )B  "&B  #! œ 'B#  #$B  #! (Choice D)

5.

)7$ Ð$7# Ñ œ )Ð$Ñ7$  # œ #%7&

7.

"%B# C$ Ð#B& CÑ œ "%Ð#ÑB#  & C$  " œ #)B( C%

9.

$BÐ#B  &Ñ œ $BÐ#BÑ  $BÐ&Ñ œ 'B#  "&B

11.

; $ Ð#  $;Ñ œ ; $ Ð#Ñ  ; $ Ð$;Ñ œ #; $  $; %

13.

'5 # Ð$5 #  #5  "Ñ œ '5 # Ð$5 # Ñ  '5 # Ð#5Ñ  '5 # Ð"Ñ œ ")5 %  "#5 $  '5 #

(a) Ð$B  (CÑÐ$B  (CÑ œ Ð$BÑ#  Ð(CÑ# œ $# B#  ( # C # œ *B#  %*C# (b) &5Ð#5  $ÑÐ#5  $Ñ œ &5ÒÐ#5Ñ#  $# Ó œ &5Ò%5 #  *Ó œ #!5 $  %&5 N6. Use ÐB  CÑ# œ B#  #BC  C# or ÐB  CÑ# œ B#  #BC  C# . (a) ÐC  "!Ñ# œ C#  # • C • "!  "!# œ C#  #!C  "!! #

#

(b) Ð%B  &CÑ œ Ð%BÑ  #Ð%BÑÐ&CÑ  Ð&CÑ œ "'B#  %!BC  #&C#

872

#

Exponents, Polynomials, and Polynomial Functions 15.

17.

19.

21.

23.

Ð#>  $ÑÐ$>#  %>  "Ñ œ #>Ð$>#  %>  "Ñ  $Ð$>#  %>  "Ñ œ #>Ð$># Ñ  #>Ð%>Ñ  #>Ð"Ñ  $Ð$># Ñ  $Ð%>Ñ  $Ð"Ñ œ '>$  )>#  #>  *>#  "#>  $ œ '>$  )>#  *>#  #>  "#>  $ œ '>$  >#  "%>  $

31.   ):  ': %  *: $  ': %  : $  $

   

$:  ' %:  " $:  ' #%:

 #(:  '

33.

Ð7  &ÑÐ7  )Ñ F O I L # œ 7  )7  &7  %! œ 7#  $7  %!

35.

Ð%5  $ÑÐ$5  #Ñ F O I L # œ "#5  )5  *5  ' œ "#5 #  5  '

37.

%B$ ÐB  $ÑÐB  #Ñ œ %B$ ÐB#  #B  $B  'Ñ œ %B$ ÐB#  B  'Ñ œ %B$ ÐB# Ñ  %B$ ÐBÑ  %B$ Ð'Ñ œ %B&  %B%  #%B$

ÐD  AÑÐ$D  %AÑ F O I L œ $D #  %DA  $DA  %A# œ $D #  DA  %A#

39.

Ð#C  $ÑÐ$C  %Ñ Rewrite vertically and multiply.

Ð'-  .ÑÐ#-  $.Ñ F O I L œ "#- #  ")-.  #-.  $. # œ "#- #  "'-.  $. #

41.

A description of the FOIL method is as follows: The product of two binomials is the sum of the product of the first terms, the product of the outer terms, the product of the inner terms, and the product of the last terms.

43.

Use the formula for the product of the sum and difference of two terms.

7Ð7  &ÑÐ7  )Ñ œ 7Ð7#  )7  &7  %!Ñ œ 7Ð7#  $7  %!Ñ œ 7Ð7# Ñ  7Ð$7Ñ  7Ð%!Ñ œ 7$  $7#  %!7 %DÐ#D  "ÑÐ$D  %Ñ œ %DÐ'D #  )D  $D  %Ñ œ %DÐ'D #  &D  %Ñ œ %DÐ'D # Ñ  %DÐ&DÑ  %DÐ%Ñ œ #%D $  #!D #  "'D

#C  $ $C  %  )C  "# o %Ð#C  $Ñ 'C#  *C o $CÐ#C  $Ñ Combine 'C#  C  "# like terms. , #  $ ,  $ #,  %

25.

ÐB  *ÑÐB  *Ñ œ B#  *# œ B#  )"

 %, #  "#,  "# #,  ', #  ',

45.

$

#, $  #, #  "),  "# 27.

"&78  *8# #&7  "&78 #&7#

 *8# $

#

#D  &D  )D %D 'D $  "&D #  #%D )D %  #!D $  $#D #  %D )D %  "%D $  "(D #  #!D

Use the formula for the product of the sum and difference of two terms. Ð#:  $ÑÐ#:  $Ñ œ Ð#:Ñ#  $# œ %:#  *

&7  $8 &7  $8 47.

Ð&7  "ÑÐ&7  "Ñ œ Ð&7Ñ#  "# œ #&7#  "

49.

Ð$+  #-ÑÐ$+  #-Ñ œ Ð$+Ñ#  Ð#-Ñ# œ *+#  %- #

51.

Ð%7  (8# ÑÐ%7  (8# Ñ œ Ð%7Ñ#  Ð(8# Ñ #3 œ "'7#  %*8%

#

29.

#: # $:# #: # "#:# "):# %: #

 "  $  $  $

873

Exponents, Polynomials, and Polynomial Functions 53.

$C &C$  # &C$  # First multiply &C$  # &C$  # .

75.

Ð&C$  #ÑÐ&C$  #Ñ œ Ð&C$ Ñ #3  Ð#Ñ# œ #&C'  % Now multiply the last polynomial times $C.

 '!BC  "#C  $'C# Square of a binomial œ #&B#  "!B  "  '!BC  "#C  $'C#

$C #&C'  % œ (&C(  "#C 55.

Use the formula for the square of a binomial. ÐC  &Ñ# œ C#  # • C • &  &# œ C#  "!C  #&

57.

77.

Use the formula for the square of a binomial. ÐB  "Ñ# œ B#  # • B • "  "# œ B#  #B  "

59.

Ð#:  (Ñ# œ #: #  #Ð#:ÑÐ(Ñ  (# œ %:#  #):  %*

61.

Ð%8  $7Ñ# œ Ð%8Ñ#  #Ð%8ÑÐ$7Ñ  Ð$7Ñ# œ "'8#  #%87  *7#

63.

To find the product "!" • ** using the special product rule

79.

81.

83.

"!" • ** œ Ð"!!  "ÑÐ"!!  "Ñ œ "!!#  "# œ "!,!!!  " œ ****. 65.

67.

Ð!Þ#B  "Þ$ÑÐ!Þ&B  !Þ"Ñ F O I L # œ !Þ"B  !Þ!#B  !Þ'&B  !Þ"$ œ !Þ"B#  !Þ'$B  !Þ"$

ˆ$A  "% D ‰ÐA  #DÑ F O I L œ $A#  'AD  "% AD  "# D # œ $A# 

69.

#$ % AD

œ "'B# 

71.

ˆ5  &( :‰# œ 5 #  #Ð5ш &( :‰  ˆ &( :‰# œ 5# 

73.

874

% *

"! ( 5:



#& # %* :

Ð!Þ#B  "Þ%CÑ# œ Ð!Þ#BÑ#  #Ð!Þ#BÑÐ"Þ%CÑ  Ð"Þ%CÑ# œ !Þ!%B#  !Þ&'BC  "Þ*'C#

c #+  ,  $dc #+  ,  $d œ #+  , #  $# Product of the sum and difference of two terms œ  #+ #  #Ð#+ÑÐ,Ñ  , # ‘  * Square of a binomial œ %+#  %+,  , #  *

cÐ#2  5Ñ  4dcÐ#2  5Ñ  4d œ Ð#2  5Ñ#  4# œ Ð#2Ñ#  #Ð#2ÑÐ5Ñ  5 #  4# œ %2#  %25  5 #  4#

ÐC  #Ñ$ œ ÐC  #Ñ# ÐC  #Ñ œ ÒC#  #ÐCÑÐ#Ñ  ## ÓÐC  #Ñ œ ÐC#  %C  %ÑÐC  #Ñ C #  %C C # #C  )C C $  %C #  %C C$  'C#  "#C

 %  #  )  )

$

85.

Ð&<  =Ñ œ Ð&<  =Ñ# Ð&<  =Ñ œ Ð#&  '>#

Exponents, Polynomials, and Polynomial Functions 60.

61.

62.

Ð#:#  ':ÑÐ&:#  %Ñ F O I L œ "!:%  ):#  $!:$  #%: œ "!:%  $!:$  ):#  #%:

&:%  "&:$  $$:#  *:  ") &:#  $

68.

&:

Ð$; #  #;  %ÑÐ;  &Ñ œ Ð$; #  #;  %ÑÐ;Ñ  Ð$; #  #;  %ÑÐ&Ñ œ $; $  #; #  %;  "&; #  "!;  #! œ $; $  #; #  "&; #  %;  "!;  #! œ $; $  "$; #  "%;  #! Ð'  %

(b) %# œ Ð%# Ñ œ "'

Now multiply the last polynomial by >.

(c) %! œ "

69.

>Ð$>  #Ñ# œ *>$  "#>#  %>

66.

[1]

" " œ Ð%Ñ# "'

 %B  '  #'B  $!  #'B  #!B 'B  $! 'B  $! !

:#  *: #  $: # "#:# "#:#

Answer: :#  ':  * 

 

':  * !:  #(

 !:  "): "):  #( "):  #( &% Remainder

&% #:  $

(A)

(E)

(g) %!  %! œ "  " œ !

(B)

(h) %!  %! œ Ð"Ñ  " œ # (i) %#  %" œ (j) %# œ "'

(H)

" " " % &  œ  œ %# % "' "' "'

(F)

(I)

'" C$ C# # C$ C% œ 'C% C" '" • 'C% C" C$  %  Ð%  "Ñ œ $' C"  & C% œ œ $' $' 3

70.

[1]

71.

[1]

&$ œ

72.

[1]

Ð$Ñ# œ Ð*Ñ œ *

73.

[4]

(:& Ð$:%  :$  #:# Ñ œ (:& Ð$:% Ñ  (:& Ð:$ Ñ  (:& Ð#:# Ñ œ #":*  (:)  "%:(

74.

[4]

Ð#B  *Ñ# œ Ð#BÑ#  #Ð#BÑÐ*Ñ  *# œ %B#  $'B  )"

#:$  *:#  #( #:  $ #:  $ #: #: $

(G)

(C)

(f) %! œ  %! œ "

Answer: B#  %B  '

$

" " œ (A) # % "'

(C)

(e) Ð%Ñ# œ

B$  *B#  #'B  $! B& B B  & B$  *B# B$  &B# %B# %B#

67.

(a) %# œ

(d) Ð%Ñ! œ "

%C$  "#C#  &C %C$ "#C# &C œ   %C %C %C %C # & œ C  $C  %

#

 ")  ") !

Answer: :#  $:  '

63.

65.

 '  ")

" " " œ œ &$ &•&•& "#&

885

Exponents, Polynomials, and Polynomial Functions 75.

[1]

ÐD # Ñ $ Ð"Ñ$ D #Ð$Ñ œ  " $ &D $Ð"Ñ &ÐD Ñ D ' œ &D $ D '  $ œ & D * " œ œ * & &D 3

82.

3

76.

77.

[1]

[5]

ÐC' Ñ &3 Ð#C$ Ñ %3 œ C$! Ð#Ñ% C"# C$!  "# œ #% C") " œ % œ # "'C") )B # B  $ )B  #$B )B#  #%B B B

78.

[1]

80.

81.

886

[4]

[5]

[2]

The area is approximately "!$,$(" mi# .

Test  "  #

1.

(a) (# œ

" " œ (C) # ( %*

(b) (! œ " (A) (c) (! œ Ð"Ñ œ " (D)

 #  $ &

(d) Ð(Ñ! œ " (A) (e) (# œ %* (E) " "  ( # # ( * œ  œ (F) "% "% "%

(f) ("  #" œ

3

3

79.

T H %Þ#"$% ‚ "!' œ %!Þ(' %Þ#"$% ‚ "!' œ %Þ!(' ‚ "!" ¸ "Þ!$$(" ‚ "!& œ "!$,$("

Ð&D # B$ Ñ # Ð#DB# Ñ " Ð"!DB$ Ñ # Ð$D " B% Ñ # &# D % B' #" D " B# œ Ð"!Ñ# D # B' $# D # B) #&Ð"!Ñ# D $ B% œ # • *D % B# #&Ð"!!ÑD ( B' œ #•* "#&!D ( B' œ * 3

See the solution for Exercise 167 in Section

Eœ

& B$

Answer: )B  " 

[1] 1.

3

cÐ$7  &8Ñ  :dcÐ$7  &8Ñ  :d œ Ð$7  &8Ñ#  Ð:Ñ# œ Ð$7Ñ#  #Ð$7ÑÐ&8Ñ  Ð&8Ñ#  :# œ *7#  $!78  #&8#  :#

(g) Ð(  #Ñ" œ *" œ (h)

Ð#5  "Ñ  Ð$5  #5  'Ñ œ #5  "  $5 #  #5  ' œ $5 #  #5  #5  "  ' œ $5 #  %5  (

(" #" # œ œ (G) #" (" (

(i) (# œ %* (I) (j) Ð(Ñ# œ 2.

#!C$ B$  "&C% B  #&CB% "!CB# $ $ #!C B "&C% B #&CB% œ   "!CB# "!CB# "!CB# $ # $C &B œ #C# B   #B # #

" (B) *

3.

" " œ (C) # Ð(Ñ %*

Ð$B# C$ Ñ #3 Ð%B$ C% Ñ œ $# B#Ð#Ñ C$Ð#Ñ %B$ C% œ $# B% C' %B$ C% %B%  $ C'  % œ $# ( "! %B C %B( œ œ "! * *C $'Ñ# œ #&B#  %>#

71.

Ð$C$  %ÑÐ#C$  $Ñ œ 'C'  *C$  )C$  "# œ 'C'  C$  "#

Factoring Trinomials

2 Now Try Exercises Step 2 Write sums of those integers. $!  " œ #* $!  Ð"Ñ œ #* "&  # œ "$ "&  Ð#Ñ œ "$ "!  $ œ ( "!  Ð$Ñ œ ( '  & œ " í '  Ð&Ñ œ "

The integers ' and & have the necessary product, $!, and sum, ", so >#  >  $!

factors as Ð>  'ÑÐ>  &Ñ.

Check Ð>  'ÑÐ>  &Ñ œ >#  >  $!

Look for two integers whose product is "" and whose sum is "#. Only two pairs of integers, "" and " and "" and ", give a product of "". Neither of these pairs has a sum of "#, so 7#  "#7  "" cannot be factored with integer coefficients and is prime.

Look for two expressions whose product is #!, # and whose sum is ,Þ The expressions &, and %, have the necessary product and sum, so +#  +,  #!, #

factors as Ð+  &,ÑÐ+  %,Ñ.

N4. &A$  %!A#  '!A Start by factoring out the GCF, &A.

To factor A#  )A  "#, look for two integers whose product is "# and whose sum is ). The necessary integers are ' and #. Remember to write the common factor &A as part of the answer. &A$  %!A#  '!A œ &AÐA  'ÑÐA  #Ñ N5. )C#  "!C  $ Since + œ ), , œ "!, and - œ $, the product +- is )Ð$Ñ œ #%. The two integers whose product is #% and whose sum is "!, are "# and #. Write "!C as "#C  #C and then factor by grouping. )C#  "!C  $ œ )C#  "#C  #C  $ œ Ð)C#  "#CÑ  #C  $ œ %CÐ#C  $Ñ  "Ð#C  $Ñ œ Ð#C  $ÑÐ%C  "Ñ N6. Use trial and error to factor the trinomial

(b) A#  "#A  $# Step 1 Find pairs of integers whose product is $#. $#Ð"Ñ $#Ð"Ñ "'Ð#Ñ "'Ð#Ñ )Ð%Ñ )Ð%Ñ

7#  "#7  ""

&A$  %!A#  '!A œ &AÐA#  )A  "#Ñ

N1. (a) >#  >  $! Step 1 Find pairs of integers whose product is $!. $!Ð"Ñ $!Ð"Ñ "&Ð#Ñ "&Ð#Ñ "!Ð$Ñ "!Ð$Ñ 'Ð&Ñ 'Ð&Ñ

factors as ÐA  )ÑÐA  %Ñ.

Step 2 Write sums of those integers. $#  " œ $$ $#  Ð"Ñ œ $$ "'  # œ ") "'  Ð#Ñ œ ") )  % œ "# í )  Ð%Ñ œ "#

The integers ) and % have the necessary product, $#, and sum, "#, so

"!  "# Ñ œ Ð>  "ÑÐ>#  >  "Ñ

Factoring (b) "#&+$  ), $ œ Ð&+Ñ$  Ð#,Ñ$ œ Ð&+  #,ÑÒÐ&+Ñ#  &+Ð#,Ñ  Ð#,Ñ# Ó œ Ð&+  #,ÑÐ#&+#  "!+,  %, # Ñ

9.

#&B#  % œ Ð&BÑ#  ## œ Ð&B  #ÑÐ&B  #Ñ

11.

")+#  *), # œ #Ð*+#  %*, # Ñ œ #ÒÐ$+Ñ#  Ð(,Ñ# Ó œ #cÐ$+  (,ÑÐ$+  (,Ñd œ #Ð$+  (,ÑÐ$+  (,Ñ

13.

'%7%  %C% œ %Ð"'7%  C% Ñ œ Ò%Ð%7# Ñ#  ÐC# Ñ# Ó Factor the difference of squares. œ %Ð%7#  C# ÑÐ%7#  C# Ñ œ %Ð%7#  C# ÑÒÐ#7Ñ#  C# Ó Factor the difference of squares again. œ %Ð%7#  C# ÑÐ#7  CÑÐ#7  CÑ

15.

ÐC  DÑ#  )" œ ÐC  DÑ#  *# œ cÐC  DÑ  *dcÐC  DÑ  *d œ ÐC  D  *ÑÐC  D  *Ñ

17.

"'  ÐB  $CÑ# œ %#  D # Let D œ ÐB  $CÑ. œ Ð%  DÑÐ%  DÑ Substitute B  $C for D . œ c%  ÐB  $CÑdc%  ÐB  $CÑd œ Ð%  B  $CÑÐ%  B  $CÑ

19.

:%  #&' œ Ð:# Ñ#  "'# œ Ð:#  "'ÑÐ:#  "'Ñ œ Ð:#  "'ÑÐ:#  %# Ñ œ Ð:#  "'ÑÐ:  %ÑÐ:  %Ñ

21.

5 #  '5  * œ Ð5Ñ#  #Ð5ÑÐ$Ñ  $# œ Ð5  $Ñ#

23.

%D #  %DA  A# œ Ð#DÑ#  #Ð#DÑÐAÑ  A# œ Ð#D  AÑ#

25.

"'7#  )7  "  8# Group the first three terms. œ Ð"'7#  )7  "Ñ  8# œ ÒÐ%7Ñ#  #Ð%7ÑÐ"Ñ  "# Ó  8# œ Ð%7  "Ñ#  8# œ cÐ%7  "Ñ  8dcÐ%7  "Ñ  8d œ Ð%7  "  8ÑÐ%7  "  8Ñ

27.

%ÑÐ'#  ' • >  ># Ñ œ Ð'  >ÑÐ$'  '>  ># Ñ

41.

B$  '% œ B$  %$ œ ÐB  %ÑÐB#  B • %  %# Ñ œ ÐB  %ÑÐB#  %B  "'Ñ

43.

"!!!  C$ œ "!$  C$ œ Ð"!  CÑÐ"!#  "! • C  C# Ñ œ Ð"!  CÑÐ"!!  "!C  C# Ñ

45.

)B$  " œ Ð#BÑ$  "$ œ Ð#B  "ÑÒÐ#BÑ#  #B • "  "# Ó œ Ð#B  "ÑÐ%B#  #B  "Ñ

47.

"#&B$  #"' œ Ð&BÑ$  '$ œ Ð&B  'ÑÒÐ&BÑ#  &B • '  '# Ó œ Ð&B  'ÑÐ#&B#  $!B  $'Ñ

49.

B$  )C$ œ B$  Ð#CÑ$ œ ÐB  #CÑÒB#  B • #C  Ð#CÑ# Ó œ ÐB  #CÑÐB#  #BC  %C# Ñ

51.

896

B#  C#  #C  " Group the last three terms. œ B#  ÐC#  #C  "Ñ œ B#  ÐC  "Ñ# œ cB  ÐC  "ÑdcB  ÐC  "Ñd œ ÐB  C  "ÑÐB  C  "Ñ

'%1$  #(2$ œ Ð%1Ñ$  Ð$2Ñ$ œ Ð%1  $2ÑÒÐ%1Ñ#  %1 Ð$2Ñ  Ð$2Ñ# Ó œ Ð%1  $2ÑÐ"'1#  "#12  *2# Ñ

Factoring ÐB#  BC  C# ÑÐB#  BC  C# Ñ œ B# ÐB#  BC  C# Ñ  BCÐB#  BC  C# Ñ  C# ÐB#  BC  C# Ñ œ B%  B$ C  B # C #  B $ C  B # C #  BC$  B# C#  BC$  C% œ B%  B# C #  C %

71.

73.

75.

77.

#

81.

:#  %:  #"

:#  %:  #" œ Ð:  (ÑÐ:  $Ñ.

Start by factoring as the difference of squares since doing so resulted in the complete factorization more directly. $

#+B  +C  #,B  ,C œ Ð#+B  +CÑ  Ð#,B  ,CÑ œ +Ð#B  CÑ  ,Ð#B  CÑ œ Ð#B  CÑÐ+  ,Ñ

By trial and error,

They are equal. 70.

79.

$

#

"#&:  #&:  );  %; œ Ð"#&:$  ); $ Ñ  Ð#&:#  %; # Ñ œ ÒÐ&:Ñ$  Ð#;Ñ$ Ó  ÒÐ&:Ñ#  Ð#;Ñ# Ó Factor within groups. œ ÒÐ&:  #; ÑÐ#&:#  "!:;  %; # ÑÓ  cÐ&:  #; ÑÐ&:  #; Ñd Factor out the GCF, &:  #; . œ Ð&:  #; Ñ # # • ÒÐ#&:  "!:;  %; Ñ  Ð&:  #; ÑÓ œ Ð&:  #; Ñ # # • Ð#&:  "!:;  %;  &:  #;Ñ #(+$  "&+  '%, $  #!, œ Ð#(+$  '%, $ Ñ  Ð"&+  #!,Ñ œ ÒÐ$+Ñ$  Ð%,Ñ$ Ó  Ð"&+  #!,Ñ Factor within groups. œ Ð$+  %,ÑÐ*+#  "#+,  "', # Ñ  &Ð$+  %,Ñ Factor out the GCF, $+  %, . œ Ð$+  %,ÑÒÐ*+#  "#+,  "', # Ñ  &Ó œ Ð$+  %,ÑÐ*+#  "#+,  "', #  &Ñ )>%  #%>$  >  $ œ Ð)>%  #%>$ Ñ  Ð>  $Ñ œ )>$ Ð>  $Ñ  "Ð>  $Ñ Factor out the GCF, >  $. œ Ð>  $ÑÐ)>$  "Ñ Factor the sum of cubes. œ Ð>  $ÑÐ#>  "ÑÐ%>#  #>  "Ñ '%7#  &"#7$  )"8#  (#*8$ œ Ð'%7#  )"8# Ñ  Ð&"#7$  (#*8$ Ñ œ ÒÐ)7Ñ#  Ð*8Ñ# Ó  ÒÐ)7Ñ$  Ð*8Ñ$ Ó Factor within groups. œ Ð)7  *8ÑÐ)7  *8Ñ  Ð)7  *8ÑÐ'%7#  (#78  )"8# Ñ Factor out the GCF, )7  *8. œ Ð)7  *8Ñ # # • ÒÐ)7  *8Ñ  Ð'%7  (#78  )"8 ÑÓ œ Ð)7  *8Ñ # # • Ð)7  *8  '%7  (#78  )"8 Ñ

4

A General Approach to Factoring

4 Now Try Exercises N1. (a) #"B$ C#  #(B# C% œ $B# C# Ð(B  *C# Ñ

GCF = 3x2 y2

(b) )CÐ7  8Ñ  &Ð7  8Ñ œ Ð7  8ÑÐ)C  &Ñ

GCF = m  n

N2. (a) %+#  %*, # œ Ð#+Ñ#  Ð(,Ñ#

Difference of squares

œ Ð#+  (,ÑÐ#+  (,Ñ (b) *B#  "!! is a sum of squares and cannot be factored. The binomial is prime. (c) #(@$  "!!! Difference of cubes œ Ð$@  "!ÑÒÐ$@Ñ#  $@ • "!  "!# Ó œ Ð$@  "!ÑÐ*@#  $!@  "!!Ñ œ Ð$@Ñ$  "!$

N3. (a) #&B#  *!B  )" œ Ð&BÑ#  #Ð&BÑÐ*Ñ  *# Perfect square trinomial œ Ð&B  *Ñ# (b) (B#  (BC  )%C# œ (ÐB#  BC  "#C# Ñ Two integer factors whose product is "# and whose sum is " are % and $. œ (ÒB#  %BC  $BC  "#C# Ó œ (cBÐB  %CÑ  $CÐB  %CÑd œ (cÐB  %CÑÐB  $CÑd œ (ÐB  %CÑÐB  $CÑ (c) "#7#  &7  #) Two integer factors whose product is "#Ð#)Ñ œ $$' and whose sum is & are #" and "'. œ "#7#  #"7  "'7  #) œ $7Ð%7  (Ñ  %Ð%7  (Ñ œ Ð%7  (ÑÐ$7  %Ñ

897

Factoring N4. (a) &+$  &+# ,  +, #  , $ œ Ð&+$  &+# ,Ñ  Ð+, #  , $ Ñ œ &+# Ð+  ,Ñ  , # Ð+  ,Ñ œ Ð+  ,ÑÐ&+#  , # Ñ (b) *?#  %)?  '%  @# œ Ð*?#  %)?  '%Ñ  @# œ Ð$?  )Ñ#  @# œ ÒÐ$?  )Ñ  @ÓÒÐ$?  )Ñ  @Ó œ Ð$?  )  @ÑÐ$?  )  @Ñ (c) B$  *C#  #(C$  B# œ ÐB$  #(C$ Ñ  ÐB#  *C# Ñ œ ÒB$  Ð$CÑ$ Ó  ÒB#  Ð$CÑ# Ó œ ˜ÐB  $CÑÒB#  B • $C  Ð$CÑ# Ó™  ÒÐB  $CÑÐB  $CÑÓ œ ÒÐB  $CÑÐB#  $BC  *C# Ó  ÒÐB  $CÑÐB  $CÑÓ œ ÐB  $CÑÐB#  $BC  *C#  B  $CÑ

4 Section Exercises 1.

5.

898

19.

*7#  $!78  #&8# œ Ð$7Ñ#  #Ð$7ÑÐ&8Ñ  Ð&8Ñ# Perfect square trinomial œ Ð$7  &8Ñ#

21.

5;  *;  5<  *< œ ;Ð5  *Ñ  # Ñ

899

Factoring 69.

71.

73.

Step 4 Solve each equation. #B œ $ or $B œ " B œ  $# B œ "$ Step 5 Check each solution in the original equation.

7#  %7  %  8#  '8  * œ Ð7#  %7  %Ñ  Ð8#  '8  *Ñ œ Ò7#  #Ð#Ñ7  ## Ó  Ò8#  #Ð$Ñ8  $# Ó Perfect square trinomials # œ Ð7  #Ñ  Ð8  $Ñ# œ cÐ7  #Ñ  Ð8  $Ñd • cÐ7  #Ñ  Ð8  $Ñd Difference of two squares œ Ð7  #  8  $ÑÐ7  #  8  $Ñ œ Ð7  8  &ÑÐ7  8  "Ñ $B  # œ ! $B œ # # B œ # $ œ $

Check B œ  $# :  #" # œ Check B œ

N4.

Solving Equations by Factoring

Check B œ &: Check B œ (% :

%B  ( œ ! %B œ ( B œ (%

!Ð#(Ñ œ ! #( % Ð!Ñ œ !

™ The solution set is ˜&ß #( % .

N2. (a) Step 1 Standard form (B œ $  'B# # 'B  (B  $ œ ! Step 2 Factor. Ð#B  $ÑÐ$B  "Ñ œ !

900

!!œ! %)  %) œ !

B&œ!

True True

True True

Factor out 4. Difference of squares

or B  & œ ! or

Check B œ „ &:

N1. ÐB  &ÑÐ%B  (Ñ œ !

or

True

B%œ! B œ %

%B#  "!! œ ! %ÐB#  #&Ñ œ ! %ÐB  &ÑÐB  &Ñ œ !

B œ &

5 Now Try Exercises or

True

Standard form Factor. Zero-factor prop. Subtract 5. Divide by 4.

The solution set is e%ß !f.

The solution set is Ö"!×.

B œ &

or

Check B œ !: Check B œ %:

œ &

B&œ!

True

The solution set is ˜ &% ™.

$B œ ! Bœ!

#Ð "# BÑ œ #Ð&Ñ B œ "!

5

#( # # $

Check B œ  &% : "'Ð #& "' Ñ  &!  #& œ !

&œ! " #B



(b) "'B#  %!B  #& œ ! Ð%B  &Ñ# œ ! %B  & œ ! %B œ & B œ  &%

&B œ ! B œ !& œ ! " #B

œ



The solution set is ˜ $# ß "$ ™.

The solution set is Ö!×. 77.

( $

" $:

' # * $

N3. $B#  "#B œ ! $BÐB  %Ñ œ !

The solution set is ˜ #$ ™. 75.

Step 3 Zero-factor property #B  $ œ ! or $B  " œ !

B&  $B%  B  $ œ B% ÐB  $Ñ  "ÐB  $Ñ œ ÐB  $ÑÐB%  "Ñ œ ÐB  $ÑÒÐB# Ñ#  "# Ó œ ÐB  $ÑÐB#  "ÑÐB#  "Ñ œ ÐB  $ÑÐB#  "ÑÒB#  "# Ó œ ÐB  $ÑÐB#  "ÑÐB  "ÑÐB  "Ñ

Zero-factor property

Bœ& %Ð#&Ñ  "!! œ !

The solution set is e&ß &f.

True

N5. ÐB  $ÑÐ#B  "Ñ œ %ÐB  %Ñ  % #B#  &B  $ œ %B  "'  % #B#  B  "& œ ! ÐB  $ÑÐ#B  &Ñ œ ! B$œ! B œ $ Check B œ $: Check B œ &# :

or or

#B  & œ ! B œ &#

!œ%% œ #'  %

"" # Ð%Ñ

The solution set is ˜$ß &# ™.

True True

Factoring N6.

"#B œ #B$  &B# #B$  &B#  "#B œ ! BÐ#B#  &B  "#Ñ œ ! BÐB  %ÑÐ#B  $Ñ œ ! Bœ!

or

or or

B%œ! B œ %

5 Section Exercises 1.

#B  $ œ ! B œ $#

!œ! Check B œ !: Check B œ %: %) œ "#)  )! %& ") œ #( Check B œ $# : %  %

The solution set is ˜%ß !ß

$™ # .

True True True

In the exercises in this section, check all solutions to the equations by substituting them back in the original equations. 3.

The triangle cannot have a negative base, so reject B œ  (# . The base is % ft and the height is #Ð%Ñ  " œ ( ft. The area of the triangle is " # # Ð%ÑÐ(Ñ œ "% ft , as required. N8.

#

5.

or or

7.

$5  ) œ ! $5 œ ) 5 œ  )$

or

B#  $B  "! œ ! ÐB  #ÑÐB  &Ñ œ ! Use the zero-factor property. B#œ! B œ #

9.

>'œ! >œ'

or or

B&œ! Bœ&

The solution set is e#ß &f. B#  *B  ") œ ! ÐB  'ÑÐB  $Ñ œ !

or or

B'œ! B œ '

11.

Solve T œ #L[  #P[  #PL for L . T  #P[ œ #L[  #PL Get the H-terms on one side. T  #P[ œ LÐ#[  #PÑ Factor out H. LÐ#[  #PÑ T  #P[ œ #[  #P #[  #P Divide by 2[  2L. T  #P[ T  #P[ œ L , or L œ #[  #P #[  #P

or

The solution set is ˜ )$ ß &# ™.

The rocket will reach a height of "*# feet after # seconds (on the way up) and again after ' seconds (on the way down). N9.

Ð#5  &ÑÐ$5  )Ñ œ ! #5  & œ ! #5 œ & 5 œ &#

2Ð>Ñ œ "'>  "#)> "*# œ "'>#  "#)> # "'>  "#)>  "*# œ ! >#  )>  "# œ ! Divide by 16. Ð>  #ÑÐ>  'Ñ œ ! Factor. >#œ! >œ#

B&œ! Bœ&

The solution set is Ö"!ß &×.

 "Ñ œ "% BÐ#B  "Ñ œ #) #B#  B  #) œ ! Ð#B  (ÑÐB  %Ñ œ ! or B  % œ ! or Bœ%

or or

Check B œ "!: !Ð"&Ñ œ ! True Check B œ &: "& Ð!Ñ œ ! True

" # BÐ#B

#B  ( œ ! B œ  (#

ÐB  "!ÑÐB  &Ñ œ ! B  "! œ ! B œ "!

N7. Let B œ the base of the triangle. Then #B  " œ the height. Use the formula "# ,2 œ T, where , œ B, 2 œ #B  ", and T œ "%.

First rewrite the equation so that one side is !. Factor the other side and set each factor equal to !. The solutions of these linear equations are solutions of the quadratic equation.

B$œ! B œ $

The solution set is e'ß $f. #B# œ (B  % Get ! on one side. #B#  (B  % œ ! Ð#B  "ÑÐB  %Ñ œ ! or

#B  " œ ! #B œ " B œ  "#

B%œ! Bœ%

The solution set is ˜ "# ß %™. 13.

"&B#  (B œ % "&B  (B  % œ ! Ð$B  "ÑÐ&B  %Ñ œ ! #

$B  " œ ! $B œ " B œ  "$ The solution set is

or or

&B  % œ ! &B œ % B œ %&

˜ "$ ß %& ™.

901

Factoring 15.

#B#  "#  %B œ B#  $B B#  B  "# œ ! ÐB  $ÑÐB  %Ñ œ ! or or

B$œ! B œ $

29.

B%œ! Bœ%

The solution set is e$ß %f. 17.

Ð&B  "ÑÐB  $Ñ œ #Ð&B  "Ñ Ð&B  "ÑÐB  $Ñ  #Ð&B  "Ñ œ ! Ð&B  "ÑcÐB  $Ñ  #d œ ! Ð&B  "ÑÐB  &Ñ œ ! or

&B  " œ ! &B œ " B œ  "&

The solution set is ˜ %$ ™. 31.

B&œ! B œ &

%:#  "': œ ! %:Ð:  %Ñ œ ! %: œ ! :œ!

21.

33. or

or or

35.

%:  "' œ ! %Ð:#  %Ñ œ ! %Ð:  #ÑÐ:  #Ñ œ !

$B#  #( œ ! $ÐB#  *Ñ œ ! $ÐB  $ÑÐB  $Ñ œ !

37.

27.

or or

B# œ *  'B ! œ B#  'B  * ! œ ÐB  $ÑÐB  $Ñ !œB$ $œB

The solution set is e$f. 902

B'œ! Bœ'

#B$  *B#  &B œ ! BÐ#B#  *B  &Ñ œ ! BÐ#B  "ÑÐB  &Ñ œ ! or

or

#B  " œ ! #B œ " B œ  "#

B&œ! Bœ&

The solution set is ˜ "# ß !ß &™.

B$œ! Bœ$

The solution set is e$ß $f.

or or

The solution set is e"ß 'f.

Bœ!

Note that the leading $ does not affect the solution set of the equation. B$œ! B œ $

B'œ! Bœ'

ÐB  $ÑÐB  'Ñ œ Ð#B  #ÑÐB  'Ñ B#  $B  ") œ #B#  "!B  "# ! œ B#  (B  ' ! œ ÐB  "ÑÐB  'Ñ B"œ! Bœ"

:#œ! :œ#

The solution set is e#ß #f.

or

The solution set is ˜ "# ß '™.

#

or or

B#œ! Bœ#

Ð#B  "ÑÐB  $Ñ œ 'B  $ #B#  'B  B  $ œ 'B  $ #B#  &B  $ œ 'B  $ #B#  ""B  ' œ ! Ð#B  "ÑÐB  'Ñ œ ! #B  " œ ! #B œ " B œ  "#

B'œ! Bœ'

The solution set is e!ß 'f.

or or

The solution set is e%ß #f.

'B#  $'B œ ! 'BÐB  'Ñ œ !

:#œ! : œ #

25.

:%œ! : œ %

The solution set is e%ß !f.

'B œ ! Bœ!

23.

ÐB  $ÑÐB  &Ñ œ ( Multiply the factors, and then add ( on both sides of the equation to get ! on the right. B#  &B  $B  "& œ ( B#  #B  ) œ ! Now factor the polynomial. ÐB  %ÑÐB  #Ñ œ ! B%œ! B œ %

The solution set is ˜&ß  "& ™. 19.

*B#  #%B  "' œ ! Ð$B  %ÑÐ$B  %Ñ œ ! $B  % œ ! $B œ % B œ  %$

39.

B$  #B# œ $B B  #B#  $B œ ! BÐB#  #B  $Ñ œ ! BÐB  $ÑÐB  "Ñ œ ! $

Bœ!

or or

B$œ! Bœ$

or

The solution set is Ö"ß !ß $×.

B"œ! B œ "

Factoring 41.

*B$ œ "'B *B$  "'B œ ! BÐ*B#  "'Ñ œ ! BÐ$B  %ÑÐ$B  %Ñ œ ! or

Bœ!

43.

˜ %$ ß !ß %$ ™.

or or

$B  % œ ! $B œ % B œ %$

#B  &B  #B  & œ ! Factor by grouping. Ð#B$  #BÑ  Ð&B#  &Ñ œ ! #BÐB#  "Ñ  &ÐB#  "Ñ œ ! ÐB#  "ÑÐ#B  &Ñ œ ! ÐB  "ÑÐB  "ÑÐ#B  &Ñ œ !

53.

B  'B  *B  &% œ ! Factor by grouping. ÐB$  'B# Ñ  Ð*B  &%Ñ œ ! B# ÐB  'Ñ  *ÐB  'Ñ œ ! ÐB  'ÑÐB#  *Ñ œ ! ÐB  'ÑÐB  $ÑÐB  $Ñ œ !

47.

49.

B$œ! B œ $

or

C&œ! Cœ&

Substitute B  " for C. B  " œ  $#

or

B"œ&

 "#

or

Bœ'

The solution set is ˜ "# ß '™. 51.

Ð#B  $Ñ# œ "'B# %B  "#B  * œ "'B# "#B#  "#B  * œ ! %B#  %B  $ œ ! Ð#B  $ÑÐ#B  "Ñ œ ! or

Divide by –3. #B  " œ ! #B œ " B œ "#

or

Let B œ the width of the garden. Then B  % œ the length of the garden.

T œ P[ $#! œ ÐB  %ÑB $#! œ B#  %B ! œ B#  %B  $#! ! œ ÐB  "'ÑÐB  #!Ñ

B$œ! Bœ$

B  "' œ ! B œ "'

#ÐB  "Ñ#  (ÐB  "Ñ  "& œ ! Let C œ B  ". #C#  (C  "& œ ! Ð#C  $ÑÐC  &Ñ œ !

Bœ

Bœ

The area of the rectangular-shaped garden is $#! ft# , so use the formula T œ P[ and substitute $#! for T, B  % for P, and B for [ .

By dividing each side by a variable expression, she "lost" the solution !. The solution set is ˜ %$ ß !ß %$ ™.

or

B œ  #$

% & % "&

The solution set is ˜ $# ß "# ™. 55.

The solution set is e$ß $ß 'f.

#C  $ œ ! #C œ $ C œ  $#

$B œ

#B  $ œ ! #B œ $ B œ  $#

#

or

$B œ #

#

The solution set is ˜ &# ß "ß "™.

B'œ! Bœ'

$B  " œ  "&

% ™ The solution set is ˜ #$ ß "& .

B  " œ ! or B  " œ ! or #B  & œ ! B œ " or B œ " or B œ  &#

45.

&C  " œ ! &C œ " C œ  "&

or

$B  " œ $

#

$

or

Substitute $B  " for C.

$B  % œ ! $B œ % B œ  %$

The solution set is $

C$œ! C œ $

&Ð$B  "Ñ#  $ œ "'Ð$B  "Ñ Let C œ $B  ". &C#  $ œ "'C # &C  "'C  $ œ ! ÐC  $ÑÐ&C  "Ñ œ !

or or

B  #! œ ! B œ #!

A rectangle cannot have a width that is a negative measure, so reject #! as a solution. The only possible solution is "'. The width of the garden is "' feet, and the length is "'  % œ #! feet. 57.

Let 2 œ the height of the parallelogram. Then 2  ( œ the base of the parallelogram. The area is '! ft# , so use the formula T œ ,2 and substitute '! for T , and 2  ( for , . T œ ,2 '! œ Ð2  (Ñ2 '! œ 2#  (2 ! œ 2#  (2  '! ! œ Ð2  "#ÑÐ2  &Ñ 2  "# œ ! 2 œ "#

or or

2&œ! 2œ&

903

Factoring A parallelogram cannot have a height that is negative, so reject "# as a solution. The only possible solution is &. The height of the parallelogram is & feet and the base is &  ( œ "# feet. 59.

Z œ P[ L ""! œ ÐA  #ÑÐA  %Ñ# ""! œ ÐA#  #A  )Ñ# && œ A#  #A  ) ! œ A#  #A  '$ ! œ ÐA  *ÑÐA  (Ñ

Let P œ the length of the rectangular area and [ œ the width.

A*œ! Aœ*

Use the formula for perimeter, T œ #P  #[ , and solve for [ in terms of P. The perimeter is $!! ft. $!! œ #P  #[ $!!  #P œ #[ "&!  P œ [

65.

>&œ! >œ&

P  "!! œ ! P œ "!!

Let B and B  " denote the two consecutive integers.

67.

B'œ! B œ '

or or

If B œ ', then B  " œ &. If B œ &, then B  " œ '. The two possible pairs of consecutive integers are ' and & or & and '. 63.

Let A œ the width of the cardboard. Then A  ' œ the length of the cardboard. If squares that measure # inches are cut from each corner of the cardboard, then the width becomes A  % and the length becomes ÐA  'Ñ  % œ A  #. Use the formula Z œ P[ L and substitute ""! for Z , A  # for P, A  % for [ , and # for L .

904

Use 0 Ð>Ñ œ "'>#  '#& with 0 Ð>Ñ œ !.

%>  #& œ ! %> œ #& > œ  #& %

Divide by 2.

B&œ! Bœ&

>"œ! > œ "

! œ "'>#  '#& ! œ "'>#  '#& Multiply by –1. ! œ Ð%>  #&ÑÐ%>  #&Ñ

The sum of their squares is '", so B#  ÐB  "Ñ# œ '". B#  B#  #B  " œ '" #B#  #B  '! œ ! B#  B  $! œ ! ÐB  'ÑÐB  &Ñ œ !

or or

Time cannot be negative, so reject " as a solution. The only possible solution is &. The object will hit the ground & seconds after it is thrown.

When P œ &!, [ œ "&!  &! œ "!!. When P œ "!!, [ œ "&!  "!! œ &!. The dimensions should be &! feet by "!! feet. 61.

When the object hits the ground, its height is given by 0 Ð>Ñ œ !. Let 0 Ð>Ñ œ ! in the given equation, and solve for >. "'>#  '%>  )! œ 0 Ð>Ñ "'>#  '%>  )! œ ! >#  %>  & œ ! Divide by –16. Ð>  &ÑÐ>  "Ñ œ !

&!!! œ PÐ"&!  PÑ &!!! œ "&!P  P# P#  "&!P  &!!! œ ! ÐP  &!ÑÐP  "!!Ñ œ ! or or

A(œ! A œ (

A box cannot have a negative width, so reject ( as a solution. The only possible solution is *. The piece of cardboard has width * inches and length *  ' œ "& inches.

Now use the formula for area, T œ P[ , substitute &!!! for T, and solve for PÞ

P  &! œ ! P œ &!

or or

or or

%>  #& œ ! %> œ #& > œ #& %

Time cannot be negative, so reject  #& % as a " solution. The only possible solution is #& % or ' % . " The ball will hit the ground after ' % seconds. 69.

Solve #5  +< œ <  $C for Ñ œ "'>#  #&'> ! œ "'>#  #&'> ! œ "'>Ð>  "'Ñ or or

"'> œ ! >œ!

>  "' œ ! > œ "'

The rock is on the ground when > œ !. It will return to the ground again after "' seconds. 54.

Equivalently, we have

The height is ! when the rock returns to the ground.

0 Ð>Ñ œ "'>#  #&'> #%! œ "'>#  #&'> ! œ "'>#  #&'>  #%! ! œ "'Ð>#  "'>  "&Ñ ! œ "'Ð>  "&ÑÐ>  "Ñ >  "& œ ! > œ "&

or or

Let f(t) œ 240.

>"œ! >œ"

The rock will be #%! ft above the ground after " second and again after "& seconds. 55.

The question in Exercise 54 has two answers because the rock will be #%! ft above the ground after " second on the way up and again after "& seconds on the way back down.

56.

0 Ð>Ñ œ "'>#  #&'> "!#% œ "'>#  #&'> ! œ "'>#  #&'>  "!#% ! œ "'Ð>#  "'>  '%Ñ ! œ "'Ð>  )Ñ#

Aœ 59.

[3]

"'  )"5 # œ %#  Ð*5Ñ# œ Ð%  *5ÑÐ%  *5Ñ

60.

[2]

$!+  +7  +7# œ +Ð$!  7  7# Ñ œ +Ð'  7ÑÐ&  7Ñ

61.

[2] *B#  "$BC  $C# is prime since it cannot be factored further.

62.

[3]

)  +$ œ #$  + $ œ Ð#  +ÑÐ##  #+  +# Ñ œ Ð#  +ÑÐ%  #+  +# Ñ

63.

[3]

#&D #  $!D7  *7# œ Ð&DÑ#  #Ð&DÑÐ$7Ñ  Ð$7Ñ# œ Ð&D  $7Ñ#

64.

[1]

"&C$  #!C# œ &C# $C  %

65.

[5]

&B#  "(B œ "# &B#  "(B  "# œ ! Ð&B  $ÑÐB  %Ñ œ ! &B  $ œ ! &B œ $ B œ  $&

Let f(t) œ 1024.

>)œ! >œ)

66.

58.

Solve D œ

$A  ( for A. A

DA œ $A  ( DA  $A œ ( AÐD  $Ñ œ ( Aœ

( D$

Multiply by w. Get y-terms on one side. Distributive property Divide by z  3.

B%œ! Bœ%

[5]

B$  B œ ! BÐB#  "Ñ œ ! BÐB  "ÑÐB  "Ñ œ ! B œ ! or B  " œ ! or B  " œ ! B œ " Bœ"

The solution set is e"ß !ß "f.

Solve $=  ,5 œ 5  #> for 5 . $=  #> œ 5  ,5 $=  #> œ 5Ð"  ,Ñ $=  #> $=  #> 5œ , or : œ ", ,"

or

The solution set is ˜ $& ß %™.

The rock will be "!#% ft above the ground after ) seconds. 57.

( . $D

67.

[5] Let B be the width of the frame. Then B  # will be the length of the frame. The area is %) in# . Use the formula P[ œ T. ÐB  #ÑB œ %) B#  #B œ %) B#  #B  %) œ ! ÐB  )ÑÐB  'Ñ œ ! B)œ! B œ )

or or

B'œ! Bœ'

The frame cannot have a negative width, so reject B œ ). The width of the frame is ' inches. 909

Factoring 68.

[5] Let B be the width of the floor. Then B  )& will be the length of the floor. The area is #(&! ft# . Use the formula P[ œ T. ÐB  )&ÑB œ #(&! B#  )&B œ #(&! B#  )&B  #(&! œ ! ÐB  ""!ÑÐB  #&Ñ œ ! B  ""! œ ! B œ ""!

or or

B  #& œ ! B œ #&

7.

"'+#  %!+,  #&, # œ Ð%+Ñ#  #Ð%+ÑÐ&,Ñ  Ð&,Ñ# œ Ð%+  &,Ñ#

8.

B#  #B  "  %D # œ ÐB#  #B  "Ñ  %D # œ ÐB  "Ñ#  Ð#DÑ# œ cÐB  "Ñ  #D dcÐB  "Ñ  #D d œ ÐB  "  #DÑÐB  "  #DÑ

9.

+$  #+#  +, #  #, # œ +# Ð+  #Ñ  , # Ð+  #Ñ œ Ð+  #ÑÐ+#  , # Ñ œ Ð+  #ÑÐ+  ,ÑÐ+  ,Ñ

10.

*5 #  "#"4 # œ Ð$5Ñ#  Ð""4Ñ# œ Ð$5  ""4ÑÐ$5  ""4Ñ

The floor cannot have a negative width, so reject B œ ""!. The width of the floor was #& feet and the length was #&  )& œ ""! feet.

Test 1.

""D #  %%D œ ""DÐD  %Ñ

11.

2.

"!B# C&  &B# C$  #&B& C$ Factor out the GCF, &B# C$ Þ œ &B# C$ Ð#C#  "  &B$ Ñ

C$  #"' œ C$  '$ œ ÐC  'ÑÐC#  'C  '# Ñ œ ÐC  'ÑÐC#  'C  $'Ñ

12.

'5 %  5 #  $& œ 'Ð5 # Ñ#  5 #  $&

3.

4.

$B  ,C  ,B  $C œ $B  $C  ,B  ,C œ $ÐB  CÑ  ,ÐB  CÑ œ ÐB  CÑÐ$  ,Ñ #

Two integer factors whose product is Ð'ÑÐ$&Ñ œ #"! and whose sum is " are "& and "%.

#B  B  $' œ "Ð#B  B  $'Ñ Two integer factors whose product is Ð#ÑÐ$'Ñ œ (# and whose sum is " are * and ). #B#  B  $' œ #B#  *B  )B  $' œ BÐ#B  *Ñ  %Ð#B  *Ñ œ Ð#B  *ÑÐB  %Ñ Thus, the final factored form is Ð#B  *ÑÐB  %ÑÞ

5.

'B#  ""B  $& Two integer factors whose product is Ð'ÑÐ$&Ñ œ #"! and whose sum is "" are #" and "!. œ 'B#  #"B  "!B  $& œ $BÐ#B  (Ñ  &Ð#B  (Ñ œ Ð#B  (ÑÐ$B  &Ñ

6.

%:#  $:;  ; # Two integer factors whose product is Ð%ÑÐ"Ñ œ % and whose sum is $ are % and ". œ %:#  %:;  :;  ; # œ %:Ð:  ;Ñ  ;Ð:  ;Ñ œ Ð:  ;ÑÐ%:  ;Ñ

910

œ 'Ð5 # Ñ#  "&5 #  "%5 #  $& œ $5 # Ð#5 #  &Ñ  (Ð#5 #  &Ñ œ Ð#5 #  &ÑÐ$5 #  (Ñ

#

13.

#(B'  " œ Ð$B# Ñ$  Ð"Ñ$ œ Ð$B#  "ÑÒÐ$B# Ñ#  Ð$B# ÑÐ"Ñ  "# Ó œ Ð$B#  "ÑÐ*B%  $B#  "Ñ

14.

A. Ð$  BÑÐB  %Ñ œ $B  "#  B#  %B œ B#  B  "# B. ÐB  $ÑÐB  %Ñ œ ÐB#  %B  $B  "#Ñ œ ÐB#  B  "#Ñ œ B#  B  "# C. ÐB  $ÑÐB  %Ñ œ B#  %B  $B  "# œ B#  B  "# D. ÐB  $ÑÐB  %Ñ œ B#  %B  $B  "# œ B#  (B  "# Therefore, only D is not a factored form of B#  B  "#.

Factoring 15.

$B#  )B œ % $B#  )B  % œ ! ÐB  #ÑÐ$B  #Ñ œ !

Length and width will be negative if B œ  "* # , so reject it as a possible solution. If B œ ", then B(œ"(œ)

or

B#œ! B œ #

The solution set is ˜#ß  #$ ™. 16.

#

$B  &B œ ! BÐ$B  &Ñ œ ! Bœ!

or

$B  & œ ! $B œ & B œ &$

The solution set is ˜!ß &$ ™. 17.

&7Ð7  "Ñ œ #Ð"  7Ñ &7#  &7 œ #  #7 &7#  $7  # œ ! Ð&7  #ÑÐ7  "Ñ œ ! &7  # œ ! &7 œ # 7 œ  #&

and

$B  # œ ! $B œ # B œ  #$

or

#B  $ œ #Ð"Ñ  $ œ &. The length is ) inches, and the width is & inches. 20.

Substitute "#) for 0 Ð>Ñ in the equation. 0 Ð>Ñ œ "'>#  *'> "#) œ "'>#  *'> # "'>  *'>  "#) œ ! "'Ð>#  '>  )Ñ œ ! "'Ð>  %ÑÐ>  #Ñ œ ! >%œ! >œ%

or or

>#œ! >œ#

The ball is "#) ft high at # seconds (on the way up) and again at % seconds (on the way down).

7"œ! 7œ"

The solution set is ˜ #& ß "™. 18.

19.

Solve +<  # œ $<  '> for #  '> #  '> $  ) ># Ð>  #ÑÐ>#  #>  %Ñ œ ># œ >#  #>  %

)"  C# C* Ð*  CÑÐ*  CÑ œ C* "Ð*  CÑÐC  *Ñ œ C*

Factor.

œ "Ð*  CÑ, or *  C

Set the denominator equal to zero, and solve the equation.

(b) 0 ÐBÑ œ

+  "! +  "! " œ œ œ " "!  + "Ð+  "!Ñ "

Factor. (b) Lowest terms

Factor the sum of cubes. Lowest terms

+7  ,7  +8  ,8 +7  ,7  +8  ,8 Ð+7  ,7Ñ  Ð+8  ,8Ñ œ Ð+7  ,7Ñ  Ð+8  ,8Ñ 7Ð+  ,Ñ  8Ð+  ,Ñ œ 7Ð+  ,Ñ  8Ð+  ,Ñ Ð+  ,ÑÐ7  8Ñ œ Ð+  ,ÑÐ7  8Ñ +, œ +,

N5. (a)

Lowest terms

)># $>  ' • # > % *> $Ð>  #Ñ )># œ • *> Ð>  #ÑÐ>  #Ñ )> œ $Ð>  #Ñ 7#  #7  "& 7#  % • 7#  &7  ' 7#  &7 Ð7  &ÑÐ7  $Ñ Ð7  #ÑÐ7  #Ñ œ • Ð7  #ÑÐ7  $Ñ 7Ð7  &Ñ 7# œ 7 #:# ; :; #:# ; ':# ; # ƒ # # œ • % $:; ': ; $:; % :; "#:% ; $ œ $:# ; & %:# œ # ; $5 #  &5  # %5 #  )5 ƒ # *5 #  " 5  (5 # $5  &5  # 5 #  (5 œ • *5 #  " %5 #  )5 Ð$5  "ÑÐ5  #Ñ 5Ð5  (Ñ œ • Ð$5  "ÑÐ$5  "Ñ %5Ð5  #Ñ 5( œ %Ð$5  "Ñ

1 Section Exercises 1.

Group the terms. Factor within the groups. Factor by grouping.

Fundamental property

B B( Set the denominator equal to zero, and solve the equation. 0 Ð BÑ œ

B(œ! Bœ( The number ( makes the rational expression undefined, so ( is not in the domain of the function. In set-builder notation, the domain is eB l B Á ( f.

From Chapter 7 of Student’s Solutions Manual for Intermediate Algebra, Eleventh Edition. Margaret L. Lial, John Hornsby, Terry McGinnis. Copyright © 2012 by Pearson Education, Inc. Publishing as Addison-Wesley. All rights reserved.

913

Rational Expressions and Functions 3.

'B  & (B  " Set the denominator equal to zero, and solve the equation. 0 Ð BÑ œ

(B  " œ ! (B œ " B œ  "(

17.

19.

The number  "( makes the rational expression undefined, so  "( is not in the domain of the function. In set-builder notation, the domain is ˜B l B Á  "( ™. 5.

7.

"#B  $ is undefined when the B denominator B equals !. So B œ ! makes the rational expression undefined and ! is not in the domain of the function. In set-builder notation, the domain is eB l B Á !f. 0 ÐBÑ œ

(c)

B$ Ð"ÑÐB  $Ñ œ B% Ð"ÑÐB  %Ñ B  $ œ (B) B  % $B Ð"ÑÐ$  BÑ œ (e) B% Ð"ÑÐB  %Ñ $  B B$ œ , or (E) B  % B  % B$ Ð"ÑÐB  $Ñ œ (f) %B Ð"ÑÐ%  BÑ B  $ B  $ œ , or (F) %  B B%

$B  " #B#  B  ' Set the denominator equal to zero and solve. #B#  B  ' œ ! ÐB  #ÑÐ#B  $Ñ œ ! or or

#B  $ œ ! #B œ $ B œ $#

The numbers # and $# are not in the domain of the function. In set-builder notation, the domain is ˜B l B Á #ß $# ™. 9.

11.

B# is never undefined, so all numbers "% are in the domain of the function. In set-builder notation, the domain is ÖB l B is a real number×.

15.

914

% ( #•# ( # # œ œ œ • • #" "! $•( #•& $•& "& $ & $ "# ƒ œ • ) "# ) & $ $•$ * #•#•$ œ œ œ • & #•& "! #•#•#

B#  %B B% The two terms in the numerator are B# and %B. The two terms in the denominator are B and %. To express the rational expression in lowest terms, factor the numerator and denominator and replace the quotient of common factors with ".

#B#  $B  % 0 ÐBÑ œ $B#  ) Set the denominator equal to zero and solve.

The square of any real number B is positive or zero, so this equation has no solution. There are no real numbers which make this rational expression undefined, so all numbers are in the domain of the function. In set-builder notation, the domain is ÖB l B is a real number×. 13.

21.

0 ÐBÑ œ

$B#  ) œ ! $B# œ ) B# œ  )$

B$ Ð"ÑÐB  $Ñ œ B% Ð"ÑÐB  %Ñ B  $ œ (D) B  %

(d)

0 ÐBÑ œ

B#œ! B œ #

# ) # * ƒ œ • $ * $ ) # $•$ $ $ œ • œ œ #•# % $ #•#•# B$ Ð"ÑÐB  $Ñ œ (a) B% Ð"ÑÐB  %Ñ B  $ $B œ , or (C) B  % B  % B$ Ð"ÑÐB  $Ñ œ (b) B% Ð"ÑÐB  %Ñ B  $ B  $ œ , or (A) B  % %B

B#  %B BÐB  %Ñ œ B% B% œ B•" œ B 23.

A.

$B "ÐB  $Ñ B$ œ œ B% "Ð%  BÑ %B

B.

B$ B$ cannot be transformed to equal . %B %B

C. 

$B Ð$  BÑ B$ œ œ %B %B %B

D. 

B$ B$ B$ œ œ B% ÐB  %Ñ %B

Only the expression in B is not equivalent to B$ . %B

Rational Expressions and Functions 25.

B# ÐB  "Ñ B BÐB  "Ñ B œ • œ •" œ B BÐB  "Ñ " BÐB  "Ñ "

51.

27.

ÐB  %ÑÐB  $Ñ B$ B% B$ œ œ • ÐB  &ÑÐB  %Ñ B& B% B&

In Exercises 53 and 55, there are other acceptable ways to express each answer.

%BÐB  $Ñ ÐB  $Ñ • %B œ )B# ÐB  $Ñ #BÐB  $Ñ • %B B$ œ #BÐB  $Ñ

53.

29.

31.

$B  ( Since the numerator and denominator $ have no common factors, the expression is already in lowest terms.

55.

(, "Ð,  (Ñ œ œ " ,( ,(

B#  C # ÐB  CÑÐB  CÑ œ CB CB "ÐC  BÑÐB  CÑ œ CB œ ÐB  CÑ Ð+  $ÑÐB  CÑ Ð+  $ÑÐB  CÑ œ Ð$  +ÑÐB  CÑ "Ð+  $ÑÐB  CÑ BC œ "ÐB  CÑ BC œ BC

33.

'7  ") 'Ð7  $Ñ ' œ œ (7  #" (Ð7  $Ñ (

35.

$D #  D DÐ$D  "Ñ D œ œ ")D  ' 'Ð$D  "Ñ '

57.

&5  "! &Ð#  5Ñ & " œ œ œ #!  "!5 "!Ð#  5Ñ "! #

37.

>#  * Ð>  $ÑÐ>  $Ñ >$ œ œ $>  * $Ð>  $Ñ $

59.

+#  , # Ð+  ,ÑÐ+  ,Ñ œ +#  , # +#  , #

39.

#>  ' #Ð>  $Ñ # œ œ >#  * Ð>  $ÑÐ>  $Ñ >$

41.

43.

45.

47.

49.

B#  #B  "& ÐB  &ÑÐB  $Ñ œ B#  'B  & ÐB  &ÑÐB  "Ñ B$ œ B" )B#  "!B  $ Ð%B  "ÑÐ#B  $Ñ œ )B#  'B  * Ð%B  $ÑÐ#B  $Ñ %B  " œ %B  $

The numerator and denominator have no common factors except ", so the original expression is already in lowest terms. 61. 63.

+$  , $ Ð+  ,ÑÐ+#  +,  , # Ñ œ +, +, œ +#  +,  , # #- #  #-.  '!. # #- #  "#-.  "!. # #Ð- #  -.  $!. # Ñ œ #Ð- #  '-.  &. # Ñ #Ð-  '.ÑÐ-  &.Ñ œ #Ð-  .ÑÐ-  &.Ñ -  '. œ -. +-  +.  ,-  ,. +-  +.  ,-  ,. +Ð-  .Ñ  ,Ð-  .Ñ œ +Ð-  .Ñ  ,Ð-  .Ñ Ð-  .ÑÐ+  ,Ñ œ Ð-  .ÑÐ+  ,Ñ +, œ +,

65.

B$ *C# $C • $B$ C $C $C œ œ # •" œ # • $C B& B# • $B$ C B B &+% , # #&+# , ƒ "'+# , '!+$ , # Multiply by the reciprocal of the divisor. &+% , # '!+$ , # œ • "'+# , #&+# , $!!+( , % œ %!!+% , # "!!+% , # • $+$ , # œ "!!+% , # • % $+$ , # œ % Ð$78Ñ# • '%Ð7# 8Ñ $ #%Ð7# 8# Ñ % ƒ Ð$7# 8$ Ñ # "'7# 8% Ð78# Ñ $ 3

3

3

3

*7# 8# • '%7' 8$ *7% 8' • "'7# 8% • 7$ 8' #%7) 8) * • % • *7"# 8"" œ $ • % • #7"$ 8") $•* #( œ œ " ( #7 8 #78(

œ

Factor by grouping.

67.

ÐB  #ÑÐB  "Ñ ÐB  $ÑÐB  %Ñ • ÐB  $ÑÐB  #Ñ ÐB  #ÑÐB  "Ñ ÐB  "ÑÐB  #ÑÐB  $Ñ B  % B% œ œ • ÐB  "ÑÐB  #ÑÐB  $Ñ B  # B#

915

Rational Expressions and Functions 69.

71.

73.

Ð#B  $ÑÐB  %Ñ ÐB  %ÑÐB  #Ñ ƒ ÐB  )ÑÐB  %Ñ ÐB  %ÑÐB  )Ñ #B  $ B  # œ ƒ B) B) #B  $ B  ) œ • B) B# #B  $ œ B# %B "%B  ( %B • (Ð#B  "Ñ œ • )B  % ' %Ð#B  "Ñ • ' (B œ ' :#  #& # Ð:  &ÑÐ:  &Ñ# œ • %: &: # • #:Ð"ÑÐ:  &Ñ :& :& œ œ Ð#:ÑÐ"Ñ #: There are several other ways to express the answer.

75.

77.

79.

81.

83.

916

%5  % & (Ð5  "Ñ & œ • " %Ð5  "Ñ (•& $& œ œ % % " ÐD #  "Ñ • "D ÐD  "ÑÐD  "Ñ " œ • " "ÐD  "Ñ D" œ œ ÐD  "Ñ or D  " " %B  #! #B  "! ƒ &B (B$ %ÐB  &Ñ (B$ œ • &B #ÐB  &Ñ # • (B# œ & "%B# œ & "#B  "!C 'B  %C • $B  #C "!C  "#B #Ð'B  &CÑ • #Ð$B  #CÑ œ Ð$B  #CÑ • #Ð&C  'BÑ #Ð"ÑÐ&C  'BÑ œ œ # Ð&C  'BÑ

85.

87.

89.

+$  , $ #+  #, ƒ # # + , #+  #, Ð+  ,Ñ +#  +,  , # œ Ð+  ,ÑÐ+  ,Ñ # +  +,  , # œ +,

B#  #& B#  (B  "# • B#  B  #! B#  #B  "& ÐB  &ÑÐB  &Ñ ÐB  $ÑÐB  %Ñ œ • ÐB  &ÑÐB  %Ñ ÐB  &ÑÐB  $Ñ B% œ B%

93.

95.

#Ð+  ,Ñ #Ð+  ,Ñ

)B$  #( #B  ' • #B#  ") )B#  "#B  ") Ð#BÑ$  $$ #ÐB  $Ñ œ • # # B  * # %B#  'B  * Ð#B  $Ñ %B#  'B  * #ÐB  $Ñ œ • #ÐB  $ÑÐB  $Ñ # %B#  'B  * #B  $ œ #ÐB  $Ñ +$  ),$ +#  +,  "#, # • #  +,  ', +#  #+,  ), # Ð+  #,Ñ +#  #+,  %, # Ð+  $,ÑÐ+  %,Ñ œ • Ð+  $,ÑÐ+  #,Ñ Ð+  #,ÑÐ+  %,Ñ # # +  #+,  %, œ +  #,

+#

Ð(5  (Ñ ƒ

91.

•

'B#  &B  ' %B#  "#B  * ƒ "#B#  ""B  # )B#  "%B  $ Ð$B  #ÑÐ#B  $Ñ Ð#B  $ÑÐ%B  "Ñ œ • Ð$B  #ÑÐ%B  "Ñ Ð#B  $ÑÐ#B  $Ñ #B  $ œ #B  $ $5 #  "(5:  "!:# '5 #  5:  #:# ƒ # # # '5  "$5:  &: '5  &5:  :# Ð$5  #:ÑÐ5  &:Ñ Ð$5  :ÑÐ#5  :Ñ œ • Ð$5  :ÑÐ#5  &:Ñ Ð$5  #:ÑÐ#5  :Ñ 5  &: œ #5  &: Œ

'5 #  "$5  & #5  & ƒ $  # 5  (5 5  '5 #  (5 5 #  &5  ' • $5 #  )5  $ Factor 5 from the denominator of the divisor; multiply by the reciprocal. '5 #  "$5  & 5 5 #  '5  ( œ” • • 5 #  (5 #5  & 5 #  &5  ' • $5 #  )5  $ Ð$5  "ÑÐ#5  &Ñ 5Ð5  (ÑÐ5  "Ñ œ” • • 5Ð5  (Ñ #5  & Ð5  #ÑÐ5  $Ñ • Ð$5  "ÑÐ5  $Ñ œ Ð5  "ÑÐ5  #Ñ

Rational Expressions and Functions 97.

99.

2

$ " * "   œ  % "# "# "# *  " ) # œ œ œ "# "# $ % " " #% "% #"   œ   ( $ # %# %# %# #%  "%  #" "( œ œ %# %#

LCD = 12

LCD = 42

Adding and Subtracting Rational Expressions

2 Now Try Exercises N1. (a) (b)

(c)

& # &# (  œ œ $B $B $B $B B# * B#  *  œ B$ B$ B$ ÐB  $ÑÐB  $Ñ œ œB$ B$ # B  B#  B  # B #  B  # #B œ # B B# B# œ ÐB  #ÑÐB  "Ñ " œ B"

N2. (a) "&7$ 8, "!7# 8 Factor the denominators. "&7$ 8 œ $ • & • 7$ • 8 "!7# 8 œ # • & • 7# • 8

# (  $B %B # The LCD for $B and %B is "#B. To write with $B % a denominator of "#B, multiply by % .

N3. (a)

#•% (•$ # (  œ  $B • % %B • $ $B %B ) #" œ  "#B "#B )  #" œ "#B #* œ "#B $ ' (b)  D D& The LCD is DÐD  &Ñ. Rewrite each rational expression with this denominator. $ ' $ÐD  &Ñ Ð'ÑD  œ  D D& DÐD  &Ñ ÐD  &ÑD $ÐD  &Ñ 'D œ  DÐD  &Ñ DÐD  &Ñ $D  "&  'D œ DÐD  &Ñ $D  "& œ DÐD  &Ñ N4. (a)

")B  ( #B  "$  %B  & %B  &

The denominator is already the same for both rational expressions, so write them as a single expression. Be careful to apply the subtraction sign to both terms of the second expression. ")B  (  Ð#B  "$Ñ %B  & ")B  (  #B  "$ œ %B  & "'B  #! œ %B  & %Ð%B  &Ñ œ %B  & œ% œ

The least common denominator (LCD) is the product of all the different factors, with each factor raised to the greatest power in any denominator. LCD œ # • $ • & • 7$ • 8 œ $!7$ 8 (b) >, >  )

& &  B$ B$

Each denominator is already factored.

(b)

LCD œ >Ð>  )Ñ

The LCD is ÐB  $ÑÐB  $Ñ.

(c) $B#  *B  $!, B#  %, B#  "!B  #& Factor the denominators. $B#  *B  $! œ $ÐB  &ÑÐB  #Ñ B#  % œ ÐB  #ÑÐB  #Ñ B#  "!B  #& œ ÐB  &Ñ# LCD œ $ÐB  #ÑÐB  #ÑÐB  &Ñ#

&ÐB  $Ñ &ÐB  $Ñ  ÐB  $ÑÐB  $Ñ ÐB  $ÑÐB  $Ñ &ÐB  $Ñ  &ÐB  $Ñ œ ÐB  $ÑÐB  $Ñ &B  "&  &B  "& œ ÐB  $ÑÐB  $Ñ $! œ ÐB  $ÑÐB  $Ñ œ

917

Rational Expressions and Functions N5.

> #  >* *>

N8.

To get a common denominator of >  *, multiply both the numerator and denominator of the second expression by ". > #Ð"Ñ  >  * Ð*  >ÑÐ"Ñ > # œ  >* >* >  Ð#Ñ œ >* ># œ >* œ

An equivalent answer is "Ð>  #Ñ >  # #> . œ œ "Ð>  *Ñ >  * *> N6.

' " $   C C  $ C#  $C ' " $ œ   C C  $ CÐC  $Ñ Factor denominators. 'ÐC  $Ñ "C $ œ   CÐC  $Ñ ÐC  $ÑC CÐC  $Ñ LCD = y(y  3) 'ÐC  $Ñ  C  $ œ CÐC  $Ñ 'C  ")  C  $ œ CÐC  $Ñ &C  "& &ÐC  $Ñ & œ œ œ CÐC  $Ñ CÐC  $Ñ C

>" #>  $  # N7. # >  #>  ) >  $>  # >" #>  $ œ  Ð>  %ÑÐ>  #Ñ Ð>  "ÑÐ>  #Ñ The LCD is Ð>  %ÑÐ>  #ÑÐ>  "Ñ. Ð>  "ÑÐ>  "Ñ œ Ð>  %ÑÐ>  #ÑÐ>  "Ñ Ð#>  $ÑÐ>  %Ñ  Ð>  %ÑÐ>  #ÑÐ>  "Ñ Ð>  "ÑÐ>  "Ñ  Ð#>  $ÑÐ>  %Ñ œ Ð>  %ÑÐ>  #ÑÐ>  "Ñ # >  "  Ð#>#  &>  "#Ñ œ Ð>  %ÑÐ>  #ÑÐ>  "Ñ # >  "  #>#  &>  "# œ Ð>  %ÑÐ>  #ÑÐ>  "Ñ >#  &>  "" œ Ð>  %ÑÐ>  #ÑÐ>  "Ñ

918

# %  7#  '7  * 7#  7  "# # % œ  Ð7  $ÑÐ7  $Ñ Ð7  %ÑÐ7  $Ñ The LCD is Ð7  $Ñ# Ð7  %Ñ. #Ð7  %Ñ %Ð7  $Ñ œ  # Ð7  $Ñ Ð7  %Ñ Ð7  %ÑÐ7  $Ñ# #Ð7  %Ñ  %Ð7  $Ñ œ Ð7  $Ñ# Ð7  %Ñ #7  )  %7  "# œ Ð7  $Ñ# Ð7  %Ñ '7  % œ Ð7  $Ñ# Ð7  %Ñ

2 Section Exercises 1. 3. 5. 7. 9. 11. 13. 15.

17.

19.

21.

) % )% "# $•% %  œ œ œ œ "& "& "& "& $•& & & ) "& "' "&  "' " "  œ  œ œ œ ' * ") ") ") ") ") & ( "! #" "!  #" $"  œ  œ œ ") "# $' $' $' $' ( # (# *  œ œ > > > > 'B C 'B  C  œ ( ( ( "" " ""  " "! #  œ œ œ &B &B &B &B B * "( *  "( ) #  $ œ œ $ œ $ %B$ %B %B$ %B B &B  % B" &B  %  B  "  œ 'B  & 'B  & 'B  & 'B  & œ œ" 'B  & B# #& B#  #&  œ B& B& B& ÐB  &ÑÐB  &Ñ œ B& œB& $:  ( ):  "$  :#  (:  "# :#  (:  "# $:  (  ):  "$ œ :#  (:  "# &:  #! œ Ð:  $ÑÐ:  %Ñ &Ð:  %Ñ & œ œ Ð:  $ÑÐ:  %Ñ :$ +$ ,$  +#  +,  , # +#  +,  , # $ $ + , œ # +  +,  , # Ð+  ,ÑÐ+#  +,  , # Ñ œ +#  +,  , # œ+,

Rational Expressions and Functions 23.

#>#  (>  "& œ Ð#>  $ÑÐ>  &Ñ >#  $>  "! œ Ð>  &ÑÐ>  #Ñ

")B# C$ , #%B% C& Factor each denominator. ")B# C$ œ # • $ • $ • B# • C$ œ # • $# • B# • C $

The LCD is Ð#>  $ÑÐ>  &ÑÐ>  #ÑÞ 37.

#%B% C& œ # • # • # • $ • B% • C& œ #$ • $ • B% • C &

Factor each denominator. #C  ' œ #ÐC  $Ñ C#  * œ ÐC  $ÑÐC  $Ñ

The least common denominator (LCD) is the product of all the different factors, with each factor raised to the greatest power in any denominator. LCD œ #$ • $# • B% • C& œ ) • * • B% • C & œ (#B% C&

Remember the factor C from the third denominator. The LCD is #CÐC  $ÑÐC  $Ñ. 39.

#B  ' œ #ÐB  $Ñ B#  B  ' œ ÐB  $ÑÐB  #Ñ ÐB  #Ñ# is already factored.

D  #, D Both D  # and D have only " and themselves for factors.

27.

LCD œ #ÐB  #Ñ# ÐB  $Ñ 41.

LCD œ DÐD  #Ñ #C  ), C  % Factor each denominator. #C  ) œ #ÐC  %Ñ The second denominator, C  %, is already factored. The LCD is #ÐC  %Ñ. 29.

B#  )", B#  ")B  )" Factor each denominator.

43.

B#  )" œ ÐB  *ÑÐB  *Ñ B#  ")B  )" œ ÐB  *ÑÐB  *Ñ LCD œ ÐB  *Ñ# ÐB  *Ñ 31.

7  8, 7  8, 7 #  8 # Both 7  8 and 7  8 have only " and themselves for factors, while 7#  8# œ Ð7  8ÑÐ7  8Ñ.

45.

B#  $B  %, B  B# Factor each denominator. B#  $B  % œ ÐB  %ÑÐB  "Ñ B  B# œ BÐ"  BÑ œ BÐB  "Ñ The LCD is BÐB  %ÑÐB  "Ñ.

35.

#>#  (>  "&, >#  $>  "! Factor each denominator.

The first step is incorrect. The third term in the numerator should be ", because the subtraction sign should be distributed to both %B and ". The correct solution follows. B %B  " B  Ð%B  "Ñ  œ B# B# B# B  %B  " œ B# $B  " œ B# ) (  The LCD is $>. > $> ) ( )•$ (  œ  > $> >•$ $> #%  ( $" œ œ $> $> & ""  The LCD is "#B# C. "#B# C 'BC & "" &  œ  # "#B C 'BC "#B# C & œ  "#B# C &  ##B œ "#B# C

LCD œ Ð7  8ÑÐ7  8Ñ 33.

#B  ', B#  B  ', ÐB  #Ñ# Factor each denominator.

The LCD is (#B% C& . 25.

#C  ', C#  *, C

47.

"" • #B 'BC • #B ##B "#B# C

% $  LCD œ '!+% , ' % & "&+ , #!+# , ' % • %, $ • $+# œ  "&+% , & • %, #!+# , ' • $+# "',  *+# œ '!+% , '

919

Rational Expressions and Functions 49.

51.

53.

55.

57.

59.

920

#< $=  LCD œ "%:% ; % (:$ ; % "%:% ; #< • #: $= • ; $ œ $ %  (: ; • #: "%:% ; • ; $ %:<  $=; $ œ "%:% ; % " # $  %  & ( LCD œ +& , ( $ # + , + , + , " • +# , & # • +, ' $ œ $ # # & %  & ( + , •+ , + , • +, ' + , +# , &  #+, '  $ œ +& , ( " "  LCD œ BÐB  "Ñ B" B "•B "• B  " œ  B" B B B" B B" œ B B" BB" œ B B" " œ B B" $+ #+  LCD œ Ð+  "ÑÐ+  $Ñ +" +$ $+Ð+  $Ñ #+Ð+  "Ñ œ  +  " Ð+  $Ñ +  $ Ð+  "Ñ $+Ð+  $Ñ  #+Ð+  "Ñ œ Ð+  "ÑÐ+  $Ñ $+#  *+  #+#  #+ œ Ð+  "ÑÐ+  $Ñ &+#  (+ œ Ð+  "ÑÐ+  $Ñ "(C  $ "!C  ")  *C  ( *C  ( "(C  $  Ð"!C  ")Ñ œ *C  ( "(C  $  "!C  ") œ *C  ( #(C  #" œ *C  ( $Ð*C  (Ñ œ *C  ( œ$ ""B  "$ $B  "  #B  $ #B  $ ""B  "$  Ð$B  "Ñ œ #B  $ ""B  "$  $B  " œ #B  $ )B  "# œ #B  $ %Ð#B  $Ñ œ œ% #B  $

61.

# &  %B B% To get a common denominator of B  %, multiply both the numerator and denominator of the first expression by ".

Ð#ÑÐ"Ñ &  Ð%  BÑÐ"Ñ B  % # & œ  B% B% #  & œ B% $ œ B% If you chose %  B for the LCD, then you should $ have obtained the equivalent answer, . %B A D  63. AD DA A  D and D  A are opposites, so factor out " from D  A to get a common denominator. A D œ  AD "ÐA  DÑ A D œ  AD AD AD A  D œ , or AD DA " "  65. LCD œ ÐB  "ÑÐB  "Ñ B" B" "ÐB  "Ñ "ÐB  "Ñ œ  ÐB  "ÑÐB  "Ñ ÐB  "ÑÐB  "Ñ B"B" œ ÐB  "ÑÐB  "Ñ # œ ÐB  "ÑÐB  "Ñ œ

67.

%B # %   B  " B  " B#  " B#  " œ ÐB  "ÑÐB  "Ñ, the LCD. %BÐB  "Ñ #ÐB  "Ñ œ  ÐB  "ÑÐB  "Ñ ÐB  "ÑÐB  "Ñ %  ÐB  "ÑÐB  "Ñ %BÐB  "Ñ  #ÐB  "Ñ  % œ ÐB  "ÑÐB  "Ñ # %B  %B  #B  #  % œ ÐB  "ÑÐB  "Ñ # %B  #B  # œ ÐB  "ÑÐB  "Ñ # #B#  B  " œ ÐB  "ÑÐB  "Ñ #Ð#B  "ÑÐB  "Ñ œ ÐB  "ÑÐB  "Ñ #Ð#B  "Ñ œ B"

Rational Expressions and Functions 69.

71.

73.

75.

"& # &   C#  $C C C  $ C#  $C œ CÐC  $Ñ, the LCD. "& #ÐC  $Ñ &C œ   CÐC  $Ñ CÐC  $Ñ ÐC  $ÑC "&  #ÐC  $Ñ  &C œ CÐC  $Ñ "&  #C  '  &C œ CÐC  $Ñ (C  #" œ CÐC  $Ñ (ÐC  $Ñ ( œ œ CÐC  $Ñ C

œ œ œ œ œ 77.

& " #   # B  # B B  #B B#  #B œ BÐB  #Ñ, the LCD. & " #   B  # B B#  #B &B "ÐB  #Ñ # œ   ÐB  #ÑB BÐB  #Ñ BÐB  #Ñ &B  B  #  # œ BÐB  #Ñ 'B ' œ œ BÐB  #Ñ B# $B % '   B  " B  " B#  " B#  " œ ÐB  "ÑÐB  "Ñ, the LCD. $BÐB  "Ñ %ÐB  "Ñ œ  ÐB  "ÑÐB  "Ñ ÐB  "ÑÐB  "Ñ '  ÐB  "ÑÐB  "Ñ $BÐB  "Ñ  %ÐB  "Ñ  ' œ ÐB  "ÑÐB  "Ñ # $B  $B  %B  %  ' œ ÐB  "ÑÐB  "Ñ $B#  B  # œ ÐB  "ÑÐB  "Ñ Ð$B  #ÑÐB  "Ñ œ ÐB  "ÑÐB  "Ñ $B  # œ B" % " "#   B  " B#  B  " B $  " B$  " œ ÐB  "ÑÐB#  B  "Ñ, the LCD. % B#  B  " œ B  " B#  B  " " • ÐB  "Ñ  # B  B  " ÐB  "Ñ "#  B  " B#  B  "

79.

81.

% B#  B  "  ÐB  "Ñ  "# ÐB  "Ñ B#  B  " # %B  %B  %  B  "  "# ÐB  "Ñ B#  B  " %B#  $B  ( ÐB  "Ñ B#  B  " Ð%B  (ÑÐB  "Ñ ÐB  "Ñ B#  B  " %B  ( # B B"

#B  % $ '   # B$ B B  $B B#  $B œ BÐB  $Ñ, the LCD. Ð#B  %ÑB $ÐB  $Ñ ' œ   ÐB  $ÑB BÐB  $Ñ BÐB  $Ñ Ð#B  %ÑB  $ÐB  $Ñ  ' œ BÐB  $Ñ #B#  %B  $B  *  ' œ BÐB  $Ñ # #B  (B  $ œ BÐB  $Ñ Ð#B  "ÑÐB  $Ñ #B  " œ œ BÐB  $Ñ B $ &  % # Ð:  #Ñ :# $ &Ð:  #Ñ %Ð:  #Ñ# œ   Ð:  #Ñ# Ð:  #Ñ# Ð:  #Ñ# LCD œ (p  2)2 $  &Ð:  #Ñ  % :#  %:  % œ Ð:  #Ñ# $  &:  "!  %:#  "':  "' œ Ð:  #Ñ# %:#  #":  #* œ Ð:  #Ñ# $ #  #  &B  ' B  %B  % $ # œ  ÐB  #ÑÐB  $Ñ ÐB  #ÑÐB  #Ñ $ÐB  #Ñ œ ÐB  #ÑÐB  $ÑÐB  #Ñ #ÐB  $Ñ  ÐB  #ÑÐB  #ÑÐB  $Ñ LCD œ (x  2)2 (x  3) $B  '  #B  ' œ ÐB  #Ñ# ÐB  $Ñ B œ ÐB  #Ñ# ÐB  $Ñ

B#

921

Rational Expressions and Functions 83.

 " ƒ $> *> >% *> œ • $> #>  " $Ð>  %Ñ œ #>  "

(b) %

œ

œ

œ œ

# C " C &C #  C C %C "  C C &C  # C %C  " C &C  # C • C %C  " &C  # %C  "

Multiply by the reciprocal of the divisor.

œ

B# ÐB  #Ñ  &ÐB  #Ñ %B# ÐB  #Ñ  B

œ

B$  #B#  &B  "! %B$  )B#  B

N3. (a) Method 1

Simplify the numerator and denominator.

Multiply by the reciprocal of the divisor.

# C N2. (a) " % C The LCD of C# and C" is C. Multiply the numerator and denominator by C. &

# Œ&   • C C œ " Œ%   • C C # &•C  •C C œ " %•C  •C C &C  # œ %C  "

& • BÐB  #Ñ B œ " %B • BÐB  #Ñ  • BÐB  #Ñ B# B • BÐB  #Ñ 

œ

&

&  • BÐB  #Ñ B œ " Œ%B   • BÐB  #Ñ B#

" :' œ & :#  $'

" :' & Ð:  'ÑÐ:  'Ñ " & œ ƒ :  ' Ð:  'ÑÐ:  'Ñ " Ð:  'ÑÐ:  'Ñ œ • :' & :' œ &

Method 2 " " :' :' œ & & # :  $' Ð:  'ÑÐ:  'Ñ The LCD of the numerator and denominator is Ð:  'ÑÐ:  'Ñ. " • Ð:  'ÑÐ:  'Ñ :' œ & • Ð:  'ÑÐ:  'Ñ Ð:  'ÑÐ:  'Ñ :' œ & (b) Method 1

Distributive property

Simplify.

" " 8# 7#   # # 7# 8# œ 7# 8# 7# 8# LCD = 7 8 8 7 " " LCD = 78   78 78 7 8 8#  7# # # œ 8787 78 923

Rational Expressions and Functions

œ

œ œ œ Method 2

Ð8  7ÑÐ8  7Ñ 7# 8# 87 78 Ð8  7ÑÐ8  7Ñ 8  7 ƒ 7# 8# 78 Ð8  7ÑÐ8  7Ñ 78 • 7# 8# 87 87 78

5.

7.

" "  # 7# 8 " "  7 8 The LCD of the numerator and the denominator is 7# 8# .

œ

œ œ œ

Œ

" "  #  • 7 # 8# 7# 8 " " Œ   • 7# 8# 7 8 8#  7# 78#  7# 8 Ð8  7ÑÐ8  7Ñ 78Ð8  7Ñ 87 78

9.

11.

# $  #C"  $C# C C# œ LCD = xy2 " $ C#  $B"  C# B # $ BC# Œ  #  C C œ " $ BC# • Œ #   C B # $ BC# •  BC# • # C C œ " $ # BC# • #  BC • C B #BC  $B œ B  $C#

N4.

"# B  " œ "# ƒ ' ' B" B B Multiply by the reciprocal of the divisor. "# B œ • B" ' #B œ B" 5" 5" %5 #5 œ • $5  " #5 $5  " %5 %5Ð5  "Ñ œ #5Ð$5  "Ñ #Ð5  "Ñ œ $5  " %D # B% %D # B% * œ *# & "#B# D & %B D "& & %D # B% %B# D & œ ƒ * & %D # B% & œ • * %B# D & # % &D B &B# œ # & œ $ *B D *D '

( Multiply the numerator and denominator by B, the LCD of all the fractions. Bˆ'  B" ‰ œ Bˆ(  B$ ‰ B • '  B • B" B • (  Bˆ B$ ‰ 'B  " œ (B  $ œ

13.

$ B $ B

 

1.

3.

924

 

" $ " '

" %

œ

# œ & % $

& * % '

 

$ * " '

) " %  % & "# % %

œ

œ

# * & '

( % "( %

œ

#

% # & # y' œ ƒ œ • œ y* & "& * ' $

œ "

œ

( "( ( %y ( ƒ œ • œ y % % "( % "( "

$ C $ C

Multiply the numerator and denominator by BC, the LCD of all the fractions.

3 Section Exercises & * # $

" B $ B

Š B$  C$ ‹BC Š B$  C$ ‹BC $ B • BC $ B • BC



$ C • BC $ C • BC

 $C  $B œ $C  $B $ÐC  BÑ CB œ œ $ÐC  BÑ CB

Rational Expressions and Functions

15.

17.

19.

21.

23.

)B  #%C )B  #%C &B "! œ • B  $C "! B  $C &B )ÐB  $CÑ&B œ "!ÐB  $CÑ %!B œ œ %B "! B#  "'C# BC " %  C B Multiply the numerator and denominator by BC, the LCD of all the fractions. B#  "'C# Œ BC BC œ " % Œ  BC C B B#  "'C# œ " % • BC  • BC C B B#  "'C# œ B  %C ÐB  %CÑÐB  %CÑ œ B  %C œ B  %C

25.

27.

' ' C#  "' C% œ • "# C% "# # C  "' ' ÐC  %ÑÐC  %Ñ œ • C% "# C% œ # " "  # # , + " "  , +

BC " "  C B

+#  , # # # œ , + +, ,+ ,+ +#  , # œ # # • +, , + Ð+  ,ÑÐ+  ,Ñ ,+ • œ +, , # +# +, œ +, BC œ BC CB B  C CB • œ BC " œ BC

C$ $ % #  * $C Multiply the numerator and denominator by *C, the LCD of all the fractions. C$ *CŒC   $ œ % # *CŒ   * $C *C#  $CÐC  $Ñ œ %C  ' *C#  $C#  *C œ %C  ' 'C#  *C œ %C  ' $CÐ#C  $Ñ $C œ œ #Ð#C  $Ñ #

C

B# "  B B# & B  B B# Multiply the numerator and denominator by BÐB  #Ñ, the LCD of all the fractions.

œ

œ

œ œ œ 29.

BÐB  #ÑŒ

B# "   B B# & B BÐB  #ÑŒ   B B# B# " BÐB  #ÑŒ   BÐB  #ш B# ‰ B & B ‰ BÐB  #ÑŒ   BÐB  #ш B# B ÐB  #ÑÐB  #Ñ  B &ÐB  #Ñ  B# # B  %B  %  B &B  "!  B# B#  & B  % B#  &B  "!

To add the fractions in the numerator, use the LCD 7Ð7  "Ñ. % 7# %Ð7  "Ñ 7Ð7  #Ñ  œ  7 7" 7Ð7  "Ñ 7Ð7  "Ñ %7  %  7#  #7 œ 7Ð7  "Ñ # 7  '7  % œ 7Ð7  "Ñ

925

Rational Expressions and Functions 30.

To subtract the fractions in the denominator, use the same LCD, 7Ð7  "Ñ.

37.

7# #  7 7" Ð7  #ÑÐ7  "Ñ 7•# œ  7Ð7  "Ñ 7Ð7  "Ñ 7#  7  #  #7 œ 7Ð7  "Ñ 7#  7  # œ 7Ð7  "Ñ 31.

32.

33.

34.

35.

926

B# C # Œ

" "  # B# C œ " " B# C # Œ   B C " " B# C # • #  B # C # • # B C œ " " # # B# C # •  B C • B C C #  B# C #  B# œ # , or BC  B# C BCÐC  BÑ

Exercise 29 Exercise 30 answer answer Æ Æ # # 7  '7  % 7  7  # ƒ 7Ð7  "Ñ 7Ð7  "Ñ Multiply by the reciprocal. 7#  '7  % 7Ð7  "Ñ œ • 7Ð7  "Ñ 7#  7  # 7#  '7  % œ 7#  7  #

39.

The LCD of all the denominators in the complex fraction is 7Ð7  "Ñ. Œ

% 7#   • 7Ð7  "Ñ 7 7" 7# #  Œ  • 7Ð7  "Ñ 7 7" %Ð7  "Ñ  7Ð7  #Ñ œ Ð7  #ÑÐ7  "Ñ  #7 %7  %  7#  #7 œ 7#  7  #  #7 7#  '7  % œ 7#  7  #

41.

Answers will vary. Because of the complicated nature of the numerator and denominator of the complex fraction, using Method 1 takes much longer to simplify the complex fraction. Method 2 is a simpler, more direct means of simplifying and is most likely the preferred method. B#

" " œ # " " C  # # B C B# C # " œ " " B# C # Œ #  #  B C B# C # œ # C  B#

B#  C# B"  C" " "  # B# C œ " "  B C Multiply the numerator and denominator by B# C# , the LCD of all the fractions.

2 2

LCD œ x y

" #  B"  #C" B C œ #C  %B #C  %B Multiply the numerator and denominator by BC, the LCD of all the fractions. " # BCŒ   B C œ BCÐ#C  %BÑ C  #B œ #BCÐC  #BÑ " œ #BC $ % )   7: : 7 (a) #7"  $:" $ % )   7: : 7 œ # $  7 : # " , not , since the exponent 7 #7 $ applies only to 7, not to #. Likewise, $:" œ , : " not . $: (b) #7" œ

$ % ) $ % )     7: : 7 7: : 7 (c) œ # $ #7"  $:"  7 : Multiply the numerator and denominator by 7:, the LCD of all the fractions.

Rational Expressions and Functions 7:Œ

$ % )    7: : 7 œ # $ 7:Œ   7 : $ % ) 7: •  7: •  7: • 7: : 7 œ # $ 7: •  7: • 7 : $  %7  ): œ #:  $7 " #B

43.



%Ð "# B

" %B

" , B  ( Á !, so the domain is B( ÖB l B Á (×. For

# , B  ( Á !, so the domain is B( ÖB l B Á (×. For

"% , B#  %* œ ÐB  (ÑÐB  (Ñ Á !, so B#  %* the domain is ÖB l B Á „ (×. For

The intersection of these three domains is ÖB l B Á „ (×.

œ *

" % BÑ

 œ %Ð*Ñ #B  B œ $' $B œ $' B œ "#

N2.

Step 1 The domain is eB l B Á !f.

The solution set is e"#f. 45.

Step 2 Multiply each side by the LCD, "#B.

B' B% œ & "! B' B% "!Œ  œ "!Œ  & "! #ÐB  'Ñ œ B  % #B  "# œ B  % B œ "'

"#BŒ

0 ÐBÑ œ

"#BŒ

" B

4 Equations with Rational Expressions and Graphs 4 Now Try Exercises " $ "  œ $B %B $

Find the domain of each term. For

" , $B Á !, so the domain is ÖB l B Á !×. $B

For

$ , %B Á !, so the domain is ÖB l B Á !×. %B

The domain of

" $

is Ð∞ß ∞Ñ.

The intersection of these three domains is ÖB l B Á !×. (b)

" $ "   "#BŒ  œ "#BŒ  $B %B $ %  * œ %B & œ %B B œ  &%

Step 4 % Check B œ  &% :  "& 

B œ ! is not in the domain. The domain is eB l B Á !f.

N1. (a)

" $ "   œ "#BŒ  $B %B $

Step 3

The solution set is e"'f. 47.

" $ "  œ $B %B $

" # "%  œ # B( B( B  %*

The solution set is ˜ &% ™.

N3.

* "&

œ

& "&

True

" # "%  œ # >( >( >  %*

The domain is e> l > Á „ (f.

Multiply each side by the LCD, Ð>  (ÑÐ>  (Ñ. Ð>  (ÑÐ>  (ÑŒ

" #   >( >( "% œ Ð>  (ÑÐ>  (ÑŒ #  >  %* " # Ð>  (ÑÐ>  (ÑŒ   Ð>  (ÑÐ>  (ÑŒ  >( >( "% œ Ð>  (ÑÐ>  (ÑŒ  Ð>  (ÑÐ>  (Ñ "Ð>  (Ñ  #Ð>  (Ñ œ "% >  (  #>  "% œ "% $>  ( œ "% $> œ #" >œ( Since ( is not in the domain, the solution set is g.

Find the domain of each term.

927

Rational Expressions and Functions N4.

# % #  œ # >#  #>  ) >#  '>  ) >  "'

Since B Á ", the equation of the vertical asymptote is B œ ". As the values of B get larger, the values of C get closer to !, so C œ ! is the equation of the horizontal asymptote.

Factor the denominators and determine the domain. >#  #>  ) œ Ð>  #ÑÐ>  %Ñ, so > Á #, %. >#  '>  ) œ Ð>  %ÑÐ>  #Ñ, so > Á %, #. >#  "' œ Ð>  %ÑÐ>  %Ñ, so > Á „ %. The domain is ÖB l B Á #ß „ %×. Multiply each side by the LCD, Ð>  #ÑÐ>  %ÑÐ>  %Ñ. # % Ð>  #ÑÐ>  %ÑÐ>  %Ñ” #  # • >  #>  ) >  '>  ) # œ Ð>  #ÑÐ>  %ÑÐ>  %Ñ # >  "' #Ð>  %Ñ  %Ð>  %Ñ œ #Ð>  #Ñ #>  )  %>  "' œ #>  % #>  #% œ #>  % %> œ #! >œ&

4 Section Exercises 1.

(b) The domain is eB l B Á !fÞ

3.

(a)

B"œ! B œ "

The domain is ÖB l B Á !ß #×.

œ BÐB  #ÑŒ

5.

' (

or B  # œ ! or Bœ#

œ#

The solution set is ˜ $# ™. N6. 0 ÐBÑ œ

" B"

B C

"" "  "!

'  &"

) (

# "

*

" &

" "!

Set the denominator equal to zero and solve for B. B"œ! B œ " 928

(b) The domain is eB l B Á „ %fÞ

7.

%

" # "  œ B#  "' B  % B% " # "  œ ÐB  %ÑÐB  %Ñ B  % B%

So „ % would have to be rejected as potential solutions.

True

! "

(a)

B#  "' equals ! if B œ „ %. B  % equals ! if B œ %. B  % equals ! if B œ %Þ

Because # is not in the domain of the equation, it is not a solution. Check B œ  $# :

B#œ! Bœ#

(b) The domain is eB l B Á "ß #fÞ

$ %   B B#

BÐ#BÑ œ $ÐB  #Ñ  %B #B# œ $B  '  %B #B#  B  ' œ ! Ð#B  $ÑÐB  #Ñ œ ! #B  $ œ ! B œ  $#

or or

Solutions of " and # would be rejected since these values would make a denominator of the original equation equal to !.

Multiply each side by the LCD, BÐB  #Ñ. #B  B#

" "  œ! B" B#

Set each denominator equal to ! and solve.

#B $ % œ  B# B B#

BÐB  #ÑŒ

" " B  œ $B #B $

Only ! would make any of the denominators equal to !, so ! would have to be rejected as a potential solution.

The solution set is Ö&×. N5.

(a)

(a)

# " %  œ B#  B B  $ B#

B#  B œ BÐB  "Ñ is ! if B œ ! or B œ ". B  $ is ! if B œ $. B  # is ! if B œ #Þ So !, ", $, and # would have to be rejected as potential solutions. (b) The domain is eB l B Á !ß "ß $ß #fÞ

Rational Expressions and Functions 9.

(a)

' $ &  œ %B  ( B 'B  "$

19.

%B  ( is ! if B œ  (% . B is ! if B œ !. 'B  "$ is ! if B œ "$ ' . So  (% , !, and "$ ' would have to be rejected as potential solutions. (b) The domain is ˜B l B Á

11.

(a)

B(œ! B œ (

™  (% ß !ß "$ ' Þ

21.

(b) The domain is ˜B l B Á %ß (# ™Þ

No, there is no possibility that the proposed solution will be rejected, because there are no variables in the denominators in the original equation.

In Exercises 15–46, check each potential solution in the original equation. $ & ( œ  %B #B % Multiply by the LCD, %B. ÐB Á !Ñ $ & ( %BŒ  œ %BŒ   %B #B % $ œ #Ð&Ñ  BÐ(Ñ $ œ "!  (B (B œ ( Bœ" Check B œ ":

$ %

œ

"! %

The solution set is e"f. 17.



( %

True

#% œ # B Multiply by the LCD, B. ÐB Á !Ñ #% BŒB   œ # • B B B#  #% œ #B B#  #B  #% œ ! ÐB  'ÑÐB  %Ñ œ ! or or

B%œ! Bœ%

Check B œ ': '  % œ # True Check B œ %: %  ' œ # True The solution set is e'ß %f.

B% #B  $ œ B' #B  " Multiply by the LCD, ÐB  'ÑÐ#B  "Ñ. Note that B Á ' and B Á "# Þ B% #B  $ ÐB  'ÑÐ#B  "ÑŒ  œ ÐB  'ÑÐ#B  "ÑŒ  B' #B  " Ð#B  "ÑÐB  %Ñ œ ÐB  'ÑÐ#B  $Ñ #B#  *B  % œ #B#  "&B  ") #%B œ "% "% ( B œ #% œ  "# A calculator check is suggested.

( "" "" Check B œ  "# :  "$ œ  "$ ( ™ The solution set is ˜ "# .

B

B'œ! B œ '

B$œ! Bœ$

The solution set is e(ß $f.

So % and (# would have to be rejected as potential solutions.

15.

or or

Check B œ (:  (%  $% œ " True Check B œ $: $%  (% œ " True

$B  " 'B  & œ B% #B  (

B  % is ! if B œ %. #B  ( is ! if B œ (# .

13.

B #"  œ " % %B Multiply by the LCD, %B. ÐB Á !Ñ B#  #" œ %B B#  %B  #" œ ! ÐB  (ÑÐB  $Ñ œ !

23.

True

$B  " 'B  & œ B% #B  ( Multiply by the LCD, ÐB  %ÑÐ#B  (Ñ. Note that B Á % and B Á (# . ÐB  %ÑÐ#B  (ÑŒ

$B  " 'B  &  œ ÐB  %ÑÐ#B  (ÑŒ  B% #B  ( Ð#B  (ÑÐ$B  "Ñ œ ÐB  %ÑÐ'B  &Ñ 'B#  "*B  ( œ 'B#  "*B  #! ( œ #! False The false statement indicates that the original equation has no solution. The solution set is g.

929

Rational Expressions and Functions 25.

$ Check C œ $:  "# 

The solution set is e$f.

27.

$ #   B  # ÐB  #ÑÐB  #Ñ " œ ÐB  #ÑÐB  #ÑŒ  B# $ÐB  #Ñ  # œ B  # $B  '  # œ B  # $B  ) œ B  # #B œ "! Bœ&

& "#

œ

# "#

True

Check B œ &:

( $



" $

The solution set is e!f.

œ#

33.

930

œ

" $

True

" $ &  œ # C# C( C  *C  "% " $ &  œ C# C( ÐC  #ÑÐC  (Ñ Multiply by the LCD, ÐC  #ÑÐC  (Ñ. ÐC Á #, (Ñ " $ ÐC  #ÑÐC  (ÑŒ   C# C( & œ ÐC  #ÑÐC  (ÑŒ  ÐC  #ÑÐC  (Ñ ÐC  (Ñ  $ÐC  #Ñ œ & C  (  $C  ' œ & %C  "$ œ & %C œ ) C œ #

* % #  œ B 'B  $ 'B  $

35.

" Multiply by the LCD, BÐ'B  $Ñ. ŒB Á !,  # * % # BÐ'B  $ÑŒ   œ BÐ'B  $ÑŒ  B 'B  $ 'B  $ *Ð'B  $Ñ  %B œ #B &%B  #(  %B œ #B &'B œ #( #( Bœ &' Check B œ

#( &' :

&' $

&'  Ð ""# $ Ñœ $

™ The solution set is ˜ #( &' .

The solution set is g. $ # "  œ B  # B#  % B# $ # "  œ B  # ÐB  #ÑÐB  #Ñ B# Multiply by the LCD, ÐB  #ÑÐB  #Ñ. ÐB Á #, #Ñ

# #"

The solution set is g.

Since ! cannot appear in the denominator, this equation has no solution.

31.



But C cannot equal # because that would make the denominator C  # equal to !. Since division by ! is undefined, the equation has no solution.

True

' & #!  œ # B% B B  %B ' & #!  œ B% B BÐB  %Ñ Multiply by the LCD, BÐB  %ÑÞ ÐB Á !ß %Ñ ' & #! BÐB  %ÑŒ   œ BÐB  %ÑŒ  B% B BÐB  %Ñ 'B  &ÐB  %Ñ œ #! 'B  &B  #! œ #! ""B œ ! Bœ!

* #"

The solution set is e&f.

( " #  œ 'B  $ $ #B  " ( " #  œ $Ð#B  "Ñ $ #B  " Multiply by the LCD, $Ð#B  "Ñ. ˆB Á  "# ‰ ( " # $Ð#B  "ÑŒ   œ $Ð#B  "ÑŒ  'B  $ $ #B  " (  "Ð#B  "Ñ œ $Ð#Ñ (  #B  " œ ' #B œ ! Bœ! Check B œ !:

29.

ÐB  #ÑÐB  #ÑŒ

" & #  œ C  " "# $C  $ " & #  œ C  " "# $ÐC  "Ñ Multiply by the LCD, "#ÐC  "Ñ. ÐC Á "Ñ " & # "#ÐC  "ÑŒ   œ "#ÐC  "ÑŒ  C  " "# $ÐC  "Ñ "#  &ÐC  "Ñ œ ) "#  &C  & œ ) &C  ( œ ) &C œ "& C œ $

37.

" " "  œ B# % %ÐB#  %Ñ " " "  œ B# % %ÐB  #ÑÐB  #Ñ Multiply by the LCD, %ÐB  #ÑÐB  #Ñ. ÐB Á #, #Ñ

True

Rational Expressions and Functions %ÐB  #ÑÐB  #ÑŒ

" "   B# %

43.

œ %ÐB  #ÑÐB  #Ñ”

" • %ÐB#  %Ñ %ÐB  #Ñ  ÐB  #ÑÐB  #Ñ œ " %B  )  B#  % œ " B#  %B  $ œ ! ÐB  $ÑÐB  "Ñ œ ! or or

B$œ! B œ $

B"œ! B œ "

" Check B œ $:  "&  "% œ #! True " " " Check B œ ":  $  % œ  "# True The solution set is e$ß "f. ' ( %)  œ # 39. A$ A& A  #A  "& ' ( %)  œ A$ A& ÐA  $ÑÐA  &Ñ Multiply by the LCD, ÐA  $ÑÐA  &Ñ. ÐA Á $, &Ñ ' ( ÐA  $ÑÐA  &ÑŒ   A$ A& %) œ ÐA  $ÑÐA  &Ñ” • ÐA  $ÑÐA  &Ñ 'ÐA  &Ñ  (ÐA  $Ñ œ %) 'A  $!  (A  #" œ %) A  &" œ %) A œ $ A œ $ But A cannot equal $ because that would make the denominator A  $ equal to !. Since division by ! is undefined, the equation has no solution.

The solution set is g. B % ")  œ # 41. B$ B$ B * B % ")  œ B$ B$ ÐB  $ÑÐB  $Ñ Multiply by the LCD, ÐB  $ÑÐB  $ÑÞ B Á $, $ B % ÐB  $ÑÐB  $ÑŒ   B$ B$ ") œ ÐB  $ÑÐB  $ÑŒ  ÐB  $ÑÐB  $Ñ BÐB  $Ñ  %ÐB  $Ñ œ ") B#  $B  %B  "# œ ") B#  (B  $! œ ! ÐB  $ÑÐB  "!Ñ œ ! B$œ! Bœ$

or or

B  "! œ ! B œ "!

But B Á $ since a denominator of ! results. The only solution to check is "!. Check B œ "!:

"! "$

 Ð %( Ñ œ

The solution set is e"!f.

") *"

True

" B )  œ # B% B% B  "' " B )  œ B% B% ÐB  %ÑÐB  %Ñ Multiply by the LCD, ÐB  %ÑÐB  %Ñ. B Á %, % " B ÐB  %ÑÐB  %ÑŒ   B% B% ) œ ÐB  %ÑÐB  %ÑŒ  ÐB  %ÑÐB  %Ñ ÐB  %Ñ  BÐB  %Ñ œ ) B  %  B#  %B œ ) B#  &B  % œ ! ÐB  %ÑÐB  "Ñ œ ! B%œ! B œ %

or

B"œ! B œ "

But B cannot equal %, so we only need to check ". ) Check B œ ": "$  "& œ "& True The solution set is e"f.

45.

# " %  œ # 5#  5  ' 5#  5  # 5  %5  $ # " %  œ Ð5  $ÑÐ5  #Ñ Ð5  #ÑÐ5  "Ñ Ð5  "ÑÐ5  $Ñ Multiply by the LCD, Ð5  $ÑÐ5  #ÑÐ5  "ÑÞ 5 Á $, #, " #Ð5  "Ñ  "Ð5  $Ñ œ %Ð5  #Ñ #5  #  5  $ œ %5  ) "$ œ 5 Check 5 œ "$:

" ))



" "&%

The solution set is e"$f.

47.

œ

" &'

True

&B  "% #B#  &B  # #B  % œ  # B * B#  * B$ &B  "% ÐB  $ÑÐB  $Ñ #B#  &B  # #B  % œ  ÐB  $ÑÐB  $Ñ B$ Multiply by the LCD, ÐB  $ÑÐB  $Ñ. ÐB Á $, $Ñ &B  "% œ #B#  &B  #  Ð#B  %ÑÐB  $Ñ &B  "% œ #B#  &B  #  #B#  "!B  "# &B  "% œ &B  "% True This equation is true for every real number value of B, but we have already determined that B Á $ or B Á $. So every real number except $ and $ is a solution. The solution set is ÖB l B Á „ $× or Ð∞ß $Ñ ∪ Ð$ß $Ñ ∪ Ð$ß ∞Ñ. 931

Rational Expressions and Functions 49.

# is not defined when B œ !, so an B equation of the vertical asymptote is B œ !. The graph approaches the B-axis as B gets very large (positive or negative), so C œ ! is the horizontal asymptote.

!Þ* # #Ð"  !Þ*Ñ !Þ)" œ œ %Þ!& ¸ %Þ" #Ð!Þ"Ñ

0 ÐBÑ œ

(c) AÐ!Þ*Ñ œ

(d) Based on the answers in (a), (b), and (c), we see that as the traffic intensity increases, the waiting time also increases. 57.

51.

53.

" is not defined when B œ !, so an B equation of the vertical asymptote is B œ !. The graph approaches the B-axis as B gets very large (positive or negative), so C œ ! is the horizontal asymptote. 1ÐBÑ œ 

" is not defined when B œ #, so an B# equation of the vertical asymptote is B œ #. The graph approaches the B-axis as B gets very large (positive or negative), so C œ ! is the horizontal asymptote.

##&,!!! < ##&,!!! %&! œ < %&!< œ ##&,!!! ##&,!!! œ "# :

"! (

œ

&# (

The solution set is ˜ "# ™.

7.

'

13.

15.

( &  'B )B No equals sign appears so this is an expression. ( &  $ • #B % • #B (Ð%Ñ &Ð$Ñ œ  $ • #B • % % • #B • $ #)  "& %$ œ œ $ • % • #B #%B

17.

LCD œ 24x

' "  B" B # %  B B" No equals sign appears so this is an expression. Multiply the numerator and denominator by the LCD of all the fractions, BÐB  "Ñ. 'ÐBÑ  "ÐB  "Ñ #ÐB  "Ñ  %ÐBÑ 'B  B  " œ #B  #  %B &B  " &B  " , or œ #B  # #ÐB  "Ñ œ

11.

, #  ,  ' , #  ),  "' • , #  #,  ) $,  "# No equals sign appears so this is an expression.

True

œ

9.

B# & • * )  %B No equals sign appears so this is an expression. B# & œ • * %ÐB  #Ñ & & œ œ $' $'

B #C  BC BC No equals sign appears so this is an expression. BÐB  CÑ #CÐB  CÑ  B  C ÐB  CÑ B  C ÐB  CÑ LCD œ (x  y)(x  y) B#  BC  #BC  #C# œ ÐB  CÑ B  C B#  BC  #C# œ ÐB  CÑ B  C

19.

œ

Ð,  $ÑÐ,  #Ñ Ð,  %ÑÐ,  %Ñ • Ð,  %ÑÐ,  #Ñ $Ð,  %Ñ

œ

,$ $

& $  B#  #B B#  % No equals sign appears so this is an expression. & $ œ  BÐB  #Ñ ÐB  #ÑÐB  #Ñ &ÐB  #Ñ $B œ  BÐB  #ÑÐB  #Ñ ÐB  #ÑÐB  #ÑB &B  "!  $B œ BÐB  #ÑÐB  #Ñ #B  "! œ BÐB  #ÑÐB  #Ñ & $  B C *B#  #&C# B# C No equals sign appears so this is an expression. Multiply the numerator and denominator by the LCD of all the fractions, B# CÞ

œ

œ œ œ

B# C Œ

& $   B C *B#  #&C# B# C Œ  B# C &BC  $B# *B#  #&C# BÐ$B  &CÑ Ð$B  &CÑÐ$B  &CÑ B $B  &C

933

Rational Expressions and Functions 21.

23.

%C#  "$C  $ %C#  ""C  $ ƒ #C#  *C  * 'C#  &C  ' No equals sign appears so this is an expression. Ð%C  "ÑÐC  $Ñ Ð#C  $ÑÐ$C  #Ñ œ • Ð#C  $ÑÐC  $Ñ Ð%C  "ÑÐC  $Ñ $C  # œ C$ $< ' œ" ÐR  8Ñ R 8 R V œ >R  >8 Get the R terms on one side. R V  >R œ >8 Factor out N. R ÐV  >Ñ œ >8 >8 8> , or Rœ >V V>

Rational Expressions and Functions N4. Step 2 Let B œ the number (in millions) of Americans who lived in poverty in 2008.

N6. (a) Let B œ the rate of the current. Use . œ , or > œ .< , to complete the table.

Step 3 Set up a proportion.

Against Current With Current

"$ B œ "!! $!# Step 4 To solve the equation, multiply by the LCD. "$ B ‹  œ $!,#!!Š "!! $!# $!# "$ œ "!! B $*#' œ "!!B B œ $*Þ#'

$!,#!!Œ

Step 5 There were $*Þ#' million Americans who lived in poverty in 2008. Step 6 The ratio of $*Þ#' to $!# equals

"$ "!! .

N5. Step 2 Let B œ the additional number of gallons of gasoline needed. Step 3 He knows that he can drive &!! miles with #) gallons of gasoline. He wants to drive %!! miles using Ð"!  BÑ gallons of gasoline. Set up a proportion. &!! %!! œ #) "!  B "#& %!! œ ( "!  B

Step 4 Find the cross products and solve for B. "#&Ð"!  BÑ œ (Ð%!!Ñ "#&!  "#&B œ #)!! "#&B œ "&&! B œ "&&! "#& B œ "#Þ%

Distributive property Subtract 1250. Divide by 125.

Step 6 Check The "! gallons plus the "#Þ% gallons equals ##Þ% gallons. We'll check the rates (miles/gallon). Note that we could also use gallons/mile. ¸ "(Þ)' mpg

$'

#!  B

%%

#!  B

> $' #!  B %% #!  B

$' %% . œ #!  B #!  B Multiply by the LCD, Ð#!  BÑÐ#!  BÑ. $' Ð#!  BÑÐ#!  BÑ #!  B %% œ Ð#!  BÑÐ#!  BÑ #!  B $'Ð#!  BÑ œ %%Ð#!  BÑ (#!  $'B œ ))!  %%B )!B œ "'! Bœ# The rate of the current is # mph. N7. Let B œ the driving rate for Pat. Then B  & œ the driving rate for James. Use . œ , or > œ .< , to complete the table. .


"$! B& ")! B

Now write an equation. Time for James Æ "$! B&

plus Æ 

time for Pat Æ ")! B

equals Æ

& hr. Æ

œ

&

Multiply by the LCD, BÐB  &Ñ.

Step 5 He will need about "#Þ% more gallons of gasoline.

&!! #)


œ

.

Step 3 At '! mph, his time at &" mph would be decreased $ hr. B B œ $ '! &"

The correct dose is #Þ% mL. 45.

"!  "#  B

'Ð"#  BÑ œ "!Ð"#  BÑ (#  'B œ "#!  "!B "'B œ %) Bœ$

œ $"Þ#& mpg

The rates are approximately equal, so the solution is correct. Note that we could have used our exact value for B to get the exact rate $!! œ $"Þ#. $  Ð"!$#Î"&'Ñ 41.

'  "#  B

Time "! "#  B ' "#  B

"(B œ #!B  $!'! $!'! œ $B "!#! œ B Step 5 The distance from Montpelier to Columbia is "!#! miles. Step 6 Check "!#! miles at &" mph takes

"!#! &" or takes "!#! '! or

#! hours; "!#! miles at '! mph "( hours; "( œ #!  $ as required.

Rational Expressions and Functions 49.

Step 5 The distance for both parts of the trip is given by

Step 2 Let B œ the one-way distance. Use > œ

.
B &!! B $&!

Step 3 The total flying time in both directions was )Þ& hours, so B B  œ )Þ&. &!! $&!

Step 6 " Check *! miles at '! mph takes *! '! or " # hours; "!! miles at &! mph takes # hours. The second part of the trip takes "# hour more than the first part, as required. 53.

Complete the table.

Step 4 Multiply by the LCD, $&!!.

Worker

(B  "!B œ #*,(&! "(B œ #*,(&! B œ "(&!

Part done by Butch " "& B " '!Ð "& B

First Part Second Part


with "# Þ =œ "'! œ

5 > 5 " #

5 œ "'!Ð "# Ñ œ )! So = œ

)! > .

=œ

Now let > œ %$ . )! $ %

)! % $#! # or "!' • œ " $ $ $ A rate of "!' #$ miles per hour is needed to go the same distance in $% minute. =œ

Let 0 œ the frequency of a string in cycles per second and = œ the length in feet. 0 varies inversely as =, so 5 0œ = for some constant 5 . Since 0 œ #&! when = œ #, substitute these values in the equation and solve for 5 . 5 0œ = 5 #&! œ # 5 œ #&!Ð#Ñ œ &!! &!! So 0 œ . Now let = œ &. = 0 œ &!! & œ "!!.

M œ 5T > #)! œ 5Ð#!!!ÑÐ%Ñ #)! 5 œ )!!! œ !Þ!$& So M œ !Þ!$&T >Þ Now let > œ '. M œ !Þ!$&Ð#!!!ÑÐ'Ñ œ %#! The interest would be $%#!Þ

943

Rational Expressions and Functions 53.

Let :" œ &!,!!!, :# œ (&,!!!, and . œ #&!.

The weight [ of a bass varies jointly as its girth K and the square of its length P, so

% &!,!!! • (&,!!! Œ  )(& #&! œ %)!(,!!! œ '),&(" ($

Rœ

[ œ 5KP# for some constant 5 . Substitute ##Þ( for [ , #" for K, and $' for P.

Rounded to the nearest hundred, there are about '),'!! calls.

#

##Þ( œ 5Ð#"ÑÐ$'Ñ ##Þ( 5œ ¸ !Þ!!!)$% #(,#"'

59.

Fœ

So [ œ !Þ!!!)$%KP# . Now let K œ ") and P œ #).

The bass would weigh about ""Þ) pounds.

61.

Let J œ the force, A œ the weight of the car, = œ the speed, and < œ the radius. The force varies inversely as the radius and jointly as the weight and the square of the speed, so

63.

J œ

5A=# .
 ) " " $  œ >% # #Ð>  %Ñ Multiply by the LCD, #Ð>  %Ñ. Ð> Á %Ñ " " $ #Ð>  %ÑŒ   œ #Ð>  %ÑŒ  >% # #Ð>  %Ñ #  Ð>  %Ñ œ $ >'œ$ > œ $ Check > œ $: "  "# œ The solution set is e$f.

$ #

True

Rational Expressions and Functions 26.

&7 7 &'  œ 7  " $7  $ '7  ' &7 7 &'  œ 7  " $Ð7  "Ñ 'Ð7  "Ñ &7 7 #)  œ 7  " $Ð7  "Ñ $Ð7  "Ñ Multiply by the LCD, $Ð7  "Ñ. Ð7 Á "Ñ &7 7 $Ð7  "ÑŒ   7  " $Ð7  "Ñ #) œ $Ð7  "ÑŒ  $Ð7  "Ñ "&7  7 œ #) "%7 œ #) 7 œ # Check 7 œ #: "!  The solution set is e#f.

27.

# $

œ  &' '

28.

Substituting $ in the original equation results in division by !, so $ is not a solution. The solution set is g.

B$ œ" B$

(b) The solution set is not Öall real numbers× because $ is not in the domain. 30.

In simplifying the expression, we are combining terms to get a single fraction with a denominator of 'B. In solving the equation, we are finding a value for B that makes the equation true.

31.

The graph in choice C has a vertical and a horizontal asymptote. The equations are B œ ! and C œ !, respectively.

32.

33.

True

& $ B  œ # B# B$ B  &B  ' & $ B  œ B# B$ ÐB  #ÑÐB  $Ñ Multiply by the LCD, ÐB  #ÑÐB  $Ñ. ÐB Á $, #Ñ & $ ÐB  #ÑÐB  $ÑŒ   B# B$ B œ ÐB  #ÑÐB  $ÑŒ  ÐB  #ÑÐB  $Ñ &ÐB  $Ñ  $ÐB  #Ñ œ B &B  "&  $B  ' œ B )B  #" œ B (B œ #" B œ $

(a)

The left side of the equation is equal to " except for B œ $, when it is undefined. Hence, the solution set is eBlB Á $fÞ

True

# %5  " "  # œ 5" 5 " 5" # %5  " "  œ 5  " Ð5  "ÑÐ5  "Ñ 5" Multiply by the LCD, Ð5  "ÑÐ5  "Ñ. # %5  " Ð5  "ÑÐ5  "ÑŒ   5  " Ð5  "ÑÐ5  "Ñ " œ Ð5  "ÑÐ5  "ÑŒ  5" #Ð5  "Ñ  Ð%5  "Ñ œ "Ð5  "Ñ #5  #  %5  " œ 5  " #5  " œ 5  " !œ5 Check 5 œ !: #  " œ " The solution set is e!f.

29.

34.

35.

" " " œ  E F G Let F œ $! and G œ "!. " " " œ  E $! "! To solve for E, multiply both sides by the LCD, $!EÞ " " " $!EŒ  œ $!EŒ   E $! "! $! œ E  $E $! œ %E $! "& Eœ œ % # KQ 7 Solve J œ for 7. .# # J . œ KQ 7 Multiply by d # Þ J .# œ7 Divide by GQ Þ KQ Q@ Solve . œ for Q . Q 7 Multiply by M + m. .ÐQ  7Ñ œ Q @ .Q  .7 œ Q @ .7 œ Q @  .Q 7. œ Q Ð@  .Ñ 7. Qœ @. Let B œ the number of passenger-kilometers per day provided by high-speed trains. Write a proportion. B #$,#!! œ "&,!!! &),!!! #$,#!!Ð"&,!!!Ñ Bœ &),!!! œ '!!! The high-speed train would provide '!!! passenger-km per day in that region.

947

Rational Expressions and Functions 36.

Part done by Jane B *

Let B œ the rate of the boat in still water. Use . œ , or > œ .< , to make a table. Distance

Rate

Upstream

#%

B%

Downstream

%!

B%

Time #% B% %! B%

$'Š

Because the times are equal,



B B  ‹ œ $' • " * ' %B  'B œ $' "!B œ $' B œ $' "! Bœ

%! #% . œ B% B% Multiply by the LCD, ÐB  %ÑÐB  %Ñ. ÐB Á %, %Ñ %! #% ÐB  %ÑÐB  %ÑŒ  œ ÐB  %ÑÐB  %ÑŒ  B% B% %!ÐB  %Ñ œ #%ÐB  %Ñ %!B  "'! œ #%B  *' "'B œ #&' B œ "'

Cold Hot

" ) " "#

Fractional Part of the Job Done B ) B "#

B B

The sink will be filled in 38.

Jane Jessica

Rate " * " '

Time Working Together B B

or $ $& ") &

or $ $&

@ œ 5/ for some constant 5 . Since @ œ #&! when / œ &, substitute these values in the equation and solve for 5 . @ œ 5/ #&! œ 5Ð&Ñ &! œ 5 So @ œ &!/. Now let / œ )Þ'. @ œ &!Ð)Þ'Ñ œ %$! It should be viewed from %$! mm. 41.

The frequency 0 of a vibrating guitar string varies inversely as its length P, so 0œ

5 P

for some constant 5 . Substitute !Þ'& for P and %Þ$ for 0 . 5 !Þ'& 5 œ %Þ$Ð!Þ'&Ñ œ #Þ(*&

%Þ$ œ So 0 œ

minutes.

Fractional Part of the Job Done B * B '

Multiply by 36.

Let @ œ the viewing distance and / œ the amount of enlargement. @ varies directly as /, so

#Þ(*& P .

Now let P œ !Þ&.

0œ

Let B œ the time to do the job working together. Make a table. Worker

948

or

% %&

"

40.

" whole Part done part done plus equals by cold by hot job. B B  œ " ) "# B B #%Š  ‹ œ #% • " Multiply by 24. ) "# $B  #B œ #% &B œ #% B œ #% or % %& & #% &

œ

If C varies inversely as B, then C œ B5 , for some constant 5 . This form fits choice C.

Make a table. Time Working Together

" whole job.

39.

Let B œ the time it takes to fill the sink with both taps open.

Rate

") &

equals

Working together, they can do the job in hours.

The rate of the boat in still water is "' km per hr. 37.

part done by Jessica B '

plus

#Þ(*& œ &Þ&* !Þ&

The frequency would be &Þ&* vibrations per second. 42.

The volume Z of a rectangular box of a given height is proportional to its width [ and length P, so Z œ 5[ P for some constant 5 . Substitute # for [ , % for P, and "# for Z .

Rational Expressions and Functions "# œ 5Ð#ÑÐ%Ñ $ 5 œ "# ) œ # So Z œ

$ # [ P.

48.

[3]

49.

[1]

50.

[3]

Now let [ œ $ and P œ &.

Z œ $# Ð$ÑÐ&Ñ œ

%& #

The volume is ##Þ& cubic feet. 43.

[1]

44.

[1]

45.

[2]

46.

47.

[2]

[3]

B  #C B  #C œ # # B  %C ÐB  #CÑÐB  #CÑ " œ B  #C B#  #B  "& ÐB  &ÑÐB  $Ñ œ # B B' ÐB  $ÑÐB  #Ñ B& œ B# # &  7 $7# The LCD is $7# . # • $7 & œ  7 • $7 $7# '7 & '7  & œ  œ # # $7 $7 $7# * #  $B B$ * #Ð"Ñ œ  $B B  $ Ð"Ñ * # œ  $B $B *  Ð#Ñ œ $B "" "" œ , or $B B$ $ B  B # B" " B

Multiply the numerator and denominator by the LCD of all the fractions, #B. $ B #BŒ   B # œ B" #BŒ"   B '  B# œ #B  #ÐB  "Ñ B#  ' œ #B  #B  # B#  ' B#  ' œ œ %B  # #Ð#B  "Ñ

$ & B " ' B Multiply the numerator and denominator by the LCD of all the fractions, B. Bˆ B$  &‰ œ Bˆ'  B" ‰ $  &B œ 'B  " %C  "' #C  ) ƒ $! & Multiply by the reciprocal. %C  "' & œ • $! #C  ) %ÐC  %Ñ & œ • $! #ÐC  %Ñ %•& # " œ œ œ # • $! ' $ " "  # # >#  =# > = œ " " >"  ="  > = Multiply the numerator and denominator by the LCD of all the fractions, ># =# . ># =# Œ

" "  # ># = œ " " ># =# Œ   > = # # = > œ # >=  ># = =#  ># œ =>Ð=  >Ñ 51.

[1]

5 #  '5  * '5 #  "(5  $ • "  #"'5 $ *  5#

Factor "  #"'5 $ as the difference of cubes, "$  Ð'5Ñ$ . Ð5  $ÑÐ5  $Ñ œ Ð"  '5Ñ "  '5  $'5 # Ð'5  "ÑÐ5  $Ñ • Ð$  5ÑÐ$  5Ñ Ð5  $ÑÐ5  $Ñ Ð"ÑÐ'5  "Ñ "  '5  $'5 # Ð'5  "ÑÐ5  $Ñ • Ð"ÑÐ5  $ÑÐ5  $Ñ 5$ 5$ œ or # # "  '5  $'5 $'5  '5  " œ

949

Rational Expressions and Functions 52.

[1]

*B#  %'B  & B#  ""B  $! ƒ $ $B#  #B  " B  &B#  'B Multiply by the reciprocal.

*B#  %'B  & B B#  &B  ' • $B#  #B  " B#  ""B  $! Ð*B  "ÑÐB  &Ñ BÐB  'ÑÐB  "Ñ œ • Ð$B  "ÑÐB  "Ñ ÐB  'ÑÐB  &Ñ BÐ*B  "Ñ œ $B  " %+ ',  +  # [2] # # +  +,  #, +  %+,  $, # %+ ',  + œ  Ð+  #,ÑÐ+  ,Ñ Ð+  $,ÑÐ+  ,Ñ The LCD is Ð+  $,ÑÐ+  #,ÑÐ+  ,Ñ. %+Ð+  $,Ñ œ Ð+  #,ÑÐ+  ,ÑÐ+  $,Ñ Ð',  +ÑÐ+  #,Ñ  Ð+  $,ÑÐ+  ,ÑÐ+  #,Ñ %+Ð+  $,Ñ  Ð',  +ÑÐ+  #,Ñ œ Ð+  $,ÑÐ+  ,ÑÐ+  #,Ñ %+#  "#+,  '+,  "#, #  +#  #+, œ Ð+  $,ÑÐ+  ,ÑÐ+  #,Ñ # %+  "#+,  '+,  "#, #  +#  #+, œ Ð+  $,ÑÐ+  ,ÑÐ+  #,Ñ # &+  %+,  "#, # œ Ð+  $,ÑÐ+  ,ÑÐ+  #,Ñ + , [2]   , . The LCD is ,-.Þ + • -. , • ,. - • ,œ   , • -. - • ,. . • ,+-.  , # .  ,- # œ ,-. B$ " #  œ # [4] B#  &B  % B B  %B B$ " #  œ ÐB  %ÑÐB  "Ñ B BÐB  %Ñ Multiply by the LCD, BÐB  %ÑÐB  "Ñ. ÐB Á !, ", %Ñ B$ " BÐB  %ÑÐB  "ÑŒ   ÐB  %ÑÐB  "Ñ B # œ BÐB  %ÑÐB  "Ñ • Œ  BÐB  %Ñ BÐB  $Ñ  ÐB  %ÑÐB  "Ñ œ #ÐB  "Ñ B#  $B  ˆB#  &B  %‰ œ #B  #

56.

[5]

57.

[4]

œ

53.

54.

55.

B#  $B  B#  &B  % œ #B  # )B  % œ #B  # 'B œ # B œ "$

Check B œ

" $:

"& ""

$œ

The solution set is ˜ "$ ™.

950

 ") ""

True

V< for $ ƒ & "! Multiply by the reciprocal of the divisor. $  > "! • œ & >$ "Ð>  $Ñ • "! œ &Ð>  $Ñ œ "Ð#Ñ œ # B#  * B#  B  "# ƒ B$  $B# B$  *B#  #!B Multiply by the reciprocal. B#  * B B#  *B  #! œ $ • B  $B# B#  B  "# ÐB  $ÑÐB  $Ñ BÐB  &ÑÐB  %Ñ œ • B# ÐB  $Ñ ÐB  %ÑÐB  $Ñ B& œ B >#  >  ', >#  $>, ># Factor each denominator. >#  >  ' œ Ð>  $ÑÐ>  #Ñ >#  $> œ >Ð>  $Ñ The third denominator is already in factored form. The LCD is ># Ð>  $ÑÐ>  #Ñ.

( "  # '> $> The LCD is '># . ( " • #> œ # '> $> • #> ( #> (  #> œ # # œ '> '> '># $ &  9. (+% , $ #"+& , # The LCD is #"+& , $ . $ • $+ &•, œ % $  (+ , • $+ #"+& , # • , *+ &, œ  #"+& , $ #"+& , $ *+  &, œ #"+& , $ * #  10. B#  'B  * B#  * Factor each denominator. 8.

B#  'B  * œ ÐB  $Ñ# B#  * œ ÐB  $ÑÐB  $Ñ

Rational Expressions and Functions The LCD is ÐB  $ÑÐB  $Ñ# . * #  ÐB  $Ñ# ÐB  $ÑÐB  $Ñ *ÐB  $Ñ #ÐB  $Ñ œ  # ÐB  $Ñ ÐB  $Ñ ÐB  $ÑÐB  $ÑÐB  $Ñ *ÐB  $Ñ  #ÐB  $Ñ œ ÐB  $ÑÐB  $Ñ# *B  #(  #B  ' œ ÐB  $ÑÐB  $Ñ# ""B  #" œ ÐB  $ÑÐB  $Ñ# 11.

12.

13.

' " $B   B  % B  # B#  'B  ) ' " $B œ   B  % B  # ÐB  %ÑÐB  #Ñ The LCD is ÐB  %ÑÐB  #Ñ. 'ÐB  #Ñ "ÐB  %Ñ œ  B  % ÐB  #Ñ B  # ÐB  %Ñ $B  ÐB  %ÑÐB  #Ñ 'ÐB  #Ñ  B  %  $B œ ÐB  %ÑÐB  #Ñ 'B  "#  B  %  $B œ ÐB  %ÑÐB  #Ñ %B  "' œ ÐB  %ÑÐB  #Ñ %ÐB  %Ñ % œ œ ÐB  %ÑÐB  #Ñ B#

14.

15.

(a) Simplify this expression. #B B ""   $ % # The LCD is "#. #B • % B • $ "" • '   $•% %•$ #•' )B $B '' œ   "# "# "# )B  $B  '' œ "# ""B  '' ""ÐB  'Ñ œ œ "# "# (b) Solve this equation. œ

#B B ""  œ $ % # Multiply by the LCD, "#.

"# <  % œ "# ƒ "" "" <  % '<  #% '<  #% Multiply by the reciprocal. "# '<  #% œ • œ .< , to make a table. .




ÐB"Î% C#Î& Ñ#! BÐ"Î%Ñ • #! CÐ#Î&Ñ • #! œ # B B# & ) B C œ # œ B$ C ) B #Î$ # %Î$ ÐB Ñ B œ "%Î$ # (Î$ B B œ B%Î$  "%Î$ œ B"!Î$ œ

# ) $ œ ˆÈ *; ‰ &  ˆÈ #B‰

45.

5 "Î$ 5 "Î$ œ œ 5 "Î$  Ð"Î$Ñ œ 5 #Î$ 5 #Î$ • 5 " 5 "Î$

$3

 Ð#BÑ#Î$ œ Ð*;Ñ"Î) ‘ &  Ð#BÑ"Î$ ‘ #

*;

65.

81.

:"Î% ; $Î#  $" :# ; #Î$ 

#3

:"Î# ; $ $# :% ; %Î$ ; $  %Î$ œ %  "Î# *: ; &Î$ œ (Î# *: œ

961

Roots, Radicals, and Root Functions 83.

85.

87.

89.

91.

103. We are to show that, in general,

:#Î$ Ð:"Î$  #:%Î$ Ñ œ :#Î$ :"Î$  :#Î$ Ð#:%Î$ Ñ œ :#Î$  "Î$  #:#Î$  %Î$ œ :$Î$  #:'Î$ œ :"  #:# œ :  #:#

È +#  , # Á +  , .

When + œ $ and , œ %,

È +#  , # œ È $ #  % # œ È*  "'

œ È#& œ &,

5 "Î% Ð5 $Î#  5 "Î# Ñ œ 5 "Î%  $Î#  5 "Î%  "Î# œ 5 "Î%  'Î%  5 "Î%  #Î% œ 5 (Î%  5 $Î%

but +  , œ $  % œ (.

Since & Á (, È+#  , # Á +  , . 105. Use X ÐHÑ œ !Þ!(H$Î# with H œ "'.

'+(Î% Ð+(Î%  $+$Î% Ñ œ '+(Î%  Ð(Î%Ñ  ")+(Î%  Ð$Î%Ñ œ '+!  ")+%Î% œ 'Ð"Ñ  ")+" œ '  ")+

X Ð"'Ñ œ !Þ!(Ð"'Ñ$Î# œ !Þ!(Ð"'"Î# Ñ$ œ !Þ!(Ð%Ñ$ œ !Þ!(Ð'%Ñ œ %Þ%) To the nearest tenth of an hour, time X is %Þ& hours.

&B(Î' ÐB&Î'  B"Î' Ñ œ &B(Î'  &Î'  &B(Î'  Ð"Î'Ñ œ &B"#Î'  &B'Î' œ &B#  &B

107. Use a calculator and the formula [ œ $&Þ(%  !Þ'#"&X  $&Þ(&Z %Î#&  !Þ%#(&X Z %Î#& . For X œ $!°F and Z œ "& mph, [ ¸ "*Þ!°F. The table gives "*Þ!°F.

& $ % È B • ÈB œ B$Î& • B"Î%

109. As in Exercise 107, for X œ #!°F and Z œ #! mph, [ ¸ %Þ#°F. The table gives %°F.

$Î&  "Î%

93.

È B& È B)

œB œ B"#Î#!  &Î#! œ B"(Î#! œ

B&

95.

The results are the same.

"Î#3

3

B) "Î# œ B&Î#  Ð)Î#Ñ 3

œB

$Î#

œ

" B$Î#

$ CD œ C"Î# • ÐCDÑ"Î$ ÈC • È

œ C"Î# C"Î$ D "Î$ œ C"Î#  "Î$ D "Î$ œ C$Î'  #Î' D "Î$ œ C&Î' D "Î$

97. 99.

101.

% % È $ 7œÈ É 7"Î$ œ ˆ7"Î$ ‰ "Î% œ 7"Î"#

ÊÉÈB œ ÉÈB"Î# œ ÉÐB"Î# Ñ "Î# $ È % ÊÉ

œ ÈB"Î% œ ˆB"Î% ‰ "Î# œ B"Î)

$ B œ ÉÈB"Î%

œ ÉÐB"Î% Ñ "Î$ œ ˆB"Î"# ‰ "Î# œ B"Î#%

962

111. È#& • È$' œ & • ' œ $! È#& • $' œ È*!! œ $!

Simplifying Radical Expressions

3 Now Try Exercises

N1. (a) È( • È"" œ È( • "" œ È((

(b) È#78 • È"& œ È#78 • "& œ È$!78

$ $ $ $ N2. (a) È %•È &œÈ %•& œ È #!

% % % % (b) È &> • È ' • ' > > œ % œ È"' "' #

(e) Ê &

- # œ +#  , # - # œ &#  )# - # œ )* - œ È)*

& È 7"& 7"& 7$ œ & œ È#%$ #%$ $

The length of the hypotenuse is È)*.

N4. (a) È&! œ È#& • # œ È#& • È# œ &È#

(b) È"*# œ È'% • $ œ È'% • È$ œ )È$

(b) Substitute ' for + and "# for - to find , .

(c) È%# No perfect square (other than ") divides into %#, so È%# cannot be simplified further. $ $ $ $ $ (d) È "!) œ È #( • % œ È #( • È % œ $È %

% % % % % (e) È )! œ È "' • & œ È "' • È & œ #È &

+#  , # ,# ,# ,#

œ -# œ - #  +# œ "##  '# œ "!) , œ È"!) , œ È$' • $

N5. (a) È$'B& œ È$'B% • B œ È$'B% • ÈB œ 'B# ÈB (b) È$#7& 8% œ È"'7% 8% • #7 œ È"'7% 8% • È#7

, œ È$' • È$ , œ 'È $

œ %7# 8# È#7

$ $ (c) È "#&5 $ :( œ È "#&5 $ :' • : $ $ : œ È"#&5 $ :' • È $ : œ &5:# È

The length of the unknown side is 'È$. N9. Use the distance formula,

. œ ÈÐB#  B" Ñ#  ÐC#  C" Ñ# .

% % (d) È "'#B( C) œ È )"B% C) • #B$ % % œ È )"B% C) • È#B$

B" ß C" œ Ð%ß $Ñ and B# ß C# œ Ð)ß 'Ñ . œ ÈÒ)  Ð%ÑÓ#  Ò'  Ð$ÑÓ# œ ÈÐ%Ñ#  *#

œ $BC# È#B$ %

' # $ N6. (a) È ( œ (#/' œ ( "/$ œ È (

œ È"'  )" œ È*(

' % $ # (b) È C œ C %/' œ C #/$ œ È C

$ N7. È $ • È'

The least common index of $ and # is '. Write each radical as a sixth root.

' $ ' È $ œ $"/$ œ $ #/' œ È $ # œ È * ' ' È ' œ '" /# œ ' $ /' œ È ' $ œ È #"'

Therefore,

' $ ' ' È $ • È' œ È *•È #"' œ È "*%%.

N8. (a) Substitute & for + and ) for , in the Pythagorean theorem to find - .

3 Section Exercises 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21.

È$ • È$ œ È$ • $ œ È*, or $

È") • È# œ È") • # œ È$', or ' È& • È' œ È& • ' œ È$!

È"% • ÈB œ È"% • B œ È"%B

È"% • È$:;< œ È"% • $:;< œ È%#:;< $ $ $ $ È #•È &œÈ #•& œ È "!

$ $ $ $ È (B • È #C œ È (B • #C œ È "%BC % % % % È "" • È $œÈ "" • $ œ È $$

% % % % È #B • È $B# œ È #B • $B# œ È 'B$

$ % È (•È $ cannot be multiplied using the product rule, because the indexes ($ and %) are different. '% É "#" œ

È'% È"#"

œ

) ""

963

Roots, Radicals, and Root Functions 23.

$ É #& œ

25.

Ê

È$ È#&

œ

È$ &

ÈB ÈB B œ œ È#& #& &

27.

È :$ Ê œ œ È)" )" *

29.

Ê

31.

Ê

33.

Ê

35. 37. 39. 41. 43. 45. 47. 49. 51. 53. 55. 57. 59. 61.

63. 65.

:'

$

$

Ê

È:'

$

#

œ

: *

$ È #( #( $ $ $ #( œÊ œ $ œ œ È'% '% '% % %

$ # $ # È È Ñ œ #" & "' < > œ

È&!B$ œ È#&B# • #B œ ÈÐ&BÑ# • È#B œ &BÈ#B

È&!! È#> N3. (a) &

È%&

 %Ê È&

È* • &

#) *

%

È% • ( È*

#È (  %  $  $È & & )È ( œ  $ $ &  )È ( œ $ È&

(b) 'Ê

"' $ *  (Ê * "# B B $ $ È È )•# * œ' $ ( $ ÈB"# È B* $ $ $ È È ' ) # (È * œ  B% B$ $ $ '•#È # (B È * œ  % B B • B$ $ $ "# È #  (B È * œ % B

$

œ (È $

(b) È'$>  $È#)>ß >   ! œ È* • È(>  $È% • È(>

966

$

œ &

œ #È $  &È $ œ Ð#  &ÑÈ$

Here the radicals differ and are already simplified, so this expression cannot be simplified further.

È&

œ&

N1. (a) È"#  È(& œ È% • È$  È#& • È$

(c) 'È(  #È$

%

$ (c) È"#)>%  #È(#>$ $ œ È'%>$ • #>  #È$'># • #>

4 Now Try Exercises

œ $È(>

%

%

Adding and Subtracting Radical Expressions

œ $È(>  $ • # • È(> œ $È(>  'È(> œ Ð$  'ÑÈ(>

%

œ Ð&+  $,Ñ È+, $

139. *; #  #;  &;  ; # œ *; #  ; #  #;  &; œ ); #  $;

4

%

œ & È+% • È+, $  È)", % • È+, $ % % œ &+ È+, $  $, È+, $

œœ #'# œ '(' œ $$) B œ È$$) ¸ ")Þ%

The length of the leg is about ")Þ% inches. Similarly, for the model MSX-83 with - œ #%, we get È#)) ¸ "(Þ! inches, and for the model MSX60 with - œ #!, we get È#!! ¸ "%Þ" inches. %

%

#

+ , B#  B # #B# B#

$

4 Section Exercises 1.

Simplify each radical and subtract.

3.

#È%)  $È(& œ #È"' • $  $È#& • $

È$'  È"!! œ '  "! œ %

œ # • %È$  $ • &È$ œ )È$  "&È$ œ (È$

Roots, Radicals, and Root Functions 5.

$ $ È "'  %È &%

23.

$ $ ) • #  %È #( • # œÈ $ $ $ $ È È È œ ) #  % #(È#

$ # $ # œ $È B C  & • #È B C $ # œ Ð$  "!ÑÈB C

œ # È #  % • $È # $ $ $ œ #È#  "#È# œ "%È# $

7.

$

% % È $#  $È #

'È")  È$#  #È&!

27.

œ 'È* • #  È"' • #  #È#& • # œ ' • $È #  % È #  # • &È # œ ")È#  %È#  "!È# œ #%È#

11.

&È'  #È"!

#È&  $È#!  %È%& œ #È &  $ È % • &  % È * • &

% œ "*È #

29.

15.

31.

œ È$' • #B  È% • #B œ 'È#B  #È#B

$È(#7#  &È$#7#  $È")7# œ $È$'7# • #  &È"'7# • #  $È*7# • # œ $ • '7È#  & • %7È#  $ • $7È# œ ")7È#  #!7È#  *7È#

%

35.

$ $ $ $ #È "'  È &% œ #È )•#  È #( • # $ $ È È œ #•# #  $ #

œ %È #  $ È # $ œ (È # $

21.

$ $ #È #(B  #È )B $ $ œ #È#( • B  #È )•B $ $ œ # • $È B  # • #È B $ $ È È œ' B% B $ œ #È B

%

$ $ È '%BC#  È #(B% C&

$ $ œÈ '% • BC#  È #(B$ C$ • BC# $ $ BC# œ %È BC#  $BCÈ

$ È "*#=>%  È#(=$ >

œ È'%>$ • $=>  È*=# • $=> $ $=>  $=È$=> œ %>È $

37.

$ % #È)B%  $È"'B& $ % œ #È)B$ • B  $È"'B% • B $ % œ # • #BÈ B  $ • #BÈ B $ % È È œ %B B  'B B

œ Ð")7  #!7  *7ÑÈ# œ ""7È#

19.

%

%

$ BC# œ Ð%  $BCÑÈ

œ %È#B

17.

#È$#+$  &È#+$ % % % œ #È "' • È#+$  &È#+$ œ # • # • È#+$  &È#+$ % % œ Ð%  &ÑÈ#+$ œ *È#+$

33.

È(#B  È)B

% & % BC $È B C  #BÈ % % BC œ $ÈB% • BC  #BÈ

% BC  #BÈ % BC œ $BÈ % % BC œ Ð$B  #BÑÈBC œ BÈ

œ # È &  $ • #È &  % • $È & œ #È&  'È&  "#È& œ #!È&

% % &È $#  $È "'# % % È œ & "' • #  $È )" • # % % œ & • #È #  $ • $È # % % È È œ "! #  * #

The radicals differ and are already simplified, so the expression cannot be simplified further. 13.

$ $ $BÈ BC#  #È )B% C# $ $ $ œ $BÈ BC#  #È)B$ • È BC#

$ $ œ $BÈ BC#  # • #B • È BC# $ $ œ Ð$B  %BÑÈBC# œ BÈ BC#

% % % œ #È #  $È # œ &È #

9.

$ # œ (È B C

25.

% % œÈ "' • #  $È # % % % È È È œ "' #  $ #

$ # $ $È B C  &È )B# C $ È $ œ $È B# C  & È ) $ B# C

39.

$

41.

È) 

È'% È"'

œ È% • # 

) %

œ È% • È#  # œ #È #  #

#È & È & %È & "È &  œ  $ ' ' ' %È &  "È & œ ' &È & œ ' 967

Roots, Radicals, and Root Functions 43.

45.

47.

49.

51.

53.

968

È) È") ) ") œ  Ê Ê È* È$' * $' È %È # È *È # œ  $ ' #È # $È # œ  $ ' È È % # $ # œ  ' ' %È #  $È # (È # œ œ ' ' È$# #È# È#   È* $ $ È"'È# #È# È# œ   $ $ $ %È #  # È #  È # &È # œ œ $ $ È È# È &! # &! œ$ ) $Ê  ) È) È* * #È # &È # "  )• œ $• $ # È œ& #% È#& È* #& *  Ê ) Ê ' œ È B) È B' B B & $ œ % $ B B & $•B œ % $ B B •B &  $B œ B% # 7& $ 7  #7Ê #( '% $ $ È & 7# $ 7 #7È œ $  $ È#( È'%

55.

Only choice B has like radical terms, so it can be simplified without first simplifying the individual radical expressions. $È'  *È' œ "#È'

57.

$ % $ % È'%  È "#&  È "' œ È)#  È&$  È#%

œ )  &  # œ "& This sum can be found easily since each radicand has a whole number power corresponding to the index of the radical; that is, each radical expression simplifies to a whole number.

59.

61.

Let P œ È"*# ¸ È"*' œ "% and [ œ È%) ¸ È%* œ (. An estimate of the perimeter is #P  #[ œ #Ð"%Ñ  #Ð(Ñ œ %# meters. (A)

The perimeter, T , of a triangle is the sum of the measures of the sides. T œ $È#!  #È%&  È(& œ $È% • &  #È* • &  È#& • $ œ $ • #È &  # • $È &  & È $ œ 'È &  ' È &  & È $ œ "#È&  &È$

The perimeter is ˆ"#È&  &È$‰ inches. 63. LCD œ x4

%È")  È"!)  #È(#  $È"# œ %È*È#  È$'È$  #È$'È#  $È%È$ œ % • $È #  ' È $  # • 'È #  $ • #È $ œ "#È#  'È$  "#È#  'È$

$Ê

œ #%È#  "#È$

$

$ $ $ 7$ È 7# 7# $È #7È œ  $ % $ $ 7È 7È 7# 7# œ  " # $ $ È È # #7 7  7 $ 7# 7È 7# œ œ # # $ $ È# È & $ # $ & % $ $Ê '  % Ê * œ $ $ È B' È B* B B $ $ È È # & œ$ # % $ B B $ $ # %È & $•B•È œ  $ # B •B B LCD œ x3 $ $ $BÈ #  %È & œ B$

To find the perimeter, add the lengths of the sides.

The perimeter is ˆ#%È#  "#È$‰ inches.

65.

&BCÐ#B# C$  %BÑ œ &BCÐ#B# C$ Ñ  &BCÐ%BÑ œ "!B$ C%  #!B# C

67.

Ð+#  ,ÑÐ+#  ,Ñ œ Ð+# Ñ#  , # œ +%  , #

69.

Ð%B$  $Ñ# œ Ð%B$ Ñ#  # • %B$ • $  $# œ "'B'  #%B$  * Now multiply by %B$  $. "'B'  #%B$ %B$ %)B'  (#B$ * '%B  *'B'  $'B$ '%B*  "%%B'  "!)B$ Thus, %B$  $

$

 *  $  #(  #(

œ '%B*  "%%B'  "!)B$  #(.

Roots, Radicals, and Root Functions 71.

5

)B#  "!B #BÐ%B  &Ñ œ # 'B #B • $B %B  & œ $B

Multiplying and Dividing Radical Expressions

5 Now Try Exercises N1. Use the FOIL method and multiply as two binomials. (a) ˆ)  È&‰ˆ*  È#‰ F O I L È È È È œ )•*  ) #  * &  &• # œ (#  )È#  *È&  È"!

(b) ˆÈ(  È&‰ˆÈ(  È&‰ œ È( • È(  È( • È&

 È& • È(  È& • È& œ(& œ#

(c) ˆÈ"&  %‰ œ ˆÈ"&  %‰ˆÈ"&  %‰ #

œ È"& • È"&  %È"&  %È"&  % • % œ "&  )È"&  "'

È"'B) • $ %)B) (b) Ë $ œ ÈC# • C C %B% È$ œ CÈC %B% È$ • ÈC œ CÈC • ÈC %B% È$C œ C•C %B% È$C œ ÐC  !Ñ C# N4. (a) Ê $

% % $ È C (B È % (B œ % • % (b) Ë ÈC ÈC$ C

œ

œ $"  )È"&

$ ‰ˆ $ ‰ (d) ˆ)  È & )È & $ $ $ $ È &È &•È & œ ) • )  ) &  )È $ #& œ '%  È

(e) For this one, we use the difference of squares. ˆÈ7  È8‰ˆÈ7  È8‰

# # œ ˆÈ7‰  ˆÈ8‰ œ 7  8, 7   ! and 8   !

) ) • È"$ )È"$ N2. (a) œ œ È"$ È"$ • È"$ "$ (b)

*È ( *È ( • È $ *È#" œ œ œ $È#" È$ È$ • È$ $

"! "! "! "! • È& œ œ œ È#! È% • & È # & #È & • È & "!È& "!È& œ œ œ È & #•& "! È#( È* • $ #( N3. (a) Ê œ œ È)! È"' • & )! È $ $ $È $ • È & œ œ %È & %È & • È & $È"& $È"& œ œ %•& #! (c)

$ È ) ) # # œ $ œ $ œ $ $ È È È È )" )" $ $ #( • $ $ $ È È #• * # * œ $ œ $ $ È È $ $• * $È#( $ $ #È * #È * œ œ $•$ *

œ N5. (a)

% È (BC$ % % È C

% È

(BC$ C

ÐB   !ß C  !Ñ

% "  È$

Multiply both the numerator and denominator by the conjugate of the denominator, "  È$. œ

%Ð"  È$Ñ ˆ"  È$‰ˆ"  È$‰

%Ð"  È$Ñ %Ð"  È$Ñ œ "$ # œ #Ð"  È$Ñ, or #  #È$

œ

(b)

% &  È(

Multiply both the numerator and denominator by the conjugate of the denominator, &  È(. œ

%Ð&  È(Ñ ˆ&  È(‰ˆ&  È(‰

%Ð&  È(Ñ #&  ( %Ð&  È(Ñ #Ð&  È(Ñ œ œ ") * œ

969

Roots, Radicals, and Root Functions (c)

È$  È(

9.

È&  È# œ œ

ˆÈ$  È(‰ˆÈ&  È#‰

(d)

È"&  È'  È$&  È"%

È$B  ÈC )

œ

œ

œ & • ' • È#  & • # • È# œ $!È#  "!È#

ˆÈ&  È#‰ˆÈ&  È#‰

&# È"&  È'  È$&  È"% œ $

)ŠÈ$B  ÈC‹

ŠÈ$B  ÈC‹ŠÈ$B  ÈC‹ )ŠÈ$B  ÈC‹

œ #!È#

11.

ˆÈ#  È$‰ˆÈ#  È$‰

15.

ˆÈ)  È#‰ˆÈ)  È#‰

17.

ˆÈ#  "‰ˆÈ$  "‰

ŠE  ÈF ‹ŠE  ÈF ‹ # œ E#  ŠÈF ‹

œ E#  F

3.

ŠÈE  ÈF ‹ŠÈE  ÈF ‹

19.

œ È##  È&&  È"%  È$&

21.

œ E  #È E F  F

7.

(D)

È ' ˆ$  È # ‰ œ $ È '  È ' • È #

œ $È'  È"# œ $È '  È % • È $ œ $È '  #È $

970

œ ˆ#È$‰ˆ$È$‰  ˆ#È$‰ˆ#È&‰  ˆÈ&‰ˆ$È$‰  ˆÈ&‰ˆ#È&‰ œ )  È"&

23.

ˆÈ &  # ‰ # œ ˆ È &‰ #  # • È & • #  # # œ &  %È &  % œ *  %È &

25.

ˆÈ#"  È&‰ #

œ ˆÈ#"‰ #  # • È#" • È&  ˆÈ&‰ # œ #"  #È"!&  & œ #'  #È"!&

27.

# # œ ŠÈE‹  #ÈEÈF  ŠÈF ‹

ˆ#È$  È&‰ˆ$È$  #È&‰

œ # • $ • $  # • #È $ • &  $ È & • $  # • & œ ")  %È"&  $È"&  "!

(A)

# ŠÈE  ÈF ‹

ˆÈ""  È(‰ˆÈ#  È&‰

F O I L È È È È È È È œ "" • #  "" • &  ( • #  ( • È&

# # œ ŠÈE‹  ŠÈF ‹

œEF

5.

(E)

œ ˆ È ) ‰ #  ˆ È #‰ # œ)#œ'

œ È'  È#  È$  "

5 Section Exercises 1.

œ ˆ È # ‰ #  ˆ È $‰ # œ #  $ œ "

F O I L È È È È œ #• $  " #  " $  "•"

Ð$B Á Cß B  !ß C  !Ñ

"&5  È&!5 # "&5  È#&5 # • # œ (b) #!5 #!5 "&5  &5 È# œ Ð5  !Ñ #!5 &5Ð$  È#Ñ œ &5 • % $  È# œ %

ˆÈ(  $‰ˆÈ(  $‰ œ ˆÈ(‰ #  $# œ (  * œ #

13.

$B  C

"&  'È# $Ð&  #È#Ñ &  #È# œ œ N6. (a) ") ") '

&ˆÈ(#  È)‰ œ &È(#  &È) œ &È$' • È#  &È% • È#

$ ‰ˆ $ ‰ œ ##  ˆ È $ ‰# ˆ#  È ' #È ' ' $ È

'• $ È œ %  $' œ%

29.

$ È

3

'

$ ‰ˆ $ $ ˆ#  È # %  #È #  È%‰

$ $ œ # • %  # • # È #  #È % $ $ $ $ $  %È #  #È # • È #  È # • È %

$ œ )  % È #  #È %  % È #  # È %  È ) œ)# œ "! $

$

$

$

Roots, Radicals, and Root Functions 31.

ˆ$ÈB  È&‰ˆ#ÈB  "‰

33.

ˆ$È<  È=‰ˆ$È<  È=‰

35.

œ ˆ$ÈB‰ˆ#ÈB‰  $ÈB  #È&ÈB  È& œ 'B  $ÈB  #È&B  È&

œ ˆ$È# Let d = 25. #& # "' œ > By the square root property, & %

? ˆ%  $ È $  % ‰ # œ #(

N6. B#  'B  # œ ! B#  'B œ #

B œ È%& B œ $È&

Solution set: š$È&ß $È&›

?   %— œ #(

È$

> œ È"!.

Check B œ „ $È&: #Ð%&Ñ  *! œ !

&

%  $È $ &

Original equation 4  3È 3 Let x = . 5

#

Solution set: š %„$&

(b) #B#  *! œ ! #B# œ *! B# œ %& By the square root property, or

%  $È $

Bœ

The check for the other solution is similar.

Solution set: šÈ"!ß È"!›

B œ È%& B œ $È &

or

%  $È $ : &

ˆ$ È $

N2. (a) ># œ "! By the square root property, > œ È"!

or

Ð&B  %Ñ# œ #(

#

&B  % œ È#( &B œ %  $È$

or

Solution set: š$ „ È""›

True

N7. B#  B  $ œ ! B#  B œ $ Complete the square.

 "# " ‘ # œ ˆ "# ‰ # œ 3

Add

" %

3

" %

to each side. B#  B  "% œ $  ˆB  "# ‰ # œ "$ %

" %

3

Use the square root property.

From Chapter 9 of Student’s Solutions Manual for Intermediate Algebra, Eleventh Edition. Margaret L. Lial, John Hornsby, Terry McGinnis. Copyright © 2012 by Pearson Education, Inc. Publishing as Addison-Wesley. All rights reserved.

995

Quadratic Equations, Inequalities, and Functions B

" #

B

" #

œ É "$ %

or

B

" #

œ

or

B

" #

È"$ #

"  È"$ Bœ #

œ

È"$ #

È š "„# "$ ›.

or or

Solution set: š% „ 3È&›

B œ %  3È&

1 Section Exercises

#

N8. $B  "#B  & œ ! $B#  "#B œ & B#  %B œ &$

1Þ

B and C are quadratic equations because they are equations with a squared term and no terms of higher degree.

Complete the square.

3.

The zero-factor property requires a product equal to !. The first step should have been to rewrite the equation with ! on one side.

Add % to each side.

5.

B#  $B  # œ ! ÐB  "ÑÐB  #Ñ œ !

 "# Ð%Ñ‘ # œ Ð#Ñ# œ %

B#  %B  % œ #

ÐB  #Ñ œ

& $  "( $

%

B"œ! B œ "

Use the square root property. B  # œ É "( $

B œ #  B œ # 

B  # œ É "( $

or

É "( $

È&"

$ '  È&" Bœ $

or

B œ # 

or

B œ # 

or

N9. (a) ># œ #%

> œ È#% > œ #3È'

or or

É "( $

È&"

›.

7.

È&"

B  % œ È$' B  % œ '3 B œ %  '3

or or or

Solution set: e% „ '3f

9.

#B# œ *B  % #B  *B  % œ ! Ð#B  "ÑÐB  %Ñ œ ! or or

#B  " œ ! B œ "#

Solution set: ˜ "# ß %™ 11.

Bœ*

or

B œ *

B œ È"(

or

B œ È"(

Solution set: e „ *f 13.

15. #

œ Ð%Ñ œ "'

Add "' to each side. B  )B  "' œ #"  "' ÐB  %Ñ# œ &

B%œ! Bœ%

B# œ )"

B# œ "(

Solution set: š „ È"(›

Complete the square.

#

$B  " œ ! B œ "$

#

(c) B#  )B  #" œ ! B#  )B œ #"  "# Ð)Ñ‘ #

or or

Solution set: ˜$ß "$ ™

> œ È#% > œ #3È'

B  % œ È$' B  % œ '3 B œ %  '3

B#œ! B œ #

$B#  )B  $ œ ! ÐB  $ÑÐ$B  "Ñ œ ! B$œ! B œ $

Solution set: š „ #3È'› (b) ÐB  %Ñ# œ $'

or or

Solution set: e#ß "f

$ '  È&" Bœ $

Check that the solution set is š '„$

996

B  % œ È& B  % œ 3È&

or

B œ %  3È&

"  È"$ Bœ #

or

Check that the solution set is

B  % œ È& B  % œ 3È&

œ É "$ %

B# œ $#

B œ È$# B œ %È #

or or

B œ È$# B œ %È#

Solution set: š „ %È#›

Quadratic Equations, Inequalities, and Functions 17.

B#  #! œ ! B# œ #!

B œ È#! B œ #È &

31.

or or

Solution set: š „ #È&› 19.

$B#  (# œ ! $B# œ (# B œ #%

B œ È#% B œ #È '

or or

ÐB  #Ñ# œ #&

B  # œ È#& B#œ& Bœ$

Solution set: e(ß $f 23.

B'

#

œ %*

B  ' œ È%* Bœ'( B œ "$

or or or

Solution set: e"ß "$f 25.

B  % œ È$

B œ %  È$

or or

Solution set: š% „ È$› 27.

>œ

B  ' œ È%* Bœ'( B œ "

> œ &  %È$

>  & œ È%) >  & œ %È$

Solution set: š& „ %È$› 29.

Ð$B  "Ñ# œ (

$B  " œ È(

$B œ "  È( "  È( Bœ $

Solution set:

È š "„$ ( ›

or

or

or or or È$

›

. œ "'># &!! œ "'># ># œ &!! "' œ $"Þ#& or or

%: œ "  #È' "  #È' :œ %

#  &> œ È"# &> œ #  È"# &> œ #  È"# #  #È $ >œ &

> œ È$"Þ#& > ¸ &Þ'

It would take the rock about &Þ' seconds (time must be positive) to fall to the ground. 37.

Solve Ð#B  "Ñ# œ & by the square root property. Solve B#  %B œ "# by completing the square.

39.

B#  'B  *** We need to add the square of half the coefficient of B to get a perfect square trinomial. " # Ð'Ñ

œ $ and $# œ *

Add * to B#  'B to get a perfect square trinomial. B#  'B  * œ ÐB  $Ñ#

or or

#  #È $ &

> œ È$"Þ#& > ¸ &Þ'

B  % œ È $ B œ %  È$

Ð>  &Ñ# œ %)

>  & œ È%) >  & œ %È $

or

Solution set: š #„#& 35.

›

Ð#  &>Ñ# œ "#

&> œ #  È"# &> œ #  È"#

#

ÐB  %Ñ œ $

È'

#  &> œ È"#

B œ È#% B œ #È'

B  # œ & B œ (

or

or %:  " œ È#% %:  " œ #È'

Solution set: š "„# %

B  # œ È#&

or

%:  " œ È#% %:  " œ #È'

%: œ "  #È' "  #È' :œ or %

33.

Solution set: š „ #È'› 21.

B œ È#! B œ #È&

Ð%:  "Ñ# œ #%

> œ &  %È$

41.

:#  "#:  *** ' and ' # œ $' :  "#:  $' œ Ð:  'Ñ#

" # Ð"#Ñ œ #

43.

; #  *;  ***

# œ *# and ˆ *# ‰ œ )" % * ‰# ˆ ; #  *;  )" œ ;  % #

" # Ð*Ñ

$B  " œ È( $B œ "  È( Bœ

"  È( $

45.

B#  "% B  ***

47.

B#  !Þ)B  ***

"ˆ"‰ " " ˆ ") ‰# œ '% # % œ ) and # " B#  "% B  '% œ ˆB  )" ‰

" # Ð!Þ)Ñ

œ !Þ% and Ð!Þ%Ñ# œ !Þ"'

B#  !Þ)B  !Þ"' œ ÐB  !Þ%Ñ# 997

Quadratic Equations, Inequalities, and Functions 49.

B#  %B  # œ ! B#  %B œ #

 "# Ð%Ñ‘ œ ## œ % B#  "!B  ") œ ! B#  "!B œ ")

 "# Ð"!Ñ‘# 53.

œ & œ #&

#

A 

" $A

Divide by 3.

œ)

 "# ˆ "$ ‰‘ œ ˆ "' ‰# œ #

55.

" $' to each side. " " œ )  $' A#  "$ A  $' " ˆA  "' ‰# œ #)) $'  $' ˆA  "' ‰# œ #)* $'

Add

#

$A#  A  #% œ ! A#  "$ A  ) œ !

#

B  #B œ #%

Add " to each side.

63.

B

Use the square root property.

B  " œ È#& B  " œ & B œ %

 "# Ð%Ñ‘# œ %

ÐB  #Ñ# œ '

B  # œ È'

B œ #  È'

or

B  # œ È '

B œ #  È'

or

Solution set: š# „ È'› 59.

B#  "!B  ") œ ! B#  "!B œ ") B#  "!B  #& œ ")  #& ÐB  &Ñ# œ (

B  & œ È(

B œ &  È(

or or

Solution set: š& „ È(›

998

" "( '  ' ") '

 "# Ð"!Ñ‘# œ #&

B  & œ È (

B œ &  È(

( #

œ É &$ %

Aœ

" '



Aœ

" '



 "# Ð(Ñ# ‘ œ

%* %

&$ %

È&$ # (È&$ #

or B 

"( '

( #

%* %

œ É &$ %

È&$ # (È&$ #

B œ  #(  or

È&$

Solution set: š („# 65.

È#)* È$'

A œ  )$

or

B œ  (#  Bœ

œ É #)* $'

A œ  "' '

B#  (B  " œ ! B#  (B œ " # B  (B  %* % œ"

ÐB  "Ñ# œ #&

B#  %B  % œ #  %

Aœ



ˆB  (# ‰# œ

Factor the left side.

B#  %B  # œ ! B#  %B œ #

" '

" '

or A 

Solution set: ˜ )$ ß $™

B#  #B  " œ #%  "

57.

È#)* È$'

Aœ

Aœ$

 "# Ð#Ñ‘# œ Ð"Ñ# œ "

Solution set: e%ß 'f

œ É #)* $'

Aœ

Complete the square by taking half of #, the coefficient of B, and squaring the result.

or or or

" '

A

" $'

B#  #B  #% œ ! Get the variable terms alone on the left side.

B  " œ È#& B"œ& Bœ'

Divide by 3.

Complete the square by taking half of "$ , the coefficient of A, and squaring the result. "  "# ˆ "$ ‰‘# œ ˆ "' ‰# œ $'

#

51.

$A#  A œ #% A#  "$ A œ )

61.

Bœ

›

#5 #  &5  # œ ! #5 #  &5 œ # 5 #  &# 5 œ " Divide by 2. Complete the square. ˆ "# • &# ‰# œ ˆ &% ‰# œ #& "' #& "' to each side. # #& 5  &# 5  #& "' œ "  "' ˆ5  &% ‰# œ %" "'

Add

5

& %

œ É %" "'

or 5 

È%" % &È%" %

5 œ  &%  5œ

& %

œ É %" "'

È%" % &È%" %

5 œ  %&  or È%"

Solution set: š &„%

›

5œ

Quadratic Equations, Inequalities, and Functions 67.

&B#  "!B  # œ ! &B#  "!B œ # B#  #B œ  #& Divide by 5. Complete the square.  "# Ð#Ñ‘# œ Ð"Ñ# œ " Add " to each side. B#  #B  " œ  #&  " #

$ &

ÐB  "Ñ œ B  " œ É $& B"œ Bœ Bœ

B"œ Bœ or

Bœ

È

Solution set: š &„ & "& › 69.

*B#  #%B œ "$ "$ B#  #% * Bœ *

"' *

ˆB 

B

œ

Bœ Bœ

É $*

"' * % ‰# $

œ œ

"$ * $ *

or

È$ % $  $ %È$ $



75.

B

œ

Bœ

É $*

79.

È$ % $  $ %È$ $

Add

% *

ˆD 

D

# $

% * # ‰# $

œ

Dœ Dœ

œ  "*  œ

É $*

% *

% *

or

D

# $

œ

Dœ or È

Solution set: š #„$ $ ›

<  & œ È% < œ &  #3

or or

<  & œ È% < œ &  #3

Ð'B  "Ñ# œ )

'B  " œ È) 'B  " œ 3È)

Dœ

É $*

È$ # $  $ #È$ $

or

'B  " œ È) 'B  " œ 3È)

'B  " œ #3È# 'B œ "  #3È#

'B œ "  #3È# "  #3È# Bœ '

81.

$ *

È$ # $  $ #È$ $

B œ #3È$

Ð<  &Ñ# œ %

Solution set: š "' „

to each side.

D #  %$ D 

or

'B  " œ #3È#

È

D #  %$ D œ  "* Complete the square.  "# ˆ %$ ‰‘# œ ˆ #$ ‰# œ

B œ È"# B œ 3È"#

or

Solution set: š „ #3È$›

Solution set: š %„$ $ › 71.

B œ È"# B œ 3È"#

Solution set: e& „ #3f

Bœ or

B# œ "#

B œ #3È$

"' *

% $

or

Solution set: š" „ È#›

77.

"' *

B  " œ È # B œ "  È#

or

B œ "  È#

to each side.

B#  )$ B 

% $

B  " œ È#

Divide by 9.

B#  )$ B œ "$ * Complete the square.  "# ˆ )$ ‰‘# œ ˆ %$ ‰# œ Add

È È  È$& • È&& È"& & &  & &È"& &

!Þ"B#  !Þ#B  !Þ" œ ! Multiply each side by "! to clear the decimals. B#  #B  " œ ! B#  #B œ " Complete the square.  "# Ð#Ñ‘# œ Ð"Ñ# œ " Add " to each side. B#  #B  " œ "  " ÐB  "Ñ# œ #

B  " œ É $&

or

È$ È& È& • È& È"& & &  & &È"& &

73.

or

È# $ 3›

Bœ

"  #3È# '

7#  %7  "$ œ ! 7#  %7 œ "$ Complete the square. ˆ "# • %‰# œ ## œ %

Add % to each side. 7#  %7  % œ "$  % Ð7  #Ñ# œ * 7  # œ È* 7 œ #  $3

or or

Solution set: e# „ $3f

7  # œ È* 7 œ #  $3

999

Quadratic Equations, Inequalities, and Functions 83.

$  %>  $ œ ! Here + œ #, , œ %, and - œ $.

Quadratic Equations, Inequalities, and Functions >œ >œ œ œ œ œ

, „ È, #  %+#+ Ð%Ñ „ ÈÐ%Ñ#  %Ð#ÑÐ$Ñ #Ð#Ñ % „ È"'  #% % „ È%! œ % % È #Š# „ "!‹ % „ #È"! œ % # • # # „ È"! " # … È"! œ • # " # È # „ "! # È"!

Solution set: š #„# 21.

, „ È, #  %+#+ ) „ È)#  %Ð$ÑÐ#&Ñ :œ #Ð$Ñ ) „ È'%  $!! ) „ È$'% œ œ ' ' È #Š% „ *"‹ ) „ #È*" % „ È*" œ œ œ ' #•$ $ :œ

È*"

Solution set: š %„$ 27.

›

Bœ

"œ! >œ"

Check > œ  #$ : $  $# œ *# Check > œ ": $"œ#

3.

7.

or or

$>  # œ ! > œ  #$

3 Section Exercises 1.

" # œ # > > Multiply each term by the LCD, ># .

$

or or

"(B  "% œ ! B œ  "% "( Check B œ  "% "( :

 "( "%  " &

Check B œ &:

™ Solution set: ˜ "% "( ß & 15.

B&œ! Bœ& "( "!



# (

œ

"( $&

True

œ

"( $&

True

# $ (  œ B" B# # Multiply each term by the LCD, #ÐB  "ÑÐB  #Ñ. #ÐB  "ÑÐB  #ÑŒ

# $   B" B# œ #ÐB  "ÑÐB  #ш (# ‰ #ÐB  #ÑÐ#Ñ  #ÐB  "ÑÐ$Ñ œ ÐB  "ÑÐB  #ÑÐ(Ñ %B  )  'B  ' œ ˆB#  $B  #‰Ð(Ñ "!B  "% œ (B#  #"B  "% ! œ (B#  ""B ! œ BÐ(B  ""Ñ

Bœ!

or

Check B œ  "" ( : Check B œ !: Solution set:

(B  "" œ ! B œ  "" (  (#  ( œ #

˜ "" ™ ( ß!

$ #

œ

( # ( #

True True

Quadratic Equations, Inequalities, and Functions 17.

% „ È"'  (# % „ È)) œ #Ð$Ñ #Ð$Ñ È È % „ # ## # „ ## œ œ #Ð$Ñ $

$ "  œ" #B #ÐB  #Ñ Multiply each term by the LCD, #BÐB  #Ñ. $ "   #B #ÐB  #Ñ œ #BÐB  #Ñ • " $ÐB  #Ñ  BÐ"Ñ œ #BÐB  #Ñ $B  '  B œ #B#  %B ! œ #B#  #B  ' ! œ B#  B  $ Use + œ ", , œ ", - œ $ in the quadratic formula.

œ

#BÐB  #ÑŒ

Bœ

Use a calculator to check both proposed solutions. Both solutions check. È

Solution set: š #„ $ ## ›

23.

, „ È, #  %+-

Ð#B  "Ñ#  "Ð#B  "Ñ  " œ ! %B#  %B  "  #B  "  " œ ! %B#  #B  " œ ! Use + œ %, , œ #, - œ " in the quadratic formula.

#+ " „ È"#  %Ð"ÑÐ$Ñ Bœ #Ð"Ñ " „ È"  "# " „ È"$ œ œ # #

, „ È, #  %+#+ # „ È##  %Ð%ÑÐ"Ñ Bœ #Ð%Ñ È # „ %  "' # „ È#! œ œ #Ð%Ñ #Ð%Ñ È È # „ # & " „ & œ œ #Ð%Ñ % Bœ

Use a calculator to check both proposed solutions. Both solutions check. È Solution set: š "„# "$ › 19.

" #  >  # Ð>  #Ñ# Multiply each term by the LCD, Ð>  #Ñ# Þ $œ

Use a calculator to check both proposed solutions. Both solutions check.

$Ð>  #Ñ# œ "Ð>  #Ñ  # $Ð>  %>  %Ñ œ >  #  # $>#  "#>  "# œ >  % $>#  "">  ) œ ! Ð$>  )ÑÐ>  "Ñ œ ! #

$>  ) œ ! > œ  )$

or

Solution set: 21.

˜ )$ ß "™

Solution set: š "„%

25.

>"œ! > œ "

Check > œ  )$ : $ œ  $#  Check > œ ": $ œ "  #

* #

'Ð:  "Ñ œ #:Ð:  "Ñ  : • : ':  ' œ #:#  #:  :# ! œ $:#  %:  ' Use + œ $, , œ %, and - œ ' in the quadratic formula. , „ È, #  %+#+ Ð%Ñ „ ÈÐ%Ñ#  %Ð$ÑÐ'Ñ :œ #Ð$Ñ

È&

›

Rate in still water: #! mph Rate of current: > mph (a) When the boat travels upstream, the current works against the rate of the boat in still water, so the rate is #!  > mph.

True True

' : œ# : :" Multiply each term by the LCD, :Ð:  "ÑÞ

:œ

" "  œ! #B  " Ð#B  "Ñ# Multiply each term by the LCD, Ð#B  "Ñ# . "

(b) When the boat travels downstream, the current works with the rate of the boat in still water, so the rate is #!  > mph. 27.

Let B œ rate of the boat in still water. With the rate of the current at "& mph, then B  "& œ rate going upstream and B  "& œ rate going downstream. Complete a table using the information in the problem, the rates given above, and the formula . œ or > œ .< . .


œ "'

Check > œ *: *  $ œ "# Check > œ "': "'  % œ "#

Solution set: e*f

True False

1011

Quadratic Equations, Inequalities, and Functions 45.

BœÊ

or or

&B  # œ ! B œ #& Check B œ #& :

# &

;# œ

% œ É #&

Solution set: ˜ #& ™

53.

True False

Check B œ #:

Solution set: e#f 49.

# œ È%

B# œ "#

B œ „ È"# B œ „ #È $

or

Bœ „#

Check B œ „ #: "'  %) œ '% Check B œ „ #È$: "%%  %) œ "*#

Solution set: š „ #ß „ #È$› 55.

False True

ÐB  $Ñ#  &ÐB  $Ñ  ' œ ! Let ? œ B  $, so ?# œ ÐB  $Ñ# . ?#  &?  ' œ ! Ð?  $ÑÐ?  #Ñ œ ! or or

?$œ! ? œ $

?#œ! ? œ #

To find B, substitute B  $ for ?.

Solution set: e „ #ß „ &f

1012

?  "# œ ! ? œ "#

B œ „ È%

Check B œ „ #: "'  ""'  "!! œ ! True Check B œ „ &: '#&  (#&  "!! œ ! True %; %  "$; #  * œ ! Let ? œ ; # , so ?# œ ; % . %?#  "$?  * œ ! Ð%?  *ÑÐ?  "Ñ œ !

or or

or or

To find B substitute B# for ?.

B%  #*B#  "!! œ ! Let ? œ B# , so ?# œ B% . ?#  #*?  "!! œ ! Ð?  %ÑÐ?  #&Ñ œ !

%?  * œ ! ? œ *%

True True

B%  %) œ "'B# B%  "'B#  %) œ ! Let ? œ B# , so ?# œ B% Þ ?#  "'?  %) œ ! Ð?  %ÑÐ?  "#Ñ œ !

B# œ %

or ?  #& œ ! ?%œ! or ?œ% ? œ #& To find B, substitute B# for ?. or B# œ % B# œ #& B œ „ # or Bœ „&

51.

; œ „" ""( %

 *œ! %  "$  * œ !

?%œ! ?œ%

B#œ! B œ #

 %$ œ É "' *

or )" %

Solution set: ˜ „ "ß „ $# ™

#

Check B œ %$ :

$ #

Check ; œ „ Check ; œ „ ":

)  #B ÐBÑ# œ ŒÊ  $ )  #B B# œ $ $B# œ )  #B $B#  #B  ) œ ! Ð$B  %ÑÐB  #Ñ œ ! or or

;# œ "

or $ #:

)  #B B œ Ê $

$B  % œ ! B œ %$

* %

;œ „

B$œ! B œ $

Check B œ $: $ œ È*

47.

To find ; , substitute ; # for ?.

'  "$B & '  "$B # B œ & &B# œ '  "$B &B#  "$B  ' œ ! Ð&B  #ÑÐB  $Ñ œ !

or or

B  $ œ $ B œ ' Check B œ ': Check B œ &:

*  "&  ' œ ! %  "!  ' œ !

Solution set: e'ß &f 57.

True True

$Ð7  %Ñ#  ) œ #Ð7  %Ñ Let ? œ 7  %, so ?# œ Ð7  %Ñ# . $?#  ) œ #? # $?  #?  ) œ ! Ð$?  %ÑÐ?  #Ñ œ ! $?  % œ ! ? œ  %$ 7  % œ  %$

?"œ! ? œ ".

B  $ œ # B œ &

7œ

 "' $

or

or

?#œ! ?œ# 7%œ# 7 œ #

True True

Quadratic Equations, Inequalities, and Functions Check 7 œ  "' $ : Check 7 œ #:

"' $

 ) œ  )$ "#  ) œ %

™ Solution set: ˜ "' $ ß # 59.

Check B œ „

True True

? ‰ ˆ*‰ %ˆ )" "'  "$ %  * œ ! )" %

$'  ""( %  % œ! Check B œ „ ": %  "$  * œ !

B#Î$  B"Î$  # œ ! Let ? œ B"Î$ , so ?# œ B#Î$ . ?#  ?  # œ ! Ð?  #ÑÐ?  "Ñ œ ! or or

?#œ! ? œ #

Solution set: ˜ „ "ß „ 65.

?"œ! ?œ"

To find B, substitute B"Î$ for ?. B"Î$ œ #

or

B"Î$ œ "

Cube each side of each equation. ˆB"Î$ ‰ $ œ Ð#Ñ$ B œ )

ˆB"Î$ ‰ $ œ "$ Bœ"

3

Check B œ ): Check B œ ":

3

or

%##œ! ""#œ!

or or

3

$B œ Bœ Check B œ

67. True True

Solution set: e'%ß #(f

%B%Î$  "$B#Î$  * œ ! Let ? œ B#Î$ , so ?# œ B%Î$ . %?#  "$?  * œ ! Ð%?  *ÑÐ?  "Ñ œ ! %?  * œ ! ? œ *%

or or

?"œ! ?œ"

* %

or

B#Î$ œ "

B#Î$ œ

ˆB#Î$ ‰ "Î# œ ˆ *% ‰ "Î# 3

or

3

B"Î$ œ „ $# ˆB"Î$ ‰ $ œ ˆ „ $# ‰ $ 3

Bœ „

3

#( )

or or or

?#œ! ? œ #

or

$B  " œ # $B œ " B œ  "$

or

#  "! œ )

True

 "#

True

 "$ :

#

Solution set: ˜ "$ ß "' ™

# >  *>#  "% œ ! ˆ>#  #‰ˆ>#  (‰ œ ! > œ „ È#

B*œ! Bœ*

Use + œ $, , œ ', and - œ % in the quadratic formula.

#B  $ œ È)

%

>#  # œ ! ># œ #

or or

#

17. Ð#B  $Ñ# œ )

#B  $ œ È) #B œ $  #È#

B#  &B  $' œ ! ÐB  %ÑÐB  *Ñ œ ! B%œ! B œ %

True

#È # œ È )

 "#! œ =Ð>Ñ "'>#  '!>  "#! œ 0 %>#  "&>  $! œ !

5. Let s(t) = 0. Divide by 4.

Since the triangle is a right triangle, use the Pythagorean theorem with legs 7 and 8 and hypotenuse :.

Use + œ %, , œ "&, and - œ $! in the quadratic formula.

, „ È, #  %+#+ Ð"&Ñ „ ÈÐ"&Ñ#  %Ð%ÑÐ$!Ñ >œ #Ð%Ñ "& „ È##&  %)! "& „ È(!& œ œ ) )

7#  8# œ :# 7 # œ : #  8# 7 œ È : #  8#

>œ

>œ

"&È(!& )

¸ &Þ# or > œ

"&È(!& )

Only the positive square root is given since 7 represents the side of a triangle. 7.

5># œ . . ># œ 5

¸ "Þ%

. 5 „ È. È5 œ • È5 È5 „ È.5 >œ 5

N6. (a) For 2005, B œ #!!&  "*)! œ #&. 0 ÐBÑ œ !Þ!'&B#  "%Þ)B  #%* 0 Ð#&Ñ œ !Þ!'&Ð#&Ñ#  "%Þ)Ð#&Ñ  #%* œ &()Þ$(& 9.

(b) Find the value of B that makes 0 ÐBÑ œ &!!.

"%Þ) „ È"%Þ)#  %Ð!Þ!'&ÑÐ#&"Ñ Bœ #Ð!Þ!'&Ñ "%Þ) „ È"&$Þ()

The solution B ¸ #!*Þ# is rejected since it corresponds to a year far beyond the period covered by the model. The solution B ¸ ")Þ& is valid for this problem. The corresponding year is 1998.

4 Section Exercises 1.

1018

œ „

The first step in solving a formula that has the specified variable in the denominator is to multiply each side by the LCD to clear the equation of fractions.

.œ 11.

Simplify.

Divide by I. 5= M

Use square root property.

„ È5=M M

Rationalize denominator. Simplify.

5E for @. @#

@# J œ 5E 5E @# œ J

@œ „Ê

@œ

Rationalize denominatorÞ

Multiply by d 2 .

È5= ÈM • ÈM ÈM

Solve J œ

œ

Use square root property.

5= for .Þ .#

. œ „Ê

Now use + œ !Þ!'&, , œ "%Þ), and - œ #&" in the quadratic formula.

!Þ"$ B ¸ ")Þ& or B ¸ #!*Þ#

Solve M œ M. # œ 5= 5= .# œ M

0 ÐBÑ œ !Þ!'&B#  "%Þ)B  #%* &!! œ !Þ!'&B#  "%Þ)B  #%* ! œ !Þ!'&B#  "%Þ)B  #&"

œ

Divide by k.

>œ „Ê

The object will hit the ground about &Þ# seconds after it is projected.

The CPI for 2005 was about &().

Solve . œ 5># for >.

Multiply by v# . Divide by F. 5E J

„ È5E ÈJ • ÈJ ÈJ „ È5EJ J

Use square root property. Rationalize denominatorÞ Simplify.

Quadratic Equations, Inequalities, and Functions 13.

Vœ

$Z œ 1#  "#)> #"$ œ "'>#  "#)> ! œ "'>#  "#)>  #"$

=Ð>Ñ œ "'>#  "'!> %!! œ "'>#  "'!> ! œ "'>#  "'!>  %!! ! œ >#  "!>  #& ! œ Ð>  &ÑÐ>  &Ñ ! œ Ð>  &Ñ# !œ>& &œ>

Let s(t) œ 400. Divide by 16.

The ball reaches a height of %!! feet after & seconds. This is its maximum height since this is the only time it reaches %!! feet.

Let s œ 213.

Here + œ "', , œ "#), and - œ #"$. , „ È, #  %+>œ #+ "#) „ È"#)#  %Ð"'ÑÐ#"$Ñ >œ #Ð"'Ñ "#) „ È"',$)%  "$,'$# œ $# "#) „ È#(&# œ $#

, „ È, #  %+#+ Ð"!!Ñ „ ÈÐ"!!Ñ#  %Ð"$ÑÐ")!Ñ #Ð"$Ñ "!! „ È"!,!!!  *$'! #Ð"$Ñ "!! „ È"*,$'! #Ð"$Ñ "!! „ %%È"! &! „ ##È"! œ #Ð"$Ñ "$ È &!  ## "! ¸ *Þ# or "$ &!  ##È"! ¸ "Þ& "$

Discard the negative solution. The car will skid ")! feet in approximately *Þ# seconds.

Let s œ 128. Divide by 16.

Let D(t) œ 180.

Here + œ "$, , œ "!!, and - œ ")!Þ

Since B represents width, discard the negative solution. The width is "# inches, and the length is #Ð"#Ñ  % œ #! inches. 43.

HÐ>Ñ œ "$>#  "!!> ")! œ "$>#  "!!> ! œ "$>#  "!!>  ")!

51.

Supply and demand are equal when $:  #!! œ

$#!! . :

Solve for :. $:#  #!!: œ $#!! $:  #!!:  $#!! œ ! #

Use the quadratic formula with + œ $, , œ #!!, and - œ $#!!.

1021

Quadratic Equations, Inequalities, and Functions :œ œ

Ð#!!Ñ „ ÈÐ#!!Ñ#  %Ð$ÑÐ$#!!Ñ #Ð$Ñ #!! „ È%!,!!!  $),%!!

59.

(b) For 2005, B œ #!!&  #!!! œ &.

' #!! „ È(),%!! #!! „ #)! œ œ ' ' )! : œ %)! œ )! : œ œ  %! or ' ' $

0 ÐBÑ œ !Þ$#"%B#  #&Þ!'B  #))Þ# 0 Ð&Ñ œ !Þ$#"%Ð&Ñ#  #&Þ!'Ð&Ñ  #))Þ# œ %#"Þ&$&

Discard the negative solution. The supply and demand are equal when the price is )! cents or $!Þ)!. 53.

(a) From the graph, the spending on physician and clinical services in 2005 appears to be $%#! billion (to the nearest $"! billion).

To the nearest $"! billion, the model gives $%#! billion, the same as the estimate in part (a). 61.

Let < œ the interest rate. Let E œ #"%#Þ%&, T œ #!!!, and solve for Ñ œ "'>#  $#> Here, + œ "'  !, so the parabola opens down. The time it takes to reach the maximum height and the maximum height are given by the vertex of the parabola. Use the vertex formula to find that

#

! œ $C  "#C  & "# „ È"%%  '! "# „ È)% Cœ œ ' ' "# „ #È#" ' „ È#" œ œ ' $

, $# $# œ œ œ ", #+ #Ð"'Ñ $# =Ð>Ñ œ "'Ð"Ñ#  $#Ð"Ñ œ "'  $# œ "'. >œ

and

The C-intercepts are approximately Ð!ß !Þ&Ñ and Ð!ß $Þ&Ñ.

The vertex is Ð"ß "'Ñ, so the maximum height is "' feet which occurs when the time is " second. The object hits the ground when = œ !.

Step 4 By symmetry, Ð&ß %Ñ is also on the graph.

! œ "'>#  $#> ! œ "'>Ð>  #Ñ "'> œ ! >œ!

or or

>#œ! >œ#

It takes # seconds for the object to hit the ground. From the graph, we see that the domain is Ò(ß ∞Ñ and the range is Ð∞ß ∞Ñ. 33.

Let B œ one number, %!  B œ the other number, and T œ the product. T œ BÐ%!  BÑ œ %!B  B# or B#  %!B This parabola opens down so the maximum occurs at the vertex. Here + œ ", , œ %!, and - œ !. , %! œ œ #! #+ #Ð"Ñ

39.

The graph of the height of the projectile, =Ð>Ñ œ "'>#  '%>  ", is a parabola that opens down since + œ "'  !. The time at which the cork reaches its maximum height and the maximum height are the > and = coordinates of the vertex. >œ

, '% œ œ# #+ #Ð"'Ñ

=Ð#Ñ œ "'Ð#Ñ#  '%Ð#Ñ  " œ '%  "#)  " œ '& The cork reaches a maximum height of '& feet after # seconds.

1031

Quadratic Equations, Inequalities, and Functions 41.

GÐBÑ œ B#  %!B  '"! Find the vertex of the graph of G . , Ð%!Ñ œ œ #! #+ #Ð"Ñ GÐ#!Ñ œ #!#  %!Ð#!Ñ  '"! œ #"! He should sell #! tacos per day to minimize his costs of $#"!.

43.

(c) The number of unsold seats B that produce the maximum revenue is #&, the B-value of the vertex.

0 ÐBÑ œ ##Þ))B#  "%"Þ$B  "!%% (a) Since the graph opens up, the C-value of the vertex is a minimum.

(d) The maximum revenue is $##,&!!, the C-value of the vertex.

(b) The B-value of the vertex is given by

49.

Ò"ß &Ó Í " Ÿ B Ÿ &

51.

Ð∞ß "Ó ∪ Ò&ß ∞Ñ Í B Ÿ " or B   &

53.

#B  "  % #B  $ B   $#

, Ð"%"Þ$Ñ Bœ œ ¸ $Þ!)) #+ #Ð##Þ))Ñ The year was #!!!  $ œ #!!$. 0 Ð$Þ!))Ñ ¸ $)#&Þ) billion, which is the minimum amount of total receipts from individual taxes. 45.

0 ÐBÑ œ #!Þ&(B#  (&)Þ*B  $"%! (a) The coefficient of B# is negative because a parabola that models the data must open down.

Solution set: ˆ $# ß ∞‰

(b) Use the vertex formula. , (&)Þ* œ ¸ ")Þ%& #+ #Ð#!Þ&(Ñ 0 Ð")Þ%&Ñ ¸ $)'! Bœ

The vertex is approximately Ð")Þ%&ß $)'!Ñ. (c) ") corresponds to #!"), so in #!") social security assets will reach their maximum value of $$)'! billion. 47.

7

Polynomial and Rational Inequalities

7 Now Try Exercises N1. (a) B#  $B  %  ! The B-intercepts are Ð"ß !Ñ and Ð%ß !Ñ. Notice from the graph that B-values less than " or greater than % result in C-values greater than !. Therefore, the solution set is

The number of people on the plane is "!!  B since B is the number of unsold seats. The price per seat is #!!  %B.

Ð∞ß "Ñ ∪ Ð%ß ∞Ñ.

(a) The total revenue received for the flight is found by multiplying the number of seats by the price per seat. Thus, the revenue is VÐBÑ œ Ð"!!  BÑÐ#!!  %BÑ œ #!,!!!  #!!B  %B# . (b) Use the formula for the vertex. Bœ

, #!! œ œ #& #+ #Ð%Ñ

VÐ#&Ñ œ ##,&!!, so the vertex is Ð#&ß ##,&!!Ñ. VÐ!Ñ œ #!,!!!, so the V -intercept is Ð!ß #!,!!!Ñ. From the factored form for V , we see that the positive B-intercept is Ð"!!ß !Ñ. (The factor #!!  %B leads to a negative B-intercept, meaningless in this problem.)

1032

(b) B#  $B  %  ! Notice from the graph that B-values between " and % result in C-values less than !. Therefore, the solution set is Ð"ß %Ñ. N2.

B#  #B  )  ! Use factoring to solve the quadratic equation B#  #B  ) œ !. ÐB  %ÑÐB  #Ñ œ ! B%œ! B œ %

or B  # œ ! or Bœ#

Locate the numbers % and # that divide the number line into three intervals A, B, and C.

Quadratic Equations, Inequalities, and Functions Interval C:

Choose a number from each interval to substitute in the inequality

?

Ð%ÑÐ$ÑÐ"Ñ Ÿ ! "# Ÿ ! Interval D: Let B œ %.

B#  #B  )  !. Interval A:

Let B œ &.

Ð)ÑÐ"ÑÐ*Ñ Ÿ ! (# Ÿ !

Ð&Ñ  #Ð&Ñ  )  ! ?

#&  "!  )  ! (! Let B œ !.

Interval B:

True

?

?

#

Let B œ !.

The numbers in Intervals A and C, including %ß  "# ß and $ because of Ÿ , are solutions.

True

?

Ð!Ñ#  #Ð!Ñ  )  ! )  ! Interval C: Let B œ $.

False

Solution set: Ð∞ß %Ó ∪ Ò "# ß $Ó False

?

Ð$Ñ#  #Ð$Ñ  )  ! ?

N5.

*')! (!

True

The numbers in Intervals A and C are solutions. The numbers % and # are excluded because of . Solution set: Ð∞ß %Ñ ∪ Ð#ß ∞

N3. (a) Ð%B  "Ñ#  $ The square of any real number is always greater than or equal to !, so any real number satisfies this inequality. Solution set: Ð∞ß ∞Ñ

$ % B" Write the inequality so that ! is on one side. $ %! B" $ %ÐB  "Ñ  ! B" B" $  %B  % ! B" %B  " ! B" The number  "% makes the numerator !, and " makes the denominator !. These two numbers determine three intervals.

(b) Ð%B  "Ñ#  $ The square of a real number is never negative. Solution set: g N4. ÐB  %ÑÐB  $ÑÐ#B  "Ñ Ÿ ! Solve the equation ÐB  %ÑÐB  $ÑÐ#B  "Ñ œ !. B%œ! or B  $ œ ! or #B  " œ ! B œ % or B œ $ or B œ  "# Locate these numbers and the intervals A, B, C, and D on a number line.

Test a number from each interval in the inequality ÐB  %ÑÐB  $ÑÐ#B  "Ñ Ÿ !. Interval A:

Let B œ &. ?

Ð"ÑÐ)ÑÐ*Ñ Ÿ ! (# Ÿ ! Interval B: Let B œ #.

True

Interval C:

Let B œ !. $ ? % !" $%

False

The solution set includes numbers in Interval B, excluding endpoints. Solution set: Ð"ß  "% Ñ

?

# & $ Ÿ ! $! Ÿ !

Test a number from each interval in the inequality $  %. B" Interval A: Let B œ #. $ ? % #  " $  % False Interval B: Let B œ  "# . $ ? %  "#  " '% True

False

1033

Quadratic Equations, Inequalities, and Functions N6.

B$ Ÿ# B$ Write the inequality so that ! is on one side. B$ #Ÿ! B$ B  $ #ÐB  $Ñ  Ÿ! B$ B$ B  $  #B  ' Ÿ! B$ B  * Ÿ! B$ The number * makes the numerator !, and $ makes the denominator !. These two numbers determine three intervals.

3.

(a) The B-intercepts determine the solutions of the equation B#  $B  "! œ !. From the graph, the solution set is Ö#ß &×. (b) The B-values of the points on the graph that are above the B-axis form the solution set of the inequality B#  $B  "!  !. From the graph, the solution set for B#  $B  "!   ! is Ò#ß &Ó. (c) The B-values of the points on the graph that are below the B-axis form the solution set of the inequality B#  $B  "!  !. From the graph, the solution set for B#  $B  "! Ÿ ! is Ð∞ß #Ó ∪ Ò&ß ∞Ñ.

5.

ÐB  "ÑÐB  &Ñ  ! Solve the equation ÐB  "ÑÐB  &Ñ œ !. B"œ! B œ "

Test a number from each interval in the inequality

or or

B&œ! Bœ&

The numbers " and & divide a number line into three intervals: A, B, and C.

B$ Ÿ #. B$ Interval A:

Let B œ "!. "!  $ ? Ÿ# "!  $ "$ True ( Ÿ# Interval B: Let B œ &. &  $ ? Ÿ# &  $ %Ÿ# False Interval C: Let B œ !. !$ ? Ÿ# !$ " Ÿ # True

Test a number from each interval in the original inequality, ÐB  "ÑÐB  &Ñ  !. Interval A:

?

Ð#  "ÑÐ#  &Ñ  ! ?

"Ð(Ñ  ! (! Interval B:

?

Interval C:

Solution set: Ð∞ß *Ó ∪ Ð$ß ∞Ñ

?

1034

True

The solution set includes the numbers in Intervals A and C, excluding " and & because of  . Solution set: Ð∞ß "Ñ ∪ Ð&ß ∞Ñ

(a) The B-intercepts determine the solutions of the equation B#  %B  $ œ !. From the graph, the solution set is Ö"ß $×.

(c) The B-values of the points on the graph that are below the B-axis form the solution set of the inequality B#  %B  $  !. From the graph, the solution set is Ð"ß $Ñ.

False

Let B œ '.

Ð'  "ÑÐ'  &Ñ  ! (!

7 Section Exercises

(b) The B-values of the points on the graph that are above the B-axis form the solution set of the inequality B#  %B  $  !. From the graph, the solution set is Ð∞ß "Ñ ∪ Ð$ß ∞Ñ.

True

Let B œ !.

Ð!  "ÑÐ!  &Ñ  ! &  !

The numbers in Intervals A and C are solutions. $ is not in the solution set (since it makes the denominator !), but * is.

1.

Let B œ #.

7.

ÐB  %ÑÐB  'Ñ  ! Solve the equation ÐB  %ÑÐB  'Ñ œ !. B%œ! B œ %

or or

B'œ! Bœ'

The numbers % and ' divide a number line into three intervals: A, B, and C.

Quadratic Equations, Inequalities, and Functions 11.

Test a number from each interval in the original inequality, ÐB  %ÑÐB  'Ñ  !. Interval A:

Let B œ &.

#B  $ œ ! B œ  $#

?

Ð&  %ÑÐ&  'Ñ  ! ?

"Ð""Ñ  ! ""  ! Interval B: Let B œ !.

or or

&B  $ œ ! B œ $&

False

?

Interval C:

"!B#  *B   * "!B#  *B  *   ! Solve the equation "!B#  *B  * œ !. Ð#B  $ÑÐ&B  $Ñ œ !

%Ð'Ñ  ! #%  ! Let B œ (.

Test a number from each interval in the original inequality, "!B#  *B   *.

True

Interval A:

?

Ð(  %ÑÐ(  'Ñ  !

Let B œ #. ?

#

"!Ð#Ñ  *Ð#Ñ   *

?

""Ð"Ñ  ! ""  !

?

False

The solution set includes Interval B, where the expression is negative.

Interval B:

Solution set: Ð%ß 'Ñ

Interval C:

%!  ")   * ##   * Let B œ !. ! * Let B œ ".

True False

?

"!Ð"Ñ#  *Ð"Ñ   * ?

9.

"!  *   * "*   *

B#  %B  $   ! Solve the equation B#  %B  $ œ !. ÐB  "ÑÐB  $Ñ œ ! B"œ! Bœ"

The solution set includes the numbers in Intervals A and C, including  $# and $& because of   . or or

Solution set: ˆ∞ß  $# ‘ ∪  $& ß ∞‰

B$œ! Bœ$ 13.

Test a number from each interval in the original inequality, B#  %B  $   !. Interval A: Interval B:

True

Let B œ !. $ ! Let B œ #.

True

%B#  * Ÿ ! Solve the equation %B#  * œ ! . Ð#B  $ÑÐ#B  $Ñ œ ! #B  $ œ ! B œ  $#

or or

#B  $ œ ! B œ $#

?

##  %Ð#Ñ  $   ! "   ! Interval C: Let B œ %.

False

?

%#  %Ð%Ñ  $   ! $ !

True

The solution set includes the numbers in Intervals A and C, including " and $ because of   . Solution set: Ð∞ß "Ó ∪ Ò$ß ∞Ñ

Test a number from each interval in the original inequality, %B#  * Ÿ !. Interval A:

Let B œ #. ?

%Ð#Ñ#  * Ÿ ! (Ÿ! Interval B: Let B œ !. * Ÿ !

False True

1035

Quadratic Equations, Inequalities, and Functions Interval C:

Let B œ #.

Interval A:

?

Let D œ ". ?

%Ð#Ñ#  * Ÿ ! (Ÿ!

Ð"Ñ#  %Ð"Ñ   ! & ! Interval B: Let D œ #.

False

The solution set includes Interval B, including the endpoints.

True

?

##  %Ð#Ñ   ! %   ! Interval C: Let D œ &.

Solution set:  $# ß $# ‘

False

?

&#  %Ð&Ñ   ! & ! 15.

#

'B  B   " 'B#  B  "   ! Solve the equation 'B#  B  " œ !. Ð#B  "ÑÐ$B  "Ñ œ ! #B  " œ ! B œ  "#

The solution set includes the numbers in Intervals A and C, including ! and % because of   . Solution set: Ð∞ß !Ó ∪ Ò%ß ∞Ñ

or or

$B  " œ ! B œ "$

19.

$B#  &B Ÿ ! Solve the equation $B#  &B œ !Þ BÐ$B  &Ñ œ ! Bœ!

Test a number from each interval in the original inequality, 'B#  B   ". Interval A:

True

or

$B  & œ ! B œ &$

Let B œ ". ?

'Ð"Ñ#  Ð"Ñ   " & " Interval B: Let B œ !. ! " Interval C: Let B œ ".

True

Test a number from each interval in the original inequality, $B#  &B Ÿ !.

False

Interval A:

'Ð"Ñ#  "   " ( "

Let B œ "Þ ?

?

$Ð"Ñ#  &Ð"Ñ Ÿ ! )Ÿ! Interval B: Let B œ "Þ

True

The solution set includes the numbers in Intervals A and C, including  "# and "$ because of   .

False

?

$Ð"Ñ#  &Ð"Ñ Ÿ ! # Ÿ ! Interval C: Let B œ #Þ

Solution set: ˆ∞ß  "# ‘ ∪  "$ ß ∞‰

True

?

$Ð#Ñ#  &Ð#Ñ Ÿ ! #Ÿ! 17.

#

D  %D   ! Solve the equation D #  %D œ ! . DÐD  %Ñ œ ! Dœ!

The solution set includes the numbers in Interval B, including the endpoints. Solution set: !ß &$ ‘

or

D%œ! Dœ% 21.

Test a number from each interval in the original inequality, D #  %D   !.

1036

False

B#  'B  '   ! Solve the equation B#  'B  ' œ ! . Since B#  'B  ' does not factor, let + œ ", , œ ', and - œ ' in the quadratic formula.

Quadratic Equations, Inequalities, and Functions Ð'Ñ „ ÈÐ'Ñ#  %Ð"ÑÐ'Ñ #Ð"Ñ È ' „ "# ' „ #È $ œ œ # # #Š$ „ È $ ‹ œ œ $ „ È$ # B œ $  È$ ¸ %Þ( or B œ $  È$ ¸ "Þ$

Interval A:

Bœ

Let B œ !. ?

Interval B:

"Ð#ÑÐ%Ñ  ! )  ! Let B œ ".&.

True

?

Ð"Þ&  "ÑÐ"Þ&  #ÑÐ"Þ&  %Ñ  ! ?

!Þ&Ð!Þ&ÑÐ#Þ&Ñ  ! !Þ'#&  ! Let B œ $. Interval C:

False

?

Ð$  "ÑÐ$  #ÑÐ$  %Ñ  ! ?

Test a number from each interval in the original inequality, B#  'B  '   !. Interval A:

Let B œ !. ' ! Let B œ $.

Interval B:

Interval D:

True

?

Ð&  "ÑÐ&  #ÑÐ&  %Ñ  ! ?

%Ð$ÑÐ"Ñ  ! "#  !

True

False

The numbers in Intervals A and C, not including ", #, or %, are solutions.

?

$#  ' $  '   ! $   ! Interval C: Let B œ &.

#Ð"ÑÐ"Ñ  ! #  ! Let B œ &Þ

False

Solution set: Ð∞ß "Ñ ∪ Ð#ß %Ñ

?

&#  'Ð&Ñ  '   ! " !

True

The solution set includes the numbers in Intervals A and C, including $  È$ and $  È$ because of   . Solution set:

23.

%  $B

#

Ð∞ß $  È$ Ó ∪ Ò$  È$ß ∞Ñ

ÐB  %ÑÐ#B  $ÑÐ$B  "Ñ œ !. These numbers divide a number line into % intervals.

#

Ð$B  &Ñ# Ÿ % Since Ð$B  &Ñ# is never negative, Ð$B  &Ñ# will never be less than or equal to a negative number. Therefore, the solution set is g.

27.

ÐB  %ÑÐ#B  $ÑÐ$B  "Ñ   ! The numbers %,  $# , and "$ are solutions of the cubic equation

  #

Since %  $B # is either ! or positive, %  $B will always be greater than #. Therefore, the solution set is Ð∞ß ∞Ñ. 25.

29.

Interval A:

?

Interval B:

'Ð"ÑÐ(Ñ   ! %#   ! Let B œ !.

Interval C:

%Ð$ÑÐ"Ñ   ! "#   ! Let B œ ".

These numbers divide a number line into four intervals.

False

?

True

?

ÐB  "ÑÐB  #ÑÐB  %Ñ  ! The numbers ", #, and % are solutions of the cubic equation ÐB  "ÑÐB  #ÑÐB  %Ñ œ !.

Let B œ #.

$Ð&ÑÐ#Ñ   ! $!   ! Interval D:

False

Let B œ &Þ ?

"Ð"$ÑÐ"%Ñ   ! ")#   !

True

The solution set includes numbers in Intervals B and D, including the endpoints. Solution set:  $# ß "$ ‘ ∪ Ò%ß ∞Ñ Test a number from each interval in the inequality ÐB  "ÑÐB  #ÑÐB  %Ñ  !. 1037

Quadratic Equations, Inequalities, and Functions 31.

B" ! B%

Interval C:

The number " makes the numerator !, and % makes the denominator !. These two numbers determine three intervals.

Let B œ '. #Ð'Ñ  $ ? Ÿ! '& "& Ÿ !

False

The solution set includes the points in Interval B. The endpoint & is not included since it makes the left side undefined. The endpoint  $# is included because it makes the left side equal to !. Solution set:  $# ß &‰

Test a number from each interval in the inequality B"  !. B% Interval A:

Interval B:

Interval C:

Let B œ !. !" ? ! !% " ! % Let B œ #. #" ? ! #% " ! # Let B œ &. &" ? ! &% %!

35.

True

False

True

The solution set includes numbers in Intervals A and C, excluding endpoints.

)  # B# Write the inequality so that ! is on one side. ) # ! B# ) #ÐB  #Ñ  !  B# B# )  #B  %  ! B# #B  "#  ! B# The number ' makes the numerator !, and # makes the denominator !. These two numbers determine three intervals.

Solution set: Ð∞ß "Ñ ∪ Ð%ß ∞Ñ Test a number from each interval in the inequality 33.

)   #. B#

#B  $ Ÿ! B& The number  $# makes the numerator !, and & makes the denominator !. These two numbers determine three intervals.

Interval A:

Interval B:

Test a number from each interval in the inequality Interval C:

#B  $ Ÿ !. B& Interval A: Let B œ #. #Ð#Ñ  $ ? Ÿ! Ð#Ñ  & " ( Ÿ! Interval B: Let B œ !. #Ð!Ñ  $ ? Ÿ! !&  $& Ÿ ! 1038

False

True

Let B œ !. ) ?  # !# %   # Let B œ $. ) ?  # $# ) # Let B œ (. ) ?  # (# ) &  #

False

True

False

The solution set includes numbers in Interval B, including ' but excluding #, which makes the fraction undefined. Solution set: Ð#ß 'Ó

Quadratic Equations, Inequalities, and Functions 37.

$ # #B  " Write the inequality so that ! is on one side. $ #! #B  " $ #Ð#B  "Ñ  ! #B  " #B  " $  %B  # ! #B  " %B  & ! #B  " & %

Test a number from each interval in the inequality B$   #Þ B# Interval A:

Interval B:

" #

The number makes the numerator !, and makes the denominator !. These two numbers determine three intervals.

Interval C:

#

# Let B œ %. ( ? #   ( #  

False

#

# Let B œ !. $ #  #

True False

Solution set: Ò(ß #Ñ

True 41.

False

True

The solution set includes numbers in Intervals A and C, excluding endpoints. Solution set: ˆ∞ß "# ‰ ∪ ˆ &% ß ∞‰

39.

"" ? '   "" '  

The solution set includes numbers in Interval B, including ( but excluding #, which makes the fraction undefined.

Test a number from each interval in the inequality $  #. #B  " Interval A: Let B œ !. $ ? # #Ð!Ñ  " $  # Interval B: Let B œ ". $ ? # #Ð"Ñ  " $# Interval C: Let B œ #. $ ? # #Ð#Ñ  " "#

Let B œ ).

B$  # B# Write the inequality so that ! is on one side. B$ # ! B# B  $ #ÐB  #Ñ  !  B# B# B  $  #B  %  ! B# B  (  ! B# The number ( makes the numerator !, and # makes the denominator !. These two numbers determine three intervals.

B) $ B% Write the inequality so that ! is on one side. B) $! B% B  ) $ÐB  %Ñ  ! B% B% B  )  $B  "# ! B% #B  % ! B% The number # makes the numerator !, and % makes the denominator !. These two numbers determine three intervals.

Test a number from each interval in the inequality B)  $Þ B% Interval A:

Interval B:

Let B œ !. ) ? $ % #$

True

Let B œ $. & ? " 

$ &$

False 1039

Quadratic Equations, Inequalities, and Functions Interval C:

Let B œ &. $ " $

45.

True

The denominator is positive for all real numbers B, so it has no effect on the solution set for the inequality.

The solution set includes numbers in Intervals A and C, excluding endpoints. Solution set: ∞ß #Ñ ∪ Ð%ß ∞

43.

%5 5 #5  " Write the inequality so that ! is on one side. %5 5 ! #5  " %5 5Ð#5  "Ñ  ! #5  " #5  " %5  #5 #  5 ! #5  " #5 #  &5 ! #5  " 5Ð#5  &Ñ ! #5  " The numbers ! and &# make the numerator !, and makes the denominator !. These three numbers determine four intervals.

#B  $  ! B#  "

#B  $   ! #B   $ B   $#

Solution set:  $# ß ∞‰

47.

$B  & B#

#

!

The numerator is positive for all real numbers B except B œ &$ , which makes it equal to !. If we solve the inequality B  #  !, then we only have to be sure to exclude &$ from that solution set to determine the solution set of the original inequality.

" #

B#! B  #

Solution set: ˆ#ß &$ ‰ ∪ ˆ &$ ß ∞‰

Test a number from each interval in the inequality %5  5. #5  " Interval A: Let 5 œ ". %Ð"Ñ ?  " #Ð"Ñ  " % $  " Interval B:

49.

51. False

Let 5 œ "% . %ˆ "% ‰ ? "  % #ˆ "% ‰  "

#  "% Interval C: Let 5 œ ". %Ð"Ñ ? " #Ð"Ñ  " %" Interval D: Let 5 œ $. %Ð$Ñ ? $ #Ð$Ñ  " "# & $

The domain is the set of B-values: e!ß "ß #ß $f. The range is the set of C-values: e"ß #ß %ß )fÞ

Using the vertical line test, we find any vertical line will intersect the graph at most once. This indicates that the graph represents a function.

Review Exercises 1.

># œ "#"

True

> œ ""

or

> œ ""

: œ È$

or

: œ È $

Solution set: e „ ""f False

2.

:# œ $

Solution set: š „ È$› True

The solution set includes numbers in Intervals B and D. None of the endpoints are included. Solution set: ˆ!ß "# ‰ ∪ ˆ &# ß ∞‰

1040

eÐ!ß "Ñß Ð"ß #Ñß Ð#ß %Ñß Ð$ß )Ñf

3.

Ð#B  &Ñ# œ "!! #B  & œ "! #B œ & B œ &#

or or

&™ Solution set: ˜ "& # ß #

#B  & œ "! #B œ "& B œ  "& #

Quadratic Equations, Inequalities, and Functions 4.

Ð$B  #Ñ# œ #&

$B  # œ È#& $B  # œ &3 $B œ #  &3 #  &3 Bœ $ B œ #$  &$ 3

or

or

Solution set: ˜ #$ „ &$ 3™ 5.

$B  # œ È#& $B  # œ &3 $B œ #  &3 #  &3 Bœ $ B œ #$  &$ 3

9.

Here + œ #, , œ ", and - œ #". , „ È, #  %+Bœ #+ " „ È"#  %Ð#ÑÐ#"Ñ Bœ #Ð#Ñ " „ È"  "') œ % " „ È"'* " „ "$ œ œ % % "  "$ "# œ œ $ or Bœ % % "  "$ "% ( œ œ Bœ % % # ( ˜ ™ Solution set:  # ß $

B#  %B œ "& Complete the square. ˆ "# • %‰# œ ## œ %

Add % to each side. B#  %B  % œ "&  % ÐB  #Ñ# œ "* B  # œ È"*

B  # œ È"* B œ #  È"*

or

B œ #  È"*

or

10.

Solution set: š# „ È"*› 6.

#B#  $B œ " B#  $# B œ  "# Complete the square.  "# ˆ $# ‰‘# œ ˆ $% ‰# œ

Divide by 2. * "'

$ %

B

$ %

" œ É "'

œ

Bœ

" % $ %

Bœ"



or " %

È&$

Solution set: ˜ "# ß "™ 7.

8.

11.

B

$ %

B

$ %

" œ É "'

œ  "%

Bœ or

B#  &B œ ( B#  &B  ( œ ! Here + œ ", , œ &, and - œ (Þ , „ È, #  %+Bœ #+ & „ È&#  %Ð"ÑÐ(Ñ Bœ #Ð"Ñ È & „ #&  #) & „ È&$ œ œ # #

Solution set: š &„#

* "' to each side. * * B#  $# B  "' œ  #"  "' ) * ˆB  $% ‰# œ  "'  "' " ˆB  $% ‰# œ "'

Add

B

#B#  B  #" œ !

Bœ

$ % " #



" %

By the square root property, the first step should be B œ È"# or B œ È"#. The solution set is š „ #È$›. %Þ*># œ . %Þ*># œ "'& Let d œ 165. "'& # > œ %Þ* "'& >œÊ t 0 %Þ* > ¸ &Þ) seconds It would take about &Þ) seconds for the wallet to fall "'& meters.

›

Ð>  $ÑÐ>  %Ñ œ # >#  >  "# œ # >#  >  "! œ ! Here + œ ", , œ ", and - œ "!. , „ È, #  %+>œ #+ Ð"Ñ „ ÈÐ"Ñ#  %Ð"ÑÐ"!Ñ >œ #Ð"Ñ È " „ "  %! " „ È%" œ œ # # È

Solution set: š "„ # %" › 12.

#B#  $B  % œ ! Here + œ #, , œ $, and - œ %Þ , „ È, #  %+Bœ #+ $ „ È$#  %Ð#ÑÐ%Ñ Bœ #Ð#Ñ $ „ È*  $# $ „ È#$ œ œ % % $ „ 3È#$ $ È#$ œ œ „ 3 % % %

Solution set: š $% „

È#$ % 3›

1041

Quadratic Equations, Inequalities, and Functions 13.

$:# œ #Ð#:  "Ñ $:# œ %:  # # $:  %:  # œ ! Here + œ $, , œ %, and - œ #. :œ :œ œ œ œ

, „ È, #  %+#+ Ð%Ñ „ ÈÐ%Ñ#  %Ð$ÑÐ#Ñ #Ð$Ñ È % „ "'  #% % „ È) œ ' ' È # Š# „ 3 # ‹ % „ #3È# œ ' ' # „ 3È# # È# œ „ 3 $ $ $

Solution set: š #$ „ 14.

È$(

Solution set: š („#

1042

%B# œ 'B  ) %B#  'B  ) œ ! Here + œ %, , œ ', and - œ ). , #  %+- œ Ð'Ñ#  %Ð%ÑÐ)Ñ œ $'  "#) œ *#

"& œ #B  " B "& BŒ  œ BÐ#B  "Ñ B "& œ #B#  B ! œ #B#  B  "& ! œ Ð#B  &ÑÐB  $Ñ or or

Multiply by the LCD, x.

B$œ! Bœ$

Check B œ  &# : ' œ &  " Check B œ $: &œ'" & ˜ Solution set:  # ß $™ 20.

B#  &B  # œ ! Here + œ ", , œ &, and - œ #. , #  %+- œ &#  %Ð"ÑÐ#Ñ œ #&  ) œ "( Since the discriminant is positive, but not a perfect square, there are two distinct irrational number solutions. The answer is C. %># œ $  %> %>  %>  $ œ ! Here + œ %, , œ %, and - œ $. , #  %+- œ %#  %Ð%ÑÐ$Ñ œ "'  %) œ '% or )# Since the discriminant is positive, and a perfect square, there are two distinct rational number solutions. The answer is A.

*D #  $!D  #& œ ! Here + œ *, , œ $!, and - œ #&. , #  %+- œ $!#  %Ð*ÑÐ#&Ñ œ *!!  *!! œ ! Since the discriminant is zero, there is exactly one rational number solution. The answer is B.

#B  & œ ! B œ  &#

›

#

17.

19.

BÐ#B  (Ñ œ $B#  $ #B#  (B œ $B#  $ ! œ B#  (B  $ Here + œ ", , œ (, and - œ $Þ

, „ È, #  %+#+ ( „ È(#  %Ð"ÑÐ$Ñ Bœ #Ð"Ñ ( „ È%*  "# ( „ È$( œ œ # #

16.

18.

È# $ 3›

Bœ

15.

Since the discriminant is negative, there are two distinct nonreal complex number solutions. The answer is D.

" #  œ# 8 8" " # Multiply by 8Ð8  "ÑŒ   the LCD, n(n+1). 8 8" œ 8Ð8  "Ñ • # 8  "  #8 œ #8#  #8 ! œ #8#  8  " ! œ Ð#8  "ÑÐ8  "Ñ or or

#8  " œ ! 8 œ  "#

8"œ! 8œ"

Check 8 œ  "# : #  % œ # Check 8 œ ": ""œ# " ˜ Solution set:  # ß "™ 21.

True True

#< œ Ê

Square each side.

True True

%)  #!< #

Ð##  %&>  %!! ! œ "'>#  %&>  #!! ! œ "'>#  %&>  #!!

Let f(t) œ 200.

Here + œ "', , œ %&, and - œ #!!Þ

, „ È, #  %+#+ Ð%&Ñ „ ÈÐ%&Ñ#  %Ð"'ÑÐ#!!Ñ >œ #Ð"'Ñ %& „ È#!#&  "#,)!! œ $# %& „ È"%,)#& œ $# >œ

1045

Quadratic Equations, Inequalities, and Functions %&  È"%,)#& ¸ &Þ# or $# %&  È"%,)#& >œ ¸ #Þ% $# >œ

37.

41.

When C œ $, B œ %, so the vertex is Ð%ß $ÑÞ 42.

C œ 0 ÐBÑ œ $B#  %B  # Use the vertex formula with + œ $ and , œ %.

Reject the negative solution since time cannot be negative. The ball will reach a height of #!! ft above the ground after about &Þ# seconds.

The B-coordinate of the vertex is

0 Ð>Ñ œ "!!>#  $!!> is the distance of the light from the starting point at > minutes. When the light returns to the starting point, the value of 0 Ð>Ñ will be !Þ

The C-coordinate of the vertex is

! œ "!!>#  $!!> ! œ >#  $> ! œ >Ð>  $Ñ >œ!

or

, % # œ œ Þ #+ #Ð$Ñ $ 0Œ

, #  œ 0Œ  #+ $

œ $ˆ #$ ‰  %ˆ #$ ‰  # #

Divide by 100.

œ  %$ 

) $

#

œ  #$ .

>$œ! >œ$

Since > œ ! represents the starting time, the light will return to the starting point in $ minutes.

The vertex is ˆ #$ ß  #$ ‰. 43.

C œ #ÐB  #Ñ#  $ The graph opens up since + œ #  !Þ

(a) Use 0 ÐBÑ œ #$!Þ&B#  #&#Þ*B  &*)( with B œ (.

38.

B œ ÐC  $Ñ#  %

The vertex is Ð#ß $Ñ and the axis is B œ #. The domain is Ð∞ß ∞Ñ. The smallest C-value is $, so the range is Ò$ß ∞ÑÞ

#

0 Ð(Ñ œ #$!Þ&Ð(Ñ  #&#Þ*Ð(Ñ  &*)( œ "&,&""Þ# ¸ "&,&"" The value of $"&,&"" million is close to the number suggested by the graph. (b) Let 0 ÐBÑ œ "%,!!!. "%,!!! œ #$!Þ&B#  #&#Þ*B  &*)( ! œ #$!Þ&B#  #&#Þ*B  )!"$ Here + œ #$!Þ&, , œ #&#Þ*, and - œ )!"$Þ Bœ Bœ œ Bœ Bœ

, „ È, #  %+#+ Ð#&#Þ*Ñ „ ÈÐ#&#Þ*Ñ#  %Ð#$!Þ&ÑÐ)!"$Ñ #Ð#$!Þ&Ñ È #&#Þ* „ (,%&",*%%Þ%" %'" #&#Þ*  È(,%&",*%%Þ%" ¸ 'Þ%( or %'" #&#Þ*  È(,%&",*%%Þ%" ¸ &Þ$( %'"

Since time cannot be negative, the correct answer is B ¸ ', which represents 2006. Based on the graph, the revenue in 2006 was closer to $"$,!!! million than $"%,!!! million. 39.

0 ÐBÑ œ ÐB  "Ñ# Write in C œ +ÐB  2Ñ#  5 form as C œ "ÐB  "Ñ#  !. The vertex Ð2ß 5Ñ is Ð"ß !Ñ.

40.

0 ÐBÑ œ ÐB  $Ñ#  ( The equation is in the form C œ +ÐB  2Ñ#  5 , so the vertex Ð2ß 5Ñ is Ð$ß (Ñ.

1046

44.

0 ÐBÑ œ #B#  )B  & Complete the square to find the vertex. 0 ÐBÑ œ #ˆB#  %B‰  & œ #ˆB#  %B  %  %‰  &

œ #ˆB#  %B  %‰  #Ð%Ñ  & œ #ÐB  #Ñ#  )  & œ #ÐB  #Ñ#  $

The equation is in the form C œ +ÐB  2Ñ#  5 , so the vertex Ð2ß 5Ñ is Ð#ß $Ñ and the axis is B œ #. Here, + œ #  !, so the parabola opens downÞ Also, k+k œ k#k œ #  ", so the graph is narrower than the graph of C œ B# . The points Ð!ß &Ñ, Ð"ß "Ñ, and Ð$ß "Ñ are on the graph.

Quadratic Equations, Inequalities, and Functions The domain is Ð∞ß ∞Ñ. The largest C-value is $, so the range is Ð∞ß $ÓÞ 45.

47.

B œ #ÐC  $Ñ#  % Since the roles of B and C are reversed, this is a horizontal parabola. B œ #cC  Ð$Ñd#  Ð%Ñ The equation is in the form

(a) Use +B#  ,B  - œ C with Ð!ß #Þ*Ñ, Ð"!ß #%Þ$Ñ, and Ð#!ß &'Þ&Ñ. Ð"Ñ Ð#Ñ Ð$Ñ

- œ #Þ* "!!+  "!,  - œ #%Þ$ %!!+  #!,  - œ &'Þ&

(b) Rewrite equations Ð#Ñ and Ð$Ñ with - œ #Þ*. "!!+  "!,  #Þ* œ #%Þ$ %!!+  #!,  #Þ* œ &'Þ&

B œ +ÐC  5Ñ#  2 , so the vertex Ð2ß 5Ñ is Ð%ß $Ñ and the axis is C œ $. Here, + œ #  !, so the parabola opens to the right and is narrower than the graph of C œ B# .

"!!+  "!, œ #"Þ% %!!+  #!, œ &$Þ'

Ð%Ñ Ð&Ñ

Now eliminate , .

Two other points on the graph are %ß " and %ß & .

# ‚ Ð%Ñ #!!+  #!, œ %#Þ) %!!+  #!, œ &$Þ' Ð&Ñ #!!+ œ "!Þ) + œ "!Þ) #!! œ !Þ!&% Use Ð%Ñ to find , . "!!+  "!, "!!Ð!Þ!&%Ñ  "!, &Þ%  "!, "!, ,

The smallest B-value is %, so the domain is Ò%ß ∞ÑÞ The range is Ð∞ß ∞Ñ. 46.

B œ  "# C#  'C  "% Since the roles of B and C are reversed, this is a horizontal parabola. Complete the square to find the vertex.

Thus, we get the quadratic function 0 ÐBÑ œ !Þ!&%B#  "Þ'B  #Þ*.

B œ  "# C#  'C  "%

(c) Use 0 ÐBÑ œ !Þ!&%B#  "Þ'B  #Þ* with B œ #" for the year 2006Þ

œ  "# C#  "#C  "% œ  "# C#  "#C  $'  $'  "% œ œ Bœ

 "# C#  "#C  $'  "# ÐC  'Ñ#  ")   "# ÐC  'Ñ#  %

œ #"Þ% Ð%Ñ œ #"Þ% œ #"Þ% œ "' œ "' "! œ "Þ'



" # Ð$'Ñ

0 Ð#"Ñ œ !Þ!&%Ð#"Ñ#  "Þ'Ð#"Ñ  #Þ* œ '!Þ$"% ¸ $'!Þ$ billion

 "%

The result using the model is close to the table value of $'"Þ% billion, but slightly low.

"%

The equation is in the form B œ +ÐC  5Ñ#  2 , so the vertex Ð2ß 5Ñ is Ð%ß 'Ñ and the axis is C œ '. Here, + œ  "#  !, so the parabola opens to the left. Also, k+k œ ¸ "# ¸ œ "#  ", so the graph is wider than the graph of C œ B# . The points Ð"%ß !Ñ, Ð#ß %Ñ, and Ð#ß )Ñ are on the graph.

48.

=Ð>Ñ œ "'>#  "'!> The equation represents a parabola. Since + œ "'  !, the parabola opens down. The time and maximum height occur at the vertex of the parabola, given by Œ

, , ß =Œ . #+ #+

Using the standard form of the equation, + œ "' and , œ "'!, so >œ and

The largest B-value is %, so the domain is Ð∞ß %Ó. The range is Ð∞ß ∞Ñ.

, "'! œ œ &, #+ #Ð"'Ñ

=Ð>Ñ œ "'Ð&Ñ#  "'!Ð&Ñ œ %!!  )!! œ %!!.

The vertex is Ð&ß %!!Ñ. The time at which the maximum height is reached is & seconds. The maximum height is %!! feet. 1047

Quadratic Equations, Inequalities, and Functions 49.

Let P œ the length of the rectangle and [ œ the width.

51.

The perimeter of the rectangle is #!! m, so #P  #[ œ #!! #[ œ #!!  #P [ œ "!!  P.

B%œ! B œ %

Since the area is length times width, substitute "!!  P for [ .

Use the vertex formula.

B#  B Ÿ "#. Interval A:

So P œ &! meters and [ œ "!!  P œ "!!  &! œ &! meters. The maximum area is

Let B œ &. ?

Interval B:

&! • &! œ #&!! m# .

Interval C:

ÐB  %ÑÐ#B  $Ñ  ! Solve the equation ÐB  %ÑÐ#B  $Ñ œ !.

#&  & Ÿ "# #! Ÿ "# Let B œ !. ! Ÿ "# Let B œ %.

False True

?

"'  % Ÿ "# #! Ÿ "#

or or

#B  $ œ ! B œ  $#

Solution set: Ò%ß $Ó

52. Test a number from each interval in the inequality

False

The numbers in Interval B, including % and $, are solutions.

The numbers  $# and % divide a number line into three intervals.

Interval A:

B$œ! Bœ$

Test a number from each interval in the inequality

, "!! Pœ œ œ &! #+ #Ð"Ñ

B%œ! Bœ%

or or

The numbers % and $ divide a number line into three intervals.

T œ P[ œ PÐ"!!  PÑ œ "!!P  P# or P#  "!!P

50.

B#  B Ÿ "# Solve the equation B#  B œ "#. B#  B  "# œ ! ÐB  %ÑÐB  $Ñ œ !

ÐB  #ÑÐB  $ÑÐB  &Ñ Ÿ ! The numbers #, $, and & are solutions of the cubic equation

ÐB  %ÑÐ#B  $Ñ  !.

ÐB  #ÑÐB  $ÑÐB  &Ñ œ !.

Let B œ #.

These numbers divide a number line into four intervals.

?

'Ð"Ñ  ! '! Interval B: Let B œ !.

True

?

Interval C:

%Ð$Ñ  ! "#  ! Let B œ &.

False

ÐB  #ÑÐB  $ÑÐB  &Ñ Ÿ !.

?

"Ð"$Ñ  ! "$  !

Test a number from each interval in the inequality Interval A:

True

The solution set includes numbers in Intervals A and C, excluding endpoints. Solution set: ˆ∞ß  $# ‰ ∪ Ð%ß ∞Ñ

Interval B: Interval C: Interval D:

1048

Let B œ '. Ð%ÑÐ*ÑÐ"Ñ Ÿ ! Let B œ $. & Ð'ÑÐ#Ñ Ÿ ! Let B œ !. Ð#ÑÐ$ÑÐ&Ñ Ÿ ! Let B œ %Þ Ð'ÑÐ"ÑÐ*Ñ Ÿ !

True False True False

Quadratic Equations, Inequalities, and Functions The number $ makes the numerator !, and # makes the denominator !. These two numbers determine three intervals.

The numbers in Intervals A and C, including &, #, and $, are solutions. Solution set: Ð∞ß &Ó ∪ Ò#ß $Ó

Ð%B  $Ñ# Ÿ % The square of a real number is never negative. So, the solution set of this inequality is g. ' # 54. #D  " Write the inequality so that ! is on one side. ' #! #D  " ' #Ð#D  "Ñ  ! #D  " #D  " '  %D  # ! #D  " %D  ) ! #D  " The number # makes the numerator !, and "# makes the denominator !. These two numbers determine three intervals. 53.

Test a number from each interval in the inequality

Interval A:

Interval B:

$>  % Ÿ ". ># Let > œ %. ) ? ' Ÿ % $ Ÿ

"

" Let > œ !.

False

% ? # Ÿ

Interval C:

" # Ÿ " Let > œ $.

True

"$ ? " Ÿ

" "$ Ÿ "

False

The numbers in Interval B, including $ but not #, are solutions. Solution set: Ò$ß #Ñ

Test a number from each interval in the inequality '  #. #D  " Interval A: Let D œ !. True '  # Interval B: Let D œ ". '# False Interval C: Let D œ $. ' # True & The solution set includes numbers in Intervals A and C, excluding endpoints. Solution set: ˆ∞ß "# ‰ ∪ #ß ∞

55.

$>  % Ÿ" ># Write the inequality so that ! is on one side. $>  % "Ÿ! ># $>  % "Ð>  #Ñ Ÿ!  ># ># $>  %  >  # Ÿ! ># #>  ' Ÿ! >#

56.

[4]

Solve Z œ œ È&% > œ $È'

#

"& "&  œ% B& B& Multiply each term by the LCD, ÐB  &ÑÐB  &ÑÞ "&ÐB  &Ñ  "&ÐB  &Ñ œ %ÐB  &ÑÐB  &Ñ "&B  (&  "&B  (& œ %B#  "!! ! œ %B#  $!B  "!! ! œ #B#  "&B  &! ! œ Ð#B  &ÑÐB  "!Ñ or or

or

(B  $ œ & (B œ # B œ #(

The total time is % hours.

#B  & œ ! B œ  &#

> œ È&% > œ $È '

Solution set: š „ $È'›

The domain is Ð∞ß ∞Ñ. The smallest C-value is $, so the range is Ò$ß ∞ÑÞ 69.

># œ &%

>œ >œ œ œ œ

, „ È, #  %+#+ Ð%Ñ „ ÈÐ%Ñ#  %Ð$ÑÐ&Ñ #Ð$Ñ % „ È%% % „ #3È"" œ ' ' #Š# „ 3È""‹ # „ 3È"" œ ' $ # È"" „ 3 $ $

Solution set: š #$ „

È"" $ 3›

Quadratic Equations, Inequalities, and Functions 6.

$B œ Ê

*B  # # *B  # # *B œ Square each side. # ")B# œ *B  # ")B#  *B  # œ ! Here + œ "), , œ *, and - œ #. Bœ Bœ œ Bœ Bœ

, „ È, #  %+#+ Ð*Ñ „ ÈÐ*Ñ#  %Ð")ÑÐ#Ñ #Ð")Ñ * „ È##& * „ "& œ $' $' *  "& #% # œ œ or $' $' $ *  "& ' " œ œ $' $' ' # œ È%

Check B œ #$ : Check B œ

 "' :

 "#

œ

Solution set: ˜ #$ ™ 7.

8.

10.

Use the quadratic formula with + œ %, , œ (, and - œ $Þ , „ È, #  %+#+ ( „ È(#  %Ð%ÑÐ$Ñ Bœ #Ð%Ñ È ( „ *( œ ) Bœ

È*(

Solution set: š („) 11.

True

É "%

False

B# œ

?%œ! ?œ%

Bœ „

"' "#  # œ! B B Multiply each term by the LCD, B# . "' "# B# Œ$   #  œ B# • ! B B # $B  "'B  "# œ ! Ð$B  #ÑÐB  'Ñ œ !

$

Solution set: ˜ #$ ß '™

) $



" $

œ!

or

12.

B œ „ È% Bœ „# True True

"# œ Ð#8  "Ñ#  Ð#8  "Ñ Let ? œ #8  ", so ?# œ Ð#8  "Ñ# . "# œ ?#  ? ! œ ?#  ?  "# ! œ Ð?  %ÑÐ?  $Ñ ?%œ! ? œ %

or or

?$œ! ?œ$

To find 8, substitute #8  " for ?. #8  " œ % #8 œ & 8 œ  &#

B'œ! Bœ'

$  #%  #( œ !

" $

B# œ %

Solution set: ˜ „ "$ ß „ #™

$

or or

or

Check B œ „ "$ : "*  % œ $( * Check B œ „ #: "%%  % œ $(Ð%Ñ

The discriminant, )), is positive but not a perfect square, so there will be two distinct irrational number solutions.

$B  # œ ! B œ  #$

" *

B œ „ É "*

#B#  )B  $ œ !

Check B œ ':

or or

To find B, substitute B# for ?.

If 5 is a negative number, then %5 is also negative, so the equation B# œ %5 will have two nonreal complex solutions. The answer is A.

Check B œ  #$ :

›

*B%  % œ $(B# *B%  $(B#  % œ ! Let ? œ B# , so ?# œ ÐB# Ñ# œ B% . *?#  $(?  % œ ! Ð*?  "ÑÐ?  %Ñ œ ! *?  " œ ! ? œ "*

, #  %+- œ Ð)Ñ#  %Ð#ÑÐ$Ñ œ '%  #% œ ))

9.

%B#  (B  $ œ !

True True

Check 8 œ  &# : Check 8 œ ":

or or

#8  " œ $ #8 œ # 8œ"

"# œ "'  % "# œ *  $

Solution set: ˜ &# ß "™

True True

1053

Quadratic Equations, Inequalities, and Functions 13.

14.

Solve W œ %1% Interval A:

?

(#  %#  "& $!  "& Let B œ !. !  "& Let B œ #.

25.

True

Interval B:

False

Interval C:

?

)  "%  "& ##  "&

True

The numbers in Intervals A and C, not including & and $# , are solutions. Solution set: Ð∞ß &Ñ ∪ ˆ $# ß ∞‰

1056

Let > œ !. & % Ÿ " Let > œ (. & $ Ÿ" Let > œ "!. & ' Ÿ"

True False True

The numbers in Intervals A and C, including * but not %, are solutions. Solution set:

∞ß % ∪ Ò*ß ∞Ñ

(b) 0 ÐBÑ œ ÐB  "Ñ#

INVERSE, EXPONENTIAL, AND LOGARITHMIC FUNCTIONS

This equation has a vertical parabola as its graph, so some horizontal lines will intersect the graph at two points. For example, both B œ # and B œ ! correspond to C œ ". Thus, the function is not a one-to-one function and does not have an inverse.

1 Inverse Functions

(c) 0 ÐBÑ œ B$  %

1 Now Try Exercises

N1. (a) J œ eÐ"ß #Ñ, Ð!ß !Ñ, Ð"ß #Ñ, Ð#ß )Ñf

Refer to Section 3 to see from its graph that a cubing function like this is a one-to-one function. To find the inverse, first replace 0 ÐBÑ with C.

Since the C-value # corresponds to B-values " and ", the function is not one-to-one. Therefore, it does not have an inverse.

Step 1

(b) K œ eÐ!ß !Ñ, Ð"ß "Ñ, Ð%ß #Ñ, Ð*ß $Ñf

Each B-value in the relation corresponds to only one C-value, so the relation is indeed a function. Furthermore, each C-value corresponds to only one B-value, so this is a one-to-one function.

Step 2

The inverse function is found by interchanging the B- and C-values in each ordered pair.

Step 3

The inverse is

K" œ eÐ!ß !Ñ, Ð"ß "Ñ, Ð#ß %Ñ, Ð$ß *Ñf.

(c) Since the height '$* corresponds to the number of stories $" and %", the function is not one-to-one. Therefore, it does not have an inverse.

Interchange B and C. C œ B$  % B œ C$  % Solve for C. B  % œ C$ Take the cube root on each side. $ È B%œC Replace C with 0 " B Þ $ 0 " B œ È B%

N4. The given graph goes through Ð$ß #Ñ, Ð#ß "Ñ, Ð"ß #Ñ, and Ð!ß (Ñ. So, the inverse will go through Ð#ß $Ñ, Ð"ß #Ñ, Ð#ß "Ñ, and Ð(ß !Ñ. Use these points and symmetry about C œ B to complete the graph of the inverse.

N2. (a) Since a horizontal line will intersect the graph in no more than one point, the function is one-toone. (b) Since a horizontal line could intersect the graph in two points, the function is not one-to-one. N3. To find the equation of the inverse of a function, Ð"Ñ determine if the function is one-to-one; Ð#Ñ replace 0 ÐBÑ with C; Ð$Ñ interchange B and C; Step 1 Ð%Ñ solve for C; Step 2 Ð&Ñ replace C with 0 " ÐBÑ. Step 3 (a) 0 ÐBÑ œ &B  ( The graph of 0 is a line. By the horizontal line test, it is a one-to-one function. To find the inverse, first replace 0 ÐBÑ with C. Step 1

Step 2

Step 3

Interchange B and C. C œ &B  ( B œ &C  ( Solve for C. B  ( œ &C B( œC & Replace C with 0 " B Þ B( " ( 0 " B œ , or 0 " ÐBÑ œ B  & & &

1 Section Exercises 1.

This function is not one-to-one because both France and the United States are paired with the same trans fat percentage, 11.

3.

The function in the table that pairs a city with a distance is a one-to-one function because for each city there is one distance and each distance has only one city to which it is paired. If the distance from Indianapolis to Denver had " mile added to it, it would be "!&)  " œ "!&* mi, the same as the distance from Los Angeles to Denver. In this case, one distance would have two cities to which it is paired, and the function would not be one-to-one.

5.

If a function is made up of ordered pairs in such a way that the same C-value appears in a correspondence with two different B-values, then the function is not one-to-one. Choice B

From Chapter 10 of Student’s Solutions Manual for Intermediate Algebra, Eleventh Edition. Margaret L. Lial, John Hornsby, Terry McGinnis. Copyright © 2012 by Pearson Education, Inc. Publishing as Addison-Wesley. All rights reserved. 1057

Inverse, Exponential, and Logarithmic Functions 7.

All of the graphs pass the vertical line test, so they all represent functions. The graph in choice A is the only one that passes the horizontal line test, so it is the one-to-one function.

9.

ÖÐ$ß 'Ñ, Ð#ß "!Ñ, Ð&ß "#Ñ× is a one-to-one function, since each B-value corresponds to only one C-value and each C-value corresponds to only one B-value. To find the inverse, interchange B and C in each ordered pair. The inverse is

19.

Replace 0 ÐBÑ with C. C œ B$  % Interchange B and C. B œ C$  % Solve for C.

ÖÐ'ß $Ñ, Ð"!ß #Ñ, Ð"#ß &Ñ×.

11.

13.

eÐ"ß $Ñ, Ð#ß (Ñ, Ð%ß $Ñ, &ß ) f is not a one-to-one function. The ordered pairs "ß $ and %ß $ have the same C-value for two different B-values. The graph of 0 ÐBÑ œ #B  % is a nonvertical, nonhorizontal line. By the horizontal line test, 0 ÐBÑ is a one-to-one function. To find the inverse, replace 0 ÐBÑ with C.

B  % œ C$ Take the cube root of each side. $ È B%œC Replace C with 0 " ÐBÑÞ

$ 0 " ÐBÑ œ È B%

In Exercises 21–24, 0 ÐBÑ œ #B is a one-to-one function. 21.

C œ #B  % Interchange B and C. B œ #C  % Solve for C. #C œ B  % B% Cœ # Replace C with 0 " ÐBÑ. B% " 0 " ÐBÑ œ , or 0 " ÐBÑ œ B  # # #

15.

Write 1ÐBÑ œ ÈB  $ as C œ ÈB  $. Since B   $, C   !. The graph of 1 is half of a horizontal parabola that opens to the right. The graph passes the horizontal line test, so 1 is oneto-one. To find the inverse, interchange B and C to get

The graph of 0 ÐBÑ œ B$  % is the graph of 1ÐBÑ œ B$ shifted down % units. (Recall that 1ÐBÑ œ B$ is the elongated S-shaped curve.) The graph of 0 passes the horizontal line test, so 0 is one-to-one.

(a) To find 0 Ð$Ñ, substitute $ for B. 0 ÐBÑ œ #B , so 0 Ð$Ñ œ #$ œ )Þ (b) Since 0 is one-to-one and 0 Ð$Ñ œ ), it follows that 0 " Ð)Ñ œ $.

23.

(a) To find 0 Ð!Ñ, substitute ! for B. 0 ÐBÑ œ #B , so 0 Ð!Ñ œ #! œ ". (b) Since 0 is one-to-one and 0 Ð!Ñ œ ", it follows that 0 " Ð"Ñ œ !.

25.

(a) The function is one-to-one since any horizontal line intersects the graph at most once. (b) In the graph, the two points marked on the line are "ß & and #ß " . Interchange B and C in each ordered pair to get &ß " and "ß # . Plot these points, then draw a dashed line through them to obtain the graph of the inverse function.

B œ ÈC  $.

Note that now C   $, so B   !. Solve for C by squaring each side. B# œ C  $ B#  $ œ C Replace C with 1" ÐBÑ. 1" ÐBÑ œ B#  $, B   ! 17.

1058

0 ÐBÑ œ $B#  # is not a one-to-one function because two B-values, such as " and ", both have the same C-value, in this case &. The graph of this function is a vertical parabola which does not pass the horizontal line test.

27.

(a) The function is not one-to-one since there are horizontal lines that intersect the graph more than once. For example, the line C œ " intersects the graph twice.

Inverse, Exponential, and Logarithmic Functions 29.

(a) The function is one-to-one since any horizontal line intersects the graph at most once.

B ! " #

(b) In the graph, the four points marked on the curve are Ð%ß #Ñ, Ð"ß "Ñ, Ð"ß "Ñ, and Ð%ß #Ñ. Interchange B and C in each ordered pair to get Ð#ß %Ñ, Ð"ß "Ñ, Ð"ß "Ñ, and Ð#ß %Ñ. Plot these points, then draw a dashed curve (symmetric to the original graph about the line C œ B) through them to obtain the graph of the inverse.

Plot these points and connect them with a dashed smooth curve.

37. 31.

33.

B ! " %

0 ÐBÑ ! " #

Plot these points and connect them with a solid smooth curve. Since 0 ÐBÑ is one-to-one, make a table of values for 0 " B by interchanging B and C.

0 ÐBÑ $ # " '

Plot these points and connect them with a solid smooth curve. Make a table of values for 0 " . B $ # " '

1ÐBÑ œ %B or C œ %B The graph is a line through Ð!ß !Ñ and Ð"ß %Ñ. For the inverse, interchange B and C in each ordered pair to get the points Ð!ß !Ñ and Ð%ß "Ñ. Draw a dashed line through these points to obtain the graph of the inverse function.

0 ÐBÑ œ ÈB, B   ! Complete the table of values.

0 ÐBÑ œ B$  # Complete the table of values. B " ! " #

0 ÐBÑ œ #B  " or C œ #B  " The graph is a line through Ð#ß &Ñ, Ð!ß "Ñ, and Ð$ß &Ñ. Plot these points and draw the solid line through them. Then the inverse will be a line through Ð&ß #Ñ, Ð"ß !Ñ, and Ð&ß $Ñ. Plot these points and draw the dashed line through them.

0 " ÐBÑ " ! " #

Plot these points and connect them with a dashed smooth curve.

39.

35.

0 " ÐBÑ ! " %

0 ÐBÑ œ %B  & Replace 0 ÐBÑ with C. C œ %B  & Interchange B and C. B œ %C  & Solve for C. B  & œ %C B& œC % Replace C with 0 " ÐBÑÞ B& œ 0 " B , % " & or 0 " B œ B  % % 1059

Inverse, Exponential, and Logarithmic Functions 40.

Replace C with 0 " ÐBÑÞ $ È B  & œ 0 " B œ Y#

Insert each number in the inverse function found in Exercise 39, B& . % %(  & &# 0 " %( œ œ œ "$ œ M, % % *&  & "!! 0 " *& œ œ œ #& œ Y, % % and so on. 0 " B œ

The decoded message is as follows: My graphing calculator is the greatest thing since sliced bread. 41.

42.

A one-to-one code is essential to this process because if the code is not one-to-one, an encoded number would refer to two different letters. Answers will vary according to the student's name. For example, Jane Doe is encoded as follows: "!!% & #(%) "#* ') $$(* "#*.

43.

Y" œ 0 ÐBÑ œ #B  ( Replace 0 ÐBÑ with C. C œ #B  ( Interchange B and C. B œ #C  ( Solve for C. B  ( œ #C B( œC # Replace C with 0 " ÐBÑÞ B( " ( œ 0 " B œ Y# Œor B   # # # Now graph Y" and Y# .

47. 49.

0 ÐBÑ œ %B , so 0 Ð$Ñ œ %$ œ '%.

0 ÐBÑ œ %B , so 0 ˆ "# ‰ œ %"Î# œ

1060

Y" œ 0 ÐBÑ œ B$  & Replace 0 ÐBÑ with C. C œ B$  & Interchange B and C. B œ C$  & Solve for C. B  & œ C$ Take the cube root of each side. $ È B&œC

œ "# .

2 Exponential Functions 2 Now Try Exercises In Now Try Exercises 1–3, choose values of B and find the corresponding values of C. Then plot the points, and draw a smooth curve through them. N1. 0 ÐBÑ œ %B B 0 ÐBÑ œ %B

" ‰ N2. 1ÐBÑ œ ˆ "!

#

"

" "'

" %

! "

" %

# "'

" "!

! "

"

#

" "!

" "!!

B

B " ‰B 1ÐBÑ œ ˆ "!

45.

" %"Î#

# "!!

N3. C œ 0 ÐBÑ œ %#B  " Make a table of values. It will help to find values for #B  " before you find C. B ! " #

" "

#B  " " ! " $

C œ %#B  " " %

" % " '%

Inverse, Exponential, and Logarithmic Functions N6. B œ #!!!  "(&! œ #&! 0 ÐBÑ œ #''Ð"Þ!!"ÑB 0 Ð#&!Ñ œ #''Ð"Þ!!"Ñ#&! 0 Ð#&!Ñ ¸ $%# parts per million N7. N4. Step 1

)B œ "' Write each side with the base #. % B #$ œ # Simplify exponents. #$B œ #% Set the exponents equal. $B œ % Solve. B œ %$

The pressure is approximately %'& millibars.

3

Step 2 Step 3 Step 4

Check B œ

% $

in the original equation.

2 Section Exercises 1.

Since the graph of J ÐBÑ œ +B always contains the point Ð!ß "Ñ, the correct response is C.

3.

Since the graph of J ÐBÑ œ +B always approaches the B-axis, the correct response is A.

5.

0 ÐBÑ œ $B Make a table of values. " " 0 Ð#Ñ œ $# œ # œ , $ * " " " 0 Ð"Ñ œ $ œ " œ , and so on. $ $

? )%/$ œ "' %3 ?

)"/$

œ "'

% ?

# œ "' "' œ "'

True

The solution set is ˜ %$ ™. N5. (a)

B 0 ÐBÑ

$#B  " œ #(B  % #B  "

0 ÐBÑ œ "!$)Ð"Þ!!!"$%ÑB 0 Ð'!!!Ñ œ "!$)Ð"Þ!!!"$%Ñ'!!! 0 Ð'!!!Ñ ¸ %'&

B%

œ $$ $ #B  " $ œ $$B  "# #B  " œ $B  "# "$ œ B

#

"

" *

" $

! "

" $

# *

Plot the points from the table and draw a smooth curve through them.

Same base Simplify exponents. Set exponents equal.

Check B œ "$: ? $#'  " œ #("$  % ? ˆ $ ‰* $#( œ $

$#( œ $#(

True

The solution set is e"$f. (b) &B œ

7.

" '#&

" &% &B œ &% B œ % &B œ

Same base Set exponents equal.

Check B œ %: &

%

œ

The solution set is e%f.

B (c) ˆ #( ‰ œ

" &%

œ

" '#&

B 1ÐBÑ œ ˆ "$ ‰ Make a table of values. # # 1Ð#Ñ œ ˆ "$ ‰ œ ˆ $" ‰ œ *,

1Ð"Ñ œ ˆ "$ ‰ B 1ÐBÑ

"

œ ˆ $" ‰ œ $, and so on. "

# *

" $

! "

"

#

" $

" *

Plot the points from the table and draw a smooth curve through them.

$%$ )

ˆ #( ‰B œ ˆ (# ‰$

ˆ #( ‰B œ ˆ #( ‰$ B œ $ Check B œ $:

Set exponents equal.

ˆ #( ‰$

The solution set is e$f.

œ ˆ (# ‰ œ $

$%$ )

1061

Inverse, Exponential, and Logarithmic Functions 9.

C œ %B This equation can be rewritten as C œ %"

B3

Check B œ $# : "!!$Î# œ "!!! The solution set is ˜ $# ™.

œ ˆ "% ‰ , B

19.

which shows that it is falling from left to right. Make a table of values. B C

# "'

" %

! "

"

#

" %

" "'

"'#B  " œ '%B  $ Write each side as a power of %. ˆ%# ‰ #B  "3 œ ˆ%$ ‰ B  $3 %%B  # œ %$B  * Set the exponents equal. %B  # œ $B  * Bœ(

Check B œ (: "'"& œ '%"! The solution set is e(f. 21. 11.

# '

" %

! #

" '%

" "'

" %

" ! "

# # %

$ % "'

15.

For an exponential function defined by 0 ÐBÑ œ + , if +  ", the graph rises from left to right. (See Example 1, 0 ÐBÑ œ #B , in your text.) If !  +  ", the graph falls from left to right. (See B Example 2, 1ÐBÑ œ ˆ "# ‰ œ #B , in your text.)

25.

$ Check B œ $: ˆ $# ‰ œ The solution set is e$f.

Bœ#

17.

"!!B œ "!!! Write each side as a power of "!. ˆ"!# ‰ B3 œ "!$ #B

$

"! œ "! For +  ! and + Á ", if +B œ +C , then B œ C. Set the exponents equal to each other. #B œ $ B œ $# 1062

ˆ $# ‰B œ

) #( ˆ #$ ‰$

ˆ $# ‰B œ ˆ $# ‰$ Set the exponents equal. B œ $

' œ $' Write each side as a power of '. 'B œ '# For +  ! and + Á ", if +B œ +C , then B œ C. Set the exponents equal to each other.

True

ˆ $# ‰B œ

True

Write each side as a power of $# .

B

Check B œ #: ' œ $' The solution set is e#f.

True

&B œ !Þ# # &B œ "! œ &" Write each side as a power of &. &B œ &" Set the exponents equal. B œ " Check B œ ": &" œ !Þ# The solution set is e"f.

B

#

" "#& ˆ "& ‰$

" Check B œ $: &$ œ "#& The solution set is e$f.

23.

13.

&B œ

True

&B œ Write each side as a power of &. &B œ &$ Set the exponents equal. B œ $

C œ ##B  # Make a table of values. It will help to find values for #B  # before you find C. B #B  # C

True

) #(

True

27.

"##Þ' ¸ '$*Þ&%&

29.

!Þ&$Þ*#" ¸ !Þ!''

31.

#Þ(")#Þ& ¸ "#Þ"(*

33.

(a) C œ Ð"Þ!%' ‚ "!$) ÑÐ"Þ!%%%B Ñ œ Ð"Þ!%' ‚ "!$) ÑÐ"Þ!%%%#!!! Ñ ¸ !Þ&( ¸ !Þ' The increase for the exponential function in the year 2000 is about !Þ'°C.

Inverse, Exponential, and Logarithmic Functions (b) C œ !Þ!!*B  "(Þ'( œ !Þ!!*Ð#!!!Ñ  "(Þ'( œ !Þ$$ ¸ !Þ$

> Z Ð>Ñ

& #*($

"! "(')

Plot the points from the table and draw a smooth curve through them.

The increase for the linear function in the year 2000 is about !Þ$°C. 35.

! &!!!

(a) C œ Ð"Þ!%' ‚ "!$) ÑÐ"Þ!%%%B Ñ œ Ð"Þ!%' ‚ "!$) ÑÐ"Þ!%%%#!#! Ñ ¸ "Þ$& ¸ "Þ% The increase for the exponential function in the year 2020 is about "Þ%°C. (b) C œ !Þ!!*B  "(Þ'( œ !Þ!!*Ð#!#!Ñ  "(Þ'( œ !Þ&" ¸ !Þ& The increase for the linear function in the year 2020 is about !Þ&°C.

37.

41.

0 ÐBÑ œ %#$"Ð"Þ!"(%ÑB 0 Ð"!Ñ œ %#$"Ð"Þ!"(%Ñ"! ¸ &!#) The answer has units in millions of metric tons. (b) 1995 corresponds to B œ #&. (c) 2000 corresponds to B œ $!. 0 Ð$!Ñ œ %#$"Ð"Þ!"(%Ñ$! ¸ (!** The actual amount, '($& millions of metric tons, is less than the (!** millions of metric tons that the model provides.

39.

Z Ð>Ñ œ &!!!Ð#Ñ!Þ"&> (a) The original value is found when > œ !. Z Ð!Ñ œ &!!!Ð#Ñ!Þ"&Ð!Ñ œ &!!!Ð#Ñ! œ &!!!Ð"Ñ œ &!!! The original value is $&!!!. (b) The value after & years is found when > œ &. Z Ð&Ñ œ &!!!Ð#Ñ!Þ"&Ð&Ñ œ &!!!Ð#Ñ!Þ(& ¸ #*($Þ!# The value after & years is about $#*($. (c) The value after "! years is found when > œ "!.

43.

"' œ # • # • # • # œ #% , so

45.

#! œ ", so

œ !.

3 Now Try Exercises N1. (a) For the exponential form '$ œ #"', the logarithmic form is log' #"' œ $. (b) For the logarithmic form log'% % œ "$ , the exponential form is '%"Î$ œ %. N2. (a) log# B œ & Write in exponential form. B œ #& " " Bœ & œ # $# The argument (the input of the logarithm) must be " a positive number, so B œ $# is acceptable. " ™ The solution set is ˜ $# .

(b) log$Î# Ð#B  "Ñ œ $

#B  " œ ˆ $# ‰

Z Ð"!Ñ œ &!!!Ð#Ñ œ &!!!Ð#Ñ"Þ& ¸ "('(Þ(( (d) Use the results of parts (a) – (c) to make a table of values.

œ %.

3 Logarithmic Functions

!Þ"&Ð"!Ñ

The value after "! years is about $"(').

Let V(t) œ 2500. Divide by 5000.

#" œ #!Þ"&> " œ !Þ"&> Equate exponents. " >œ ¸ 'Þ'( !Þ"& The value of the machine will be $#&!! in approximately 'Þ'( years after it was purchased.

(a) 1980 corresponds to B œ "!.

0 Ð#&Ñ œ %#$"Ð"Þ!"(%Ñ#& ¸ '&"#

Z Ð>Ñ œ &!!!Ð#Ñ!Þ"&> #&!! œ &!!!Ð#Ñ!Þ"&> " !Þ"&> # œ Ð#Ñ

#B  " œ #B œ Bœ

$& ) $& "'

#( )

$

Exponential form Apply the exponent.

Add 1. Divide by 2.

1063

Inverse, Exponential, and Logarithmic Functions Substitute $& "' for B in the argument, #B  ", $& ‰ #( $& ˆ # "'  " œ $& )  " œ )  !, so "' is acceptable. ™ The solution set is ˜ $& "' .

(c) logB "! œ # B# œ "!

B œ „ È"!

Exponential form Take square roots.

Reject B œ È"! since the base of a logarithm must be positive and not equal to ".

N6. KÐ>Ñ œ "&Þ!  #Þ!! log"! > (a) KÐ"Ñ œ "&Þ!  #Þ!! log"! " œ "&Þ!  #Þ!!Ð!Ñ œ "&Þ!

The solution set is šÈ"!›.

$ (d) log"#& È &œB $ "#&B œ È &

ˆ&$ ‰B œ &"Î$

Bœ

The GNP in 2004 was $"&Þ! million. (b) KÐ"0Ñ œ "&Þ!  #Þ!! log"! "0 œ "&Þ!  #Þ!!Ð"Ñ œ "(Þ!

Exponential form Same base

&$B œ &"Î$ $B œ "$

Power rule Equate exponents.

" *

The solution set is ˜ "* ™.

Divide by 3.

3 Section Exercises 1.

(c) log # È# œ (D)

with ,  ! and , Á ".

$ (e) log ) È )œ (A)

(c) log!Þ" " œ ! N4. C œ 0 ÐBÑ œ log' B is equivalent to B œ ' . Choose values for C and find B. C

B C

#

"

" !

' "

$' #

Plot the points, and draw a smooth curve through them.

N5. C œ 1ÐBÑ œ log"/% B is equivalent to B œ

% "

" !

" %

" "'

"

#

ˆ "% ‰C .

Plot the points, and draw a smooth curve through them. 1064

is equivalent to #"/# œ È#.

" $

$ is equivalent to )"/$ œ È ).

(f) log % % œ " is equivalent to %" œ %.

(C)

3.

The base is %, the exponent (logarithm) is &, and the number is "!#%, so %& œ "!#% becomes log% "!#% œ & in logarithmic form.

5.

" #

ˆ "# ‰$

7.

The base is "!, the exponent (logarithm) is $, and the number is !Þ!!", so "!$ œ !Þ!!" becomes log"! !Þ!!" œ $ in logarithmic form.

9.

is the base and $ is the exponent, so œ ) becomes log"Î# ) œ $ in logarithmic form.

% È '#& œ '#&"Î% œ &

The base is '#&, the exponent (logarithm) is "% , and % the number is &, so È '#& œ & becomes log'#& & œ "% in logarithmic form.

Choose values for C and find B. B C

" #

(E)

(d) log "! "!!! œ $ is equivalent to "!$ œ "!!!. (F)

(b) log) " œ !

" '

(a) log "/$ $ œ " is equivalent to Ð "$ Ñ" œ $. (B) (b) log & " œ ! is equivalent to &! œ ".

log, , œ " and log, " œ !,

" $'

Let t œ 10. log10 10 œ 1

The GNP in 2013 is predicted to be $"(Þ! million.

N3. For each expression, use the properties of logarithms,

(a) log"! "! œ "

Let t œ 1. log10 1 œ 0

" %

becomes log)

" %

œ  #$ in logarithmic

11.

)#Î$ œ form.

13.

&! œ " becomes log& " œ ! in logarithmic form.

Inverse, Exponential, and Logarithmic Functions 15.

In log% '% œ $, % is the base and $ is the logarithm (exponent), so log% '% œ $ becomes %$ œ '% in exponential form.

17.

" In log"! "!,!!! œ %, the base is "!, the logarithm (exponent) is %, and the number is "!,"!!! , so log"! "!,"!!! œ % becomes "!% œ "!,"!!! in exponential form.

19.

In log' " œ !, ' is the base and ! is the logarithm (exponent), so log' " œ ! becomes '! œ " in exponential form.

21.

In log* $ œ "# , the base is *, the logarithm (exponent) is "# , and the number is $, so log* $ œ becomes *"Î# œ $ in exponential form.

23.

log"Î% "# form.

œ

" #

becomes

Ð "% Ñ"Î#

œ

" #

33.

" #

B œ "& is an acceptable base since it is a positive number (not equal to ").

in exponential

25.

log& &" œ " becomes &" œ &" in exponential form.

27.

Use the properties of logarithms,

The solution set is ˜ "& ™.

35.

log, , œ " and log, " œ !, (a) log) ) œ " (C) 37.

(c) log!Þ$ " œ ! (B)

(d) logÈ( È( œ " (C)

29.

B œ log#( $ Write in exponential form. #(B œ $ Write each side as a power of $. B $$ œ $ $B $ œ $" Set the exponents equal. $B œ " B œ "$ " $

39.

œ log#( $ since #("Î$ œ $.

logB * œ "# Change to exponential form. B"/# œ * # B"/# # œ * Square. B" œ )" B œ )" B œ )" is an acceptable base since it is a positive number (not equal to "). Check B œ )": log)" * œ The solution set is e)"f.

" #

since )""/# œ *

The solution set is e"f. logB B œ " B" œ B

Write in exponential form.

The solution set is eB l B  !, B Á "f. " œ # #& " B# œ #& " " œ # B #& B# œ #& Bœ „&

logB

Write in exponential form.

Denominators must be equal.

Reject B œ & since the base of a logarithm must be positive and not equal to ".

The solution set is ˜ "$ ™.

31.

Write in exponential form.

This equation is true for all the numbers B that are allowed as the base of a logarithm; that is, all positive numbers B, B Á ".

3

Check B œ "$ :

log"# B œ ! "#! œ B "œB

The argument (the input of the logarithm) must be a positive number, so B œ " is acceptable.

for ,  !, , Á ". (b) log"' " œ ! (B)

logB "#& œ $ Write in exponential form. B$ œ "#& " œ "#& B$ " œ "#&ˆB$ ‰ " œ B$ "#& Take the cube root of each side. " $ $ œ È B$ Ê "#& " $ " BœÊ $ œ & &

41.

The solution set is e&f.

log) $# œ B )B œ $# Exponential form Write each side as a power of #. ˆ#$ ‰ B3 œ #& #$B œ #& $B œ & B œ &$

Equate exponents.

Check B œ &$ : log) $# œ The solution set is ˜ &$ ™.

& $

since )&Î$ œ #& œ $#.

1065

Inverse, Exponential, and Logarithmic Functions 43.

B about the line C œ B to the graph of C œ ˆ "$ ‰ . The graph can be plotted by reversing the ordered pairs in the table of values belonging to B 1ÐBÑ œ ˆ "$ ‰ .

log1 1% œ B 1B œ 1% Exponential form Bœ% Equate exponents. Check B œ %: log1 1% œ % since 1% œ 1% . The solution set is e%f.

45.

B C

log' È#"' œ B log' #"'"/# œ B 'B œ #"'"/# 'B œ ˆ'$ ‰ "Î#3 B

' œ' B œ $#

" $

" *

"

#

Same base Equate exponents.

log% Ð#B  %Ñ œ $ #B  % œ %$ #B œ '%  % #B œ '! B œ $!

53.

The number " is not used as a base for a logarithmic function since the function would look like B œ "C in exponential form. Then, for any real value of C, the statement " œ " would always be the result since every power of " is equal to ".

55.

The range of 0 ÐBÑ œ +B is the domain of 1ÐBÑ œ log + B, that is, Ð!ß ∞Ñ . The domain of 0 ÐBÑ œ +B is the range of 1ÐBÑ œ log + B, that is, Ð∞ß ∞Ñ .

57.

The values of > are on the horizontal axis, and the values of 0 Ð>Ñ are on the vertical axis. Read the value of 0 Ð>Ñ from the graph for the given value of >. At > œ !, 0 Ð!Ñ œ ).

59.

To find 0 Ð'!Ñ, find '! on the >-axis, then go up to the graph and across to the 0 Ð>Ñ axis to read the value of 0 Ð'!Ñ. At > œ '!, 0 Ð'!Ñ œ #%.

61.

0 ÐBÑ œ $)!!  &)& log# B

Exponential form

The solution set is e$!f. C œ log$ B $C œ B

Exponential form

Refer to Section 2, Exercise 5, for the graph of 0 ÐBÑ œ $B . Since C œ log$ B or $C œ B is the inverse of 0 ÐBÑ œ C œ $B , its graph is symmetric about the line C œ B to the graph of 0 ÐBÑ œ $B . The graph can be plotted by reversing the ordered pairs in the table of values belonging to 0 ÐBÑ œ $B . B C

" !

$/#

Check B œ $!: log% Ð# • $!  %Ñ œ log% '% œ $.

49.

$ "

Plot the points, and draw a smooth curve through them.

Equivalent form Exponential form

Check B œ $# : log' È#"' œ $# since '$Î# œ È'$ œ È#"'. The solution set is ˜ $# ™. 47.

* #

" *

" $

#

"

" !

$ "

* #

Plot the points, and draw a smooth curve through them.

(a) B œ "*)#  "*)! œ # 0 Ð#Ñ œ $)!!  &)& log# # œ $)!!  &)&Ð"Ñ œ %$)& The model gives an approximate withdrawal of %$)& billion ft$ of natural gas from crude oil wells in the United States for 1982. (b) B œ "*))  "*)! œ )

51.

C œ log"Î$ B ˆ "$ ‰C œ B

Exponential form

Refer to Section 2, Exercise 7, for the graph of B C 1ÐBÑ œ ˆ "$ ‰ . Since C œ log"Î$ B ˆor ˆ "$ ‰ œ B‰ is B " the inverse of C œ ˆ $ ‰ , its graph is symmetric 1066

0 Ð)Ñ œ $)!!  &)& log # ) œ $)!!  &)&Ð$Ñ œ &&&& The model gives an approximate withdrawal of &&&& billion ft$ of natural gas from crude oil wells in the United States for 1988.

Inverse, Exponential, and Logarithmic Functions The ratio of B# to B" is

(c) B œ "**'  "*)! œ "'

B# B! "!(Þ$ œ œ "!!Þ' ¸ $Þ*). B" B! "!'Þ(

0 Ð"'Ñ œ $)!!  &)& log# "' œ $)!!  &)&Ð%Ñ œ '"%! The model gives an approximate withdrawal of '"%! billion ft$ of natural gas from crude oil wells in the United States for 1996. 63.

The Landers earthquake was about % times more powerful than the Northridge earthquake. 67.

WÐ>Ñ œ "!!  $! log$ Ð#>  "Ñ

1ÐBÑ œ log$ B On a TI-83/4, assign

to Y" . Then enter

DrawInv Y"

(a) WÐ"Ñ œ "!!  $! log$ Ð# • "  "Ñ œ "!!  $! log$ Ð$Ñ œ "!!  $!Ð"Ñ œ "$!

on the home screen to obtain the figure that follows. DrawInv is choice 8 under the DRAW menu. Y" is choice 1 under VARS, Y-VARS, Function.

After " year, the sales were "$! thousand units. (b) WÐ"$Ñ œ "!!  $! log$ Ð# • "$  "Ñ œ "!!  $! log$ Ð#(Ñ œ "!!  $!Ð$Ñ œ "*! After "$ years, the sales were "*! thousand units. (c) Make a table of values, plot the points they represent, and draw a smooth curve through them. To make the table, find values of > such that #>  " œ $5 , where 5 œ !, ", #, $, %Þ !

"

#

$

%

" " ! ! "!!

$ $ # " "$!

* * ) % "'!

#( #( #' "$ "*!

)" )" )! %! ##!

5 5

$ #>  " #> > WÐ>Ñ

69.

1ÐBÑ œ log"Î$ B Assign to

"

and enter DrawInv Y" .

71.

%( • %# œ % (  # œ % *

73.

() œ ()  Ð%Ñ œ ("# (%

4 Properties of Logarithms 4 Now Try Exercises 65.

V œ log"!

B B!

N1. Use the product rule for logarithms.

Change to exponential form. "!V œ

B , so B œ B! "!V . B!

Let V œ 'Þ( for the Northridge earthquake, with intensity B" . B" œ B! "!'Þ( Let V œ (Þ$ for the Landers earthquake, with intensity B# . B# œ B! "!(Þ$

log, BC œ log, B  log, C (a) log"! Ð( • *Ñ œ log"! (  log"! * (b) log& ""  log& ) œ log& Ð"" • )Ñ œ log& )) (c) log& Ð&BÑ œ log& &  log& B œ "  log& B ÐB  !Ñ (d) log# >$ œ log# Ð> • > • >Ñ œ log# >  log# >  log# > œ $ log# > Ð>  !Ñ 1067

Inverse, Exponential, and Logarithmic Functions N2. Use the quotient rule for logarithms. log, (a) log"!

( *

B œ log, B  log, C C

#& #(

œ log& ÒÐB  "!ÑÐB  "!ÑÓ  $& log& B Product rule # œ log& ÐB  "!!Ñ  log& B$Î& Power rule # B  "!! œ log& Quotient rule B$Î&

œ log"! (  log"! *

(b) log% B  log% "# œ log% (c) log&

(d) log& ÐB  "!Ñ  log& ÐB  "!Ñ  $& log& B ÐB  "!Ñ

B "#

ÐB  !Ñ

œ log& #&  log& #( œ #  log& #(

(e) log( Ð%*  #BÑ cannot be written as a sum of logarithms, since

N3. Use the power rule for logarithms.

log, ÐQ  R Ñ Á log, Q  log, R .

log , B< œ < log , B

There is no property of logarithms to rewrite the logarithm of a sum.

(a) log( &$ œ $ log( &

(b) log+ È"! œ log+ "!"/# œ "# log+ "! Ð+  !Ñ

N6. (a) log# (! œ log# Ð( • "!Ñ œ log# (  log# "! œ #Þ)!(%  $Þ$#"* œ 'Þ"#*$

(c) log$ ÈB$ œ log$ B$/% œ $% log$ B ÐB  !Ñ %

( (b) log# !Þ( œ log# "! œ log# (  log# "! œ #Þ)!(%  $Þ$#"* œ !Þ&"%&

N4. Use the Special Properties, , log, B œ B or log, , B œ B. (a) By the second property,

(c) log# %* œ log# (# œ # log# ( œ #Ð#Þ)!(%Ñ œ &Þ'"%)

log% %( œ (. (b) Using the second property, log"! "!,!!! œ log"! "!% œ %.

N7. (a) log# "'  log# "' œ log# $#

(c) By the first property,

Evaluate each side.

)log) & œ &.

LS œ log# "'  log# "' œ log# #%  log# #% œ %  % œ ) RS œ log# $# œ log# #& œ &

N5. Use the properties of logarithms. (a) log$ *D % œ log$ *  log$ D % œ #  % log$ D

The statement is false because ) Á &. Product rule log3 32 œ 2; Power rule

(b) log# % log$ * œ log' $' Evaluate each side.

8 (b) log' É $7

LS œ log# % log$ * œ ˆlog# ## ‰ˆlog$ $# ‰ œ Ð#ÑÐ#Ñ œ % RS œ log' $' œ log' '# œ #

8 "/ # œ log' Ð $7 Ñ

œ œ œ œ

" 8 # log' Ð $7 Ñ " # log' 8  log' $7 " # clog' 8  Ðlog' $  log' 7Ñd " # log' 8  log' $  log' 7

(c) log# B  $ log# C  log# D œ log# B  log# C$  log# D œ log# ÐBC$ Ñ  log# D BC$ œ log# D

1068

Po. rule Qu. rule

The statement is false because % Á #.

4 Section Exercises 1.

By the product rule, log"! Ð( • )Ñ œ log"! (  log"! ).

Power rule Product rule

3.

Quotient rule

5.

By a special property, $log$ % œ %. By a special property, log$ $* œ *.

Inverse, Exponential, and Logarithmic Functions 7.

Use the product rule for logarithms.

27.

log( Ð% • &Ñ œ log( %  log( & 9.

Use the quotient rule for logarithms. log&

11.

13.

) $

œ log& )  log& $ 29.

Use the power rule for logarithms. $ È %

log% '# œ # log% '

%"Î$ B# C B# C Use the quotient rule for logarithms. log$

œ log$

31.

œ log$ %"Î$  log$ ˆB# C‰

œ log$ %"Î$  Š log$ B#  log$ C ‹

œ œ œ 17.

" # " # " #

33.

Power rule

clog$ ÐBCÑ  log$ &d

log$ B  log$ C  log$ &

log#

$ B•È & C È

œ "!. E œ "!,!!!Œ" 

!Þ!#&  "# "#! œ "!,!!!ˆ"Þ!!#!)$‰ ¸ "#,)$'Þ*#

Check B œ "*Þ")): /!Þ"#Ð"*Þ"))Ñ ¸ "! The solution set is e"*Þ"))f.

N3. log& ÐB  "Ñ$ œ # ÐB  "Ñ$ œ &# ÐB  "Ñ$ œ #& $ B"œÈ #&

$ Bœ"È #&

$ Check B œ "  È #&:

log& ˆ" 

$ È

There will be $"#,)$'Þ*# in the account. N7.

Exponential form Cube root Add 1.

$ #&  "‰ œ log& ŠÈ#&‹

$

$

œ log& #& œ log& &# œ #

$ The solution set is ˜"  È #&™.

N4. log4 Ð#B  "$Ñ  log% ÐB  "Ñ œ log% "! #B  "$ log% œ log% "! B" #B  "$ œ "! B" #B  "$ œ "!ÐB  "Ñ #B  "$ œ "!B  "! $ œ )B $ ) œB

Given equation Quotient rule Property 4 Multiply by x  1. Distributive prop. Subtract 2x; 10. Divide by 8.

E œ T Š" 

< 8> ‹ 8 %•> !Þ!% #T œ T Œ"   % # œ "Þ!"

log%

&& %

 log%

"" )

œ log%

The solution set is ˜ $) ™.

B%œ! B œ %

or or

B#œ! Bœ#

Reject B œ % since it yields an equation in which the logarithm of a negative number must be found. Check B œ #: log4 %  log4 % œ "  " œ # The solution set is e#f.

Double, so A = 2P Quarterly, n = 4 Divide by P. Property 3 Power rule Divide by log 1.01. Divide by 4.

It will take about "(Þ%# years for any amount of money in an account paying %% interest compounded quarterly to double. N8. (a) E œ T / E œ %!!!/!Þ!$Ð#Ñ E ¸ %#%(Þ$& It will grow to $%#%(Þ$&. (b)

œ log% "!

N5. log4 ÐB  #Ñ  log4 #B œ # log4 c#B B  # d œ # Product rule #B B  # œ %# Exponential form #B#  %B œ "' # #B  %B  "' œ ! B#  #B  ) œ ! ÐB  %ÑÐB  #Ñ œ !

%>

log # œ log "Þ!" %> log # œ %> log "Þ!" log # %> œ log "Þ!" log # >œ % log "Þ!" > ¸ "(Þ%#

Check B œ $) : && % "" )

"#Ð"!Ñ

E œ T / #Ð%!!!Ñ œ %!!!/!Þ!$> # œ /!Þ!$> ln # œ ln /!Þ!$> ln # œ !Þ!$> ln # >œ !Þ!$ > ¸ #$Þ"

Divide by 4000. Take natural logs. ln /5 œ 5 Divide by 0.03.

It would take about #$Þ" years for the initial investment to double. N9. C œ C! /!Þ!!!%$> (a) C œ %Þ&/!Þ!!!%$Ð"&!Ñ C ¸ %Þ##

Let t = 150.

After "&! years, there will be about %Þ# grams. (b)

" #

%Þ& œ %Þ&/!Þ!!!%$> ln

" # " #

œ /!Þ!!!%$> œ !Þ!!!%$>

>œ

ln "# ¸ "'"# !Þ!!!%$

The half-life is about "'"# years. 1073

Inverse, Exponential, and Logarithmic Functions Get B-terms on one side. B log #  B log $ œ $ log #  % log $ Factor out B. BÐlog #  log $Ñ œ $ log #  % log $ $ log #  % log $ Bœ log #  log $ ¸ "&Þ*'(

6 Section Exercises 1.

3.

(B œ & Take the logarithm of each side. log (B œ log & Use the power rule for logarithms. B log ( œ log & log & Bœ ¸ !Þ)#( log (

The solution set is e!Þ)#(f.

The solution set is e"&Þ*'(f. 11.

B  #

* œ "$ B  # œ log "$ log * ÐB  #Ñ log * œ log "$ (‡) B log *  # log * œ log "$ B log * œ log "$  # log * B log * œ # log *  log "$ # log *  log "$ Bœ log * ¸ !Þ)$$ (‡) Alternative solution steps:

The solution set is e'Þ!'(f.

ÐB  #Ñ log * œ log "$ B  # œ #

5.

7.

log "$ log *

log "$ log *

The solution set is e!Þ)$$f.

1074

15.

#B  $ œ &B

The solution set is e#Þ#'*f.

#B  $ œ $B  % log #B  $ œ log $B  % ÐB  $Ñ log # œ ÐB  %Ñ log $ B log #  $ log # œ B log $  % log $ Distributive property

/!Þ!"#B œ #$ Take base / logarithms on each side. ln /!Þ!"#B œ ln #$ !Þ!"#B ln / œ ln #$ ln e œ 1 !Þ!"#B œ ln #$ ln #$ Bœ ¸ #'"Þ#*" !Þ!"#

The solution set is e#'"Þ#*"f.

The solution set is e"Þ#!"f.

log #B  $ œ log &B ÐB  $Ñ log # œ B log & B log #  $ log # œ B log & Get B-terms on one side. B log #  B log & œ $ log # B Ðlog #  log &Ñ œ $ log # Factor out x. $ log # Bœ log #  log & ¸ #Þ#'*

9.

13.

œB

$#B œ "% log $#B œ log "% #B log $ œ log "% log "% Bœ ¸ "Þ#!" # log $

%#B  $ œ 'B  " log %#B  $ œ log 'B  " Ð#B  $Ñ log % œ ÐB  "Ñ log ' #B log %  $ log % œ B log '  " log ' Distributive property Get B-terms on one side. #B log %  B log ' œ $ log %  log ' Factor out B. BÐ# log %  log 'Ñ œ $ log %  log ' $ log %  log ' Bœ # log %  log ' ¸ 'Þ!'(

/!Þ#!&B œ * Take base / logarithms on each side. ln /!Þ#!&B œ ln * !Þ#!&B ln / œ ln * ln e œ 1 !Þ#!&B œ ln * ln * Bœ ¸ "!Þ(") !Þ#!&

The solution set is e"!Þ(")f. 17.

ln /$B œ * $B œ * Bœ$

Special property

The solution set is e$f. 19.

ln /!Þ%&B œ È( !Þ%&B œ È( Bœ

È(

!Þ%&

¸ &Þ)(*

The solution set is e&Þ)(*f.

Inverse, Exponential, and Logarithmic Functions 21.

ln /B œ 1 B œ 1 B œ 1 ¸ $Þ"%#

37.

The solution set is e1f, or e$Þ"%#f. 23.

25.

/ln #B œ /ln ÐB  "Ñ #B œ B  " Bœ"

Special property

The solution set is e"f.

Check > œ #: log& )  log& # œ log& The solution set is e#f.

Let's try Exercise 14.

/!Þ!!'B œ $! log /!Þ!!'B œ log $! !Þ!!'B log / œ log $! log $! Bœ ¸ &''Þ)'' !Þ!!' log /

39.

The natural logarithm is a better choice because ln / œ ", whereas log / needs to be calculated. 27.

log$ Ð'B  &Ñ œ # 'B  & œ $# 'B  & œ * 'B œ % B œ %' œ

Exponential form

Check B œ

$$ # :

41.

Exponential form

log( ÐB  "Ñ$ œ # ÐB  "Ñ$ œ (# $ B"œÈ %*

$ B œ "  È %*

log# B  log# ÐB  (Ñ œ $ log# cBÐB  (Ñd œ $ BÐB  (Ñ œ #$ B#  (B œ ) B#  (B  ) œ ! ÐB  )ÑÐB  "Ñ œ ! or or

Exponential form

B"œ! B œ "

Exponential form

Reject B œ ", because it yields an equation in which the logarithm of a negative number must be found.

Cube root

Check B œ ): log# )  log# " œ log# #$  ! œ $

$ Check B œ "  È %*: log( %* œ log( (# œ # $ The solution set is š"  È %*›.

33.

# cannot be a solution because log Ð#  $Ñ œ log Ð"Ñ, and " is not in the domain of log B.

35.

log Ð'B  "Ñ œ log $ 'B  " œ $ 'B œ # B œ #' œ "$

Property 4

Check B œ "$ : log Ð#  "Ñ œ log $ The solution set is ˜ "$ ™.

log %B  log ÐB  $Ñ œ log # %B œ log # log B$ %B œ# B$ %B œ #ÐB  $Ñ %B œ #B  ' #B œ ' B œ $

B)œ! Bœ)

log# $# œ log# #& œ &

™ The solution set is ˜ $$ # .

31.

œ log& %

The solution set is g.

The solution set is ˜ #$ ™. log# Ð#B  "Ñ œ & #B  " œ #& #B  " œ $# #B œ $$ B œ $$ #

) #

Reject B œ $, because %B œ "#, which yields an equation in which the logarithm of a negative number must be found.

# $

Check B œ #$ : log$ * œ log$ $# œ # 29.

log& Ð$>  #Ñ  log& > œ log& % $>  # log& œ log& % > $>  # œ% > $>  # œ %> #œ>

The solution set is e)f. 43.

log &B  log Ð#B  "Ñ œ log % &B log œ log % #B  " &B œ% #B  " &B œ )B  % % œ $B % $ œB

True

#!Î$ & Check B œ %$ : log #! $  log $ œ log &Î$ œ log % The solution set is ˜ %$ ™.

1075

Inverse, Exponential, and Logarithmic Functions 45.

log# B  log# ÐB  'Ñ œ % log# cBÐB  'Ñd œ % BÐB  'Ñ œ #% B#  'B œ "' # B  'B  "' œ ! ÐB  )ÑÐB  #Ñ œ ! B)œ! Bœ)

or or

E œ &!!!ˆ" 

œ &!!!Ð"Þ!(Ñ

Exponential form

(b) If the interest is compounded semiannually, 8 œ #. ‰# • "# E œ &!!!ˆ"  !Þ!( #

B#œ! B œ #

œ &!!!Ð"Þ!$&Ñ#% ¸ "",%"'Þ'%

There will be $"",%"'Þ'% in the account. (c) If the interest is compounded quarterly, 8 œ %. E œ &!!!ˆ" 

Check B œ ): log# )  log# # œ log# "' œ log# #% œ %

(d) If the interest is compounded daily, 8 œ $'&. ‰ E œ &!!!ˆ"  !Þ!( $'& ¸ "",&)!Þ*!

‰ E œ #!!!ˆ"  !Þ!% % #% œ #!!!Ð"Þ!"Ñ ¸ #&$*Þ%( %•'

(e) Use the continuous compound interest formula.

%>

log ˆ $# ‰ ¸ "!Þ# % log Ð"Þ!"Ñ

E œ T / E œ &!!!/!Þ!(Ð"#Ñ œ &!!!/!Þ)% ¸ "",&)"Þ)$ There will be $"",&)"Þ)$ in the account. 53.

(a) Use the formula E œ T / with T œ %!!!, < œ !Þ!$&, and > œ 'Þ E œ %!!!/Ð!Þ!$&ÑÐ'Ñ œ %!!!/!Þ#" ¸ %*$%Þ(" There will be $%*$%Þ(" in the account. (b) If the initial amount doubles, then E œ #T , or $)!!!Þ )!!! œ %!!!/!Þ!$&> # œ /!Þ!$&> ln # œ ln /!Þ!$&> ln # œ !Þ!$&> ln # >œ ¸ "*Þ) !Þ!$&

Divide by 4000.

The initial amount will double in about "*Þ) years.

51.

8> Use E œ T ˆ"  8< ‰ with T œ &!!!, < œ !Þ!(, and > œ "#.

(a) If the interest is compounded annually, 8 œ ". 1076

In the continuous compound interest formula, let E œ ")&!, < œ !Þ!'&, and > œ %!. E œ T / ")&! œ T /!Þ!'&Ð%!Ñ ")&! œ T /#Þ' ")&! T œ #Þ' ¸ "$(Þ%" / Deposit $"$(Þ%" today.

It will take about "!Þ# years for the account to grow to $$!!!. 49.

$'& • "#

There will be $"",&)!Þ*! in the account.

The account will contain $#&$*Þ%(.

>œ

¸ "",%*(Þ**

There will be $"",%*(Þ** in the account.

8> (a) Use the formula E œ T ˆ"  8< ‰ with T œ #!!!, < œ !Þ!%, 8 œ %, and > œ '.

‰ $!!! œ #!!!ˆ"  !Þ!% % $!!! œ Ð"Þ!"Ñ%> #!!! log ˆ $# ‰ œ log Ð"Þ!"Ñ%> log ˆ $# ‰ œ %> log Ð"Þ!"Ñ

!Þ!( ‰% • "# % %)

œ &!!!Ð"Þ!"(&Ñ

The solution set is e)f.

(b)

¸ "",#'!Þ*'

There will be $"",#'!Þ*' in the account.

Reject B œ #, because it yields an equation in which the logarithm of a negative number must be found.

47.

!Þ!( ‰" • "# " "#

55.

0 ÐBÑ œ "&Þ*%/!Þ!'&'B (a) 1980 corresponds to B œ !. 0 Ð!Ñ œ "&Þ*%/!Þ!'&'Ð!Ñ œ "&Þ*%Ð"Ñ œ "&Þ*% The approximate volume of materials recovered from municipal solid waste collections in the United States in 1980 was "&Þ* million tons. (b) 1990 corresponds to B œ "!. 0 Ð"!Ñ ¸ $!Þ( million tons (c) 2000 corresponds to B œ #!. 0 Ð#!Ñ ¸ &*Þ# million tons (d) 2007 corresponds to B œ #(. 0 Ð#(Ñ ¸ *$Þ( million tons

Inverse, Exponential, and Logarithmic Functions 57.

59.

WÐBÑ œ ""#,!%(/!Þ!)#(B 2007 corresponds to B œ $. WÐ$Ñ œ ""#,!%(/!Þ!)#(Ð$Ñ ¸ "%$,&*) The approximate value of software-publisher revenues for 2007 was "%$,&*) million dollars. EÐ>Ñ œ #Þ!!/!Þ!&$>

"$! !Þ!'&'B œ ln "&Þ*% "$! ln "&Þ*% ¸ $# !Þ!'&' Since B œ ! corresponds to 1980, B œ $# corresponds to 2012.

Bœ

65.

The display means that after #&! years, approximately #Þ*") (%) ( grams of the original sample of radium 226 remain.

67.

0 ÐBÑ œ #B#

(a) Let > œ %. EÐ%Ñ œ #Þ!!/!Þ!&$Ð%Ñ œ #Þ!!/!Þ#"# ¸ "Þ'#

The graph of 0 is narrower than the graph of C œ B# . B C

About "Þ'# grams would be present.

! !

„" #

„# )

(b) EÐ"!Ñ œ #Þ!!/!Þ!&$Ð"!Ñ ¸ "Þ") About "Þ") grams would be present. (c) EÐ#!Ñ œ #Þ!!/!Þ!&$Ð#!Ñ ¸ !Þ'* About !Þ'* grams would be present.

69.

(d) The initial amount is the amount EÐ>Ñ present at time > œ !.

0 ÐBÑ œ ÐB  "Ñ# The graph of 0 is the graph of C œ B# shifted left " unit.

EÐ!Ñ œ #Þ!!/!Þ!&$Ð!Ñ œ #Þ!!/! œ #Þ!!Ð"Ñ œ #Þ!!

B C

$ %

# "

" 0

! "

" %

Initially, #Þ!! grams were present. 61.

(a) Find EÐ>Ñ œ %!!/!Þ!$#> when > œ #&. EÐ#&Ñ œ %!!/!Þ!$#Ð#&Ñ œ %!!/!Þ) ¸ "(*Þ($ About "(*Þ($ grams of lead will be left. (b) Use EÐ>Ñ œ %!!/!Þ!$#> , with EÐ>Ñ œ "# Ð%!!Ñ œ #!!. #!! œ %!!/!Þ!$#> #!! !Þ!$#> %!! œ /

Review Exercises 1.

Since a horizontal line intersects the graph in two points, the function is not one-to-one.

2.

Since every horizontal line intersects the graph in no more than one point, the function is one-to-one.

3.

The function 0 ÐBÑ œ $B  ( is a linear function. By the horizontal line test, it is a one-to-one function. To find the inverse, replace 0 ÐBÑ with C.

!Þ!$#>

!Þ& œ / ln !Þ& œ ln /!Þ!$#> ln !Þ& œ !Þ!$#>Ðln /Ñ ln !Þ& >œ ¸ #"Þ'' !Þ!$# It would take about #"Þ'' years for the sample to decay to half its original amount. 63.

0 ÐBÑ œ "&Þ*%/!Þ!'&'B "$! œ "&Þ*%/!Þ!'&'B !Þ!'&'B "$! / œ "&Þ *% "$! ln /!Þ!'&'B œ ln "&Þ *% "$! !Þ!'&'B Ðln /Ñ œ ln "&Þ*%

C œ $B  ( Interchange B and C. B œ $C  ( Solve for C. B  ( œ $C B( (B œ C, or œC $ $ Replace C with 0 " B . B( " ( 0 " B œ , or 0 " B œ  B  $ $ $ 1077

Inverse, Exponential, and Logarithmic Functions 4.

$ 0 ÐBÑ œ È 'B  % The function is one-to-one since each 0 ÐBÑ-value corresponds to exactly one B-value. To find the inverse, replace 0 ÐBÑ with C.

Plot the points from the table and draw a smooth curve through them.

$ CœÈ 'B  % Interchange B and C. $ BœÈ 'C  %

Solve for C. B$ œ 'C  % $ B  % œ 'C B$  % œC ' Replace C with 0 " B B$  % œ 0 " B '

Cube each side.

B 0 ÐBÑ œ ˆ "$ ‰ Make a table of values.

B 0 ÐBÑ

# *

" $

0 ÐBÑ œ B#  $ This is an equation of a vertical parabola which opens down. Since a horizontal line will intersect the graph in two points, the function is not one-to-one.

6.

This function is not one-to-one because two sodas in the list have %" mg of caffeine.

7.

The graph is a linear function through !ß " and $ß ! . The graph of 0 " B will include the points Ð"ß !Ñ and Ð!ß $Ñ, found by interchanging B and C. Plot these points, and draw a straight line through them.

The graph is a curve through "ß # , !ß " , and ˆ"ß "# ‰. Interchange B and C to get Ð#ß "Ñ, Ð"ß !Ñ, and ˆ "# ß "‰, which are on the graph of 0 " B . Plot these points, and draw a smooth curve through them.

11.

" *

#

 $#

"

!

" #

C

" #

"

#

)

"'

Plot the points from the table and draw a smooth curve through them.

12.

13.

1078

#

" $

B

&#B  " œ #& Write each side as a power of &. & #B  " œ & # #B  " œ # Equate exponents. #B œ " B œ "# The solution set is ˜ "# ™.

True

%$B œ )B  % Write each side as a power of #. ˆ## ‰ $B3 œ ˆ#$ ‰ B  % 3

0 ÐBÑ œ $B Make a table of values. B 0 ÐBÑ

"

C œ ##B  $ Make a table of values.

Check B œ "# : &# œ #&

9.

! "

Plot the points from the table and draw a smooth curve through them.

5.

8.

10.

'B

#

"

" *

" $

! "

" $

# *

$B  "#

# œ# 'B œ $B  "# $B œ "# Bœ%

Equate exponents.

Inverse, Exponential, and Logarithmic Functions Check B œ %: %"# œ )) The solution set is e%f.

14.

17.

True

" ‰B" ˆ #( œ *#B

’ˆ "$ ‰ “

$ B"

C œ log"Î$ B ˆ "$ ‰C œ B

œ ˆ$# ‰ #B3

C Make a table of values. Since B œ ˆ "$ ‰ is the B inverse of 0 ÐBÑ œ C œ ˆ "$ ‰ in Exercise 10, simply reverse the ordered pairs in the table of B values belonging to 0 ÐBÑ œ ˆ "$ ‰ .

Write each side as a power of $. ˆ$$ ‰ B"3 œ ˆ$# ‰ #B3

$$B$ œ $%B $B  $ œ %B Equate exponents. $ œ (B $ ( œB " ‰%Î( Check B œ $( : ˆ #( œ *'Î( True

B C

The solution set is ˜ $( ™.

15.

1ÐBÑ œ log"Î$ B Replace 1ÐBÑ with C, and write in exponential form.

* #

$ "

" !

" $

" *

"

#

Plot the points from the table and draw a smooth curve through them.

WÐBÑ œ $$Þ!(Ð"Þ!#%"ÑB (a) B œ "*(&  "*(! œ & W Ð&Ñ œ $$Þ!(Ð"Þ!#%"Ñ& ¸ #*Þ% The approximate amount of sulfur dioxide emissions in the United States in 1975 was #*Þ% million tons.

18.

(b) B œ "**&  "*(! œ #& WÐ#&Ñ ¸ ")Þ# million tons (c) B œ #!!&  "*(! œ $&

The solution set is e#f.

WÐ$&Ñ ¸ "%Þ% million tons 16.

1ÐBÑ œ log$ B Replace 1ÐBÑ with C, and write in exponential form.

19.

Make a table of values. Since B œ $ is the inverse of 0 ÐBÑ œ C œ $B in Exercise 9, simply reverse the ordered pairs in the table of values belonging to 0 ÐBÑ œ $B . " $

#

"

" !

B

"Î#

B

$Î#

Exponential form

3

C

" *

log# È) œ B #B œ È )

# œ) Write each side as a power of #. #B œ ˆ#$ ‰ "Î#

C œ log$ B $C œ B

B C

log) '% œ B )B œ '% Exponential form Write each side as a power of ). )B œ )# Bœ# Equate exponents.

$ "

* #

Plot the points from the table and draw a smooth curve through them.

# œ# B œ $#

Equate exponents.

The solution set is ˜ $# ™. 20.

logB Œ

"  œ # %* " B# œ %* " " œ B# %* B# œ %* Bœ „(

Exponential form

Since B is the base, we cannot have a negative number. The solution set is e(f. 1079

Inverse, Exponential, and Logarithmic Functions 21.

log% B œ

$ #

B œ %$Î# Exponential form B œ ˆÈ % ‰ $ œ # $ œ )

28.

3

22.

log5 % œ " 5" œ % 5œ%

log, , # œ # ,# œ ,#

Quotient rule

#

Product rule Power rule

29.

log, $  log, B  # log, C Use the product and power rules for logarithms. œ log, Ð$ • BÑ  log, C# $B œ log, # Quotient rule C

30.

log$ ÐB  (Ñ  log$ Ð%B  'Ñ B( œ log$ Œ  %B  '

The solution set is e%f.

23.

D œ log% ˆÈB • A# ‰  log% D

œ log% B  log% A  log% D œ "# log% B  # log% A  log% D

Exponential form

5 œ % is an acceptable base since it is a positive number (not equal to ").

È B • A#

"Î#

The argument (the input of the logarithm) must be a positive number, so B œ ) is acceptable. The solution set is e)f.

log%

Quotient rule

31.

log #)Þ* ¸ "Þ%'!*

32.

log !Þ#&( ¸ !Þ&*!"

33.

ln #)Þ* ¸ $Þ$'$)

34.

ln !Þ#&( ¸ "Þ$&)(

24.

The solution set is e, l ,  !, , Á "f.

log, + is the exponent to which , must be raised to obtain +.

35.

log"' "$ œ

25.

From Exercise 24,

36.

log% "# œ

37.

Milk has a hydronium ion concentration of %Þ! ‚ "!( .

Exponential form

This is an identity. Thus, , can be any real number, ,  ! and , Á ".

, log, + œ +. 26.

WÐBÑ œ "!! log# ÐB  #Ñ (a) When B œ ', WÐ'Ñ œ "!! log# Ð'  #Ñ œ "!!Ð$Ñ œ $!!. After ' weeks the sales were $!! thousand dollars or $$!!,!!!. (b) To graph the function, make a table of values that includes the ordered pair from above. B WÐBÑ

! "!!

# #!!

' $!!

log "$ ¸ !Þ*#&" log "' log "# ¸ "Þ(*#& log %

pH œ log cH$ O d pH œ log ˆ%Þ! ‚ "!( ‰ ¸ 'Þ%

38.

Crackers have a hydronium ion concentration of $Þ) ‚ "!* . pH œ log cH$ O d pH œ log ˆ$Þ) ‚ "!* ‰ ¸ )Þ%

39.

Plot the ordered pairs and draw the graph through them. 40.

Orange juice has a pH of %Þ'. pH œ log cH$ O d %Þ' œ log cH$ O d log"! cH$ O d œ %Þ' cH$ O d œ "!%Þ' ¸ #Þ& ‚ "!&

Since we want to determine a ratio of intensities, we'll solve the given formula for the intensity M . Q œ '  #Þ& log #Þ& log log"!

27.

1080

log# $BC# œ log# $  log# B  log# C# œ log# $  log# B  # log# C

Product rule Power rule

M œ'Q M! M 'Q œ M! #Þ&

M œ "!Ð'  Q ÑÎ#Þ& M! M œ "!!Þ%Ð'  Q Ñ M!

M M!

Common log Exponential form 1/2.5 = 0.4

Inverse, Exponential, and Logarithmic Functions We now have M as a function of Q . The ratio of intensities between stars of magnitude " and $ is

44.

MÐ"Ñ "!!Þ%Ð'  "Ñ M! "!# œ !Þ%Ð'  $Ñ œ "Þ# œ "!!Þ) ¸ 'Þ$. MÐ$Ñ "! "! M! Thus, a star of magnitude " is about 'Þ$ times as intense as a star of magnitude $. 41.

>ÐÐ!Þ!'Ñ œ

ln # ¸ "# ln Ð"  !Þ!'Ñ 46.

ln # ¸' ln Ð"  !Þ"#Ñ

42.

$B œ *Þ%# log $B œ log *Þ%# B log $ œ log *Þ%# log *Þ%# Bœ ¸ #Þ!%# log $ Check B œ #Þ!%#: $#Þ!%# ¸ *Þ%#& The solution set is e#Þ!%#f.

43.

#B  " œ "& log #B  " œ log "& ÐB  "Ñ log # œ log "& log "& B"œ log # log "& Bœ  " ¸ %Þ*!( log # Check B œ %Þ*!(: #$Þ*!( ¸ "&Þ! The solution set is e%Þ*!(f.

Exponential form

log& ÐB  'Ñ$ œ # Change to exponential form. ÐB  'Ñ$ œ &# ÐB  'Ñ$ œ #& Take the cube root of each side. $ B'œÈ #& $ BœÈ #&  '

$ Check B œ '  È #&: log& #& œ log& &# œ #

At "#%, it would take about ' years. (e) Each comparison shows approximately the same number. For example, in part (a) the doubling time is ") yr (rounded) and (# % œ "). (# (called the rule of 72) Thus, the formula > œ "!!< is an excellent approximation of the doubling time formula.

log$ Ð*B  )Ñ œ # *B  ) œ $# *B  ) œ * *B œ " B œ "*

The solution set is ˜ "* ™.

At "!%, it would take about ( years. (d) "#% œ !Þ"#; >Ð!Þ"#Ñ œ

The solution set is e")Þ$"!f.

Check B œ "* : log$ * œ log$ $# œ #

At '%, it would take about "# years. ln # ¸( (c) "!% œ !Þ"!; >Ð!Þ"!Ñ œ ln Ð"  !Þ"!Ñ

/!Þ!'B œ $ Take base / logarithms on each side. ln /!Þ!'B œ ln $ !Þ!'B ln / œ ln $ !Þ!'B œ ln $ ln e œ 1 ln $ Bœ ¸ ")Þ$"! !Þ!' Check B œ ")Þ$"!: /"Þ!*)' ¸ $Þ!

$ The solution set is š'  È #&›.

47.

log$ ÐB  #Ñ  log$ B œ log$ # B# log$ œ log$ # B B# œ# B B  # œ #B #œB

Quotient rule Property 4

Check B œ #: log$ %  log$ # œ log$ 48.

The solution set is e#f.

% #

œ log$ #

log Ð#B  $Ñ œ "  log B log Ð#B  $Ñ  log B œ " #B  $ log"! œ" Quotient rule B #B  $ Exponential "!" œ form B "!B œ #B  $ )B œ $ B œ $)

Check B œ $) : LS œ log ˆ $%  $‰ œ log "& %

"& RS œ log "!  log $) œ log $! ) œ log %

The solution set is ˜ $) ™.

1081

Inverse, Exponential, and Logarithmic Functions 49.

log% B  log% Ð)  BÑ œ # log% cBÐ)  BÑd œ # BÐ)  BÑ œ %# )B  B# œ "' # B  )B  "' œ ! ÐB  %ÑÐB  %Ñ œ ! B%œ! Bœ%

Product rule Exponential form

54.

E œ "!!!ˆ" 

E œ "!!!ˆ" 

55.

log# B  log# ÐB  "&Ñ œ log# "'

log# cBÐB  "&Ñd œ log# "'

or or

Product rule Property 4

Check B œ ": log# "  log# "' œ !  log# "' œ log# "'

The solution set is e"f.

When the power rule was applied in the second step, the domain was changed from eB l B Á !f to eB l B  !f. Instead of using the power rule for logarithms, we can change the original equation to the exponential form B# œ "!# and get B œ „ "!. As you can see in the erroneous solution, the valid solution "! was "lost." The solution set is e „ "!f. E œ T Š" 

< 8> ‹ 8 Let T œ #!,!!!, < œ !Þ!%, and > œ &. For 8 œ % (quarterly compounding), E œ #!,!!!Œ" 

!Þ!%  %

%•&

¸ #%,%!$Þ)!.

There will be $#%,%!$Þ)! in the account after & years. 53.

In the continuous compounding formula, let T œ "!,!!!, < œ !Þ!$(&, and > œ $. E œ T / E œ "!,!!!/!Þ!$(&Ð$Ñ ¸ "","*!Þ(# There will be $"","*!Þ(# in the account after $ years.

1082

!Þ!$* ‰"# • $ "#

¸ ""#$Þ*"

Let UÐ>Ñ œ "# E! to find the half-life of the radioactive substance. UÐ>Ñ œ E! /!Þ!&> " !Þ!&> # E! œ E! / !Þ& œ /!Þ!&> ln !Þ& œ ln /!Þ!&> ln !Þ& œ !Þ!&> ln / !Þ!&> œ ln !Þ& ln !Þ& >œ ¸ "$Þ* !Þ!&

B"œ! Bœ"

Reject B œ "', because it yields an equation in which the logarithm of a negative number must be found.

52.

¸ ""#'Þ)$

Plan A is the better plan by $#Þ*#.

B#  "&B œ "' B#  "&B  "' œ ! ÐB  "'ÑÐB  "Ñ œ !

51.

!Þ!% ‰% • $ %

Plan B: Let T œ "!!!, < œ !Þ!$*, 8 œ "#, and > œ $.

The solution set is e%f.

B  "' œ ! B œ "'

< 8> ‹ . 8

Plan A: Let T œ "!!!, < œ !Þ!%, 8 œ %, and > œ $.

Check B œ %: log% %  log% % œ "  " œ #

50.

Use E œ T Š" 

The half-life is about "$Þ* days. 56.

W œ GÐ" 