Beginning Algebra (13th Edition) [13 ed.] 013499499X, 9780134994994

Table of contents :
Beginning Algebra 13th Edition [Margaret L. Lial]
Title Page
Copyright
Contents
PREFACE
PHOTO CREDITS
STUDY SKILLS
Chapter 0 PREALGEBRA REVIEW
Chapter 1 THE REAL NUMBER SYSTEM
Chapter 2 LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
Chapter 3 LINEAR EQUATIONS AND INEQUALITIES IN TWO VARIABLES; FUNCTIONS
Chapter 4 SYSTEMS OF LINEAR EQUATIONS AND INEQUALITIES
Chapter 5 EXPONENTS AND POLYNOMIALS
Chapter 6 FACTORING AND APPLICATIONS
Chapter 7 RATIONAL EXPRESSIONS AND APPLICATIONS
Chapter 8 ROOTS AND RADICALS
Chapter 9 QUADRATIC EQUATIONS
ANSWERS TO SELECTED EXERCISES
Index

Citation preview

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EDITION

13

Beginning Algebra Margaret L. Lial American River College

John Hornsby University of New Orleans

Terry McGinnis

@ Pearson

Vice President, Courseware Portfolio Management: Chris Hoag Director, Courseware Portfolio Management: Michael Hirsch Courseware Portfolio Manager: Karen Montgomery Courseware Portfolio Assistant: Kayla Shearns Managing Producer: Scott Disanno Content Producer: Lauren Morse Producers: Vicki Dreyfus and Stacey Miller Associate Content Producer, TestGen: Rajinder Singh Content Managers, MathXL: Eric Gregg and Dominick Franck Manager, Courseware QA: Mary Durnwald Senior Product Marketing Manager: Alicia Frankel Product Marketing Assistant: Brooke lmbornone Senior Author Supportffechnology Specialist: Joe Vetere Full Service Vendor, Cover Design, Composition: Pearson CSC Full Service Project Management: Pearson CSC (Carol Merrigan) Cover Image: Borchee/E+/Getty Images Copyright © 2020, 201 6, 2012 by Pearson Education, Inc. 221 River Street, Hoboken, NJ 07030. All Rights Reserved. Printed in the United States of America. This publication is protected by copyright, and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise. For information regarding permissions, request forms, and the appropriate contacts within the Pearson Education Global Rights & Permissions department, please visit www.pearsoned.com/permissions/. Acknowledgments of third-party content appear on page xiv, which constitutes an extension of this copyright page. PEARSON, ALWAYS LEARNING, MyLabTMMath, MathXL, and TestGen are exclusive trademarks in the U.S. and/or other countries owned by Pearson Education, Inc. or its affiliates. Unless otherwise indicated herein, any third-party trademarks that may appear in this work are the property of their respective owners and any references to third-party trademarks, logos or other trade dress are for demonstrative or descriptive purposes only. Such references are not intended to imply any sponsorship, endorsement, authorization, or promotion of Pearson's products by the owners of such marks, or any relationship between the owner and Pearson Education, Inc. or its affiliates, authors, licensees or distributors.

Library of Congress Cataloging-in-Publication Data Names: Lia!, Margaret L. , author. I Hornsby, John, 1949- author. I McGinnis, Terry, author. Title: Beginning algebra I Margaret L. Lia! (American River College), John Hornsby (University of New Orleans), Terry McGinnis. Description: 13th edition. I Boston : Pearson, [2020] I Includes index. Identifiers: LCCN 201 80422251 ISBN 97801 34994994 (student edition) I ISBN 01 3499499X (student edition) Subjects: LCSH: Algebra. I Algebra--Textbooks. Classification: LCC QA152.3 .LS 2020 I DDC 512.9--dc23 LC record available at https://lccn.loc.gov/201 8042225

@ Pearson

ISBN 13: 978-0-134-99499-4 ISBN 10: 0-134-99499-X

CONTENTS Preface

vii

Photo Credits

xiv

Study Skills

S-1 S-1

STUDY SKILL 1

Using Your Math Text

STUDY SKILL 2

Reading Your Math Text

S-2

S-3

STUDY SKILL 3

Taking Lecture Notes

STUDY SKILL 4

Completing Your Homework

STUDY SKILL 5

Using Study Cards

STUDY SKILL 6

Managing Your Time

S-6

STUDY SKILL 7

Reviewing a Chapter

S-7

STUDY SKILL 8 Taking Math Tests

S-4

STUDY SKILL 9

S-5

Prealgebra Review

R.1 Fractions

1

R.2 Decimals and Percents

The Real Number System

27 1.7 Simplifying Expressions

1.2 Variables, Expressions, and Equations 1.3 Real Numbers and the Number Line

16

1.6 Properties of Real Numbers

1.1 Exponents, Order of Operations, and Inequality 28 36 42

1.4 Adding and Subtracting Real Numbers

51

1.5 Multiplying and Dividing Real Numbers

65

SUMMARY EXERCISES Performing Operations with Real

2

S-10

1

1

Numbers

77

Chapter 1 Summary 94 Chapter 1 Review Exercises 97 Chapter 1 Mixed Review Exercises 100 Chapter 1 Test 100 Chapters R and 1 Cumulative Review Exercises 1 02

112

2.3 Solving Linear Equations Using Both Properties of Equality 117 2.4 Clearing Fractions and Decimals When Solving Linear Equations 125 SUMMARY EXERCISES Applying Methods for Solving

Linear Equations 2.5 Applications of Linear Equations

131 132

2.6 Formulas and Additional Applications from Geometry 146

103

2.7 Ratio, Proportion, and Percent

104

2.2 The Multiplication Property of Equality

78

88

Linear Equations and Inequalities in One Variable

2.1 The Addition Property of Equality

S-9

STUDY SKILL 10 Preparing for Your Math Final

Exam

R

S-8

Analyzing Your Test Results

157

2.8 Further Applications of Linear Equations 169 2.9 Solving Linear Inequalities

182

Chapter 2 Summary 196 Chapter 2 Review Exercises 200 Chapter 2 Mixed Review Exercises Chapter 2 Test 204 Chapters R-2 Cumulative Review Exercises 205

203

iii

iv

Contents

3

Linear Equations and Inequalities in Two Variables; Functions 207

3.1 Linear Equations and Rectangular Coordinates 208

3.6 Graphing Linear Inequalities in Two Variables 262

3.2 Graphing Linear Equations in Two Variables 219

3.7 Introduction to Functions

3.3 The Slope of a Line

231

3.4 Slope-Intercept Form of a Linear Equation

268

Chapter 3 Summary 277 Chapter 3 Review Exercises 281 Chapter 3 Mixed Review Exercises Chapter 3 Test 284 Chapters R-3 Cumulative Review Exercises 285

245

3.5 Point-Slope Form of a Linear Equation and Modeling 253 SUMMARY EXERCISES Applying Graphing and

283

Equation-Writing Techniques for Lines 261

4

Systems of Linear Equations and Inequalities

4.4 Applications of Linear Systems

4.1 Solving Systems of Linear Equations by Graphing 288

Chapter 4 Summary 331 Chapter 4 Review Exercises 334 Chapter 4 Mixed Review Exercises Chapter 4 Test 337 Chapters R-4 Cumulative Review Exercises 338

4.3 Solving Systems of Linear Equations by Elimination 305 SUMMARY EXERCISES Applying Techniques for Solving

5

312

Exponents and Polynomials

SUMMARY EXERCISES Applying the Rules for

Exponents 5.3 Scientific Notation

360

361

5.4 Adding, Subtracting, and Graphing Polynomials 369 5.5 Multiplying Polynomials

380

336

387

5.7 Dividing Polynomials 350

326

341 5.6 Special Products

5.1 The Product Rule and Power Rules for Exponents 342 5.2 Integer Exponents and the Quotient Rule

314

4.5 Solving Systems of Linear Inequalities

4.2 Solving Systems of Linear Equations by Substitution 297

Systems of Linear Equations

287

393

Chapter 5 Summary 402 Chapter 5 Review Exercises 405 Chapter 5 Mixed Review Exercises Chapter 5 Test 408 Chapters R-5 Cumulative Review Exercises 410

408

Contents

6

Factoring and Applications

6.1 Greatest Common Factors; Factoring by Grouping 414 6.2 Factoring Trinomials

423

6.3 More on Factoring Trinomials

430

6.4 Special Factoring Techniques

439

SUMMARY EXERCISES Recognizing and Applying

Factoring Strategies

449

413 6.6 Applications of Quadratic Equations Chapter 6 Summary 472 Chapter 6 Review Exercises 475 Chapter 6 Mixed Review Exercises Chapter 6 Test 478 Chapters R-6 Cumulative Review Exercises 479

460

477

6.5 Solving Quadratic Equations Using the Zero-Factor Property 452

7

Rational Expressions and Applications

7.7 Applications of Rational Expressions

7.1 The Fundamental Property of Rational Expressions 482

7.8 Variation

7.2 Multiplying and Dividing Rational Expressions 492 7.3 Least Common Denominators

499

514

537

547

Chapter 7 Summary 555 Chapter 7 Review Exercises 560 Chapter 7 Mixed Review Exercises Chapter 7 Test 563 Chapters R-7 Cumulative Review Exercises 565

7.4 Adding and Subtracting Rational Expressions 506 7.5 Complex Fractions

481

562

7.6 Solving Equations with Rational Expressions 523 SUMMARY EXERCISES Simplifying Rational Expressions

vs. Solving Rational Equations 535

8

Roots and Radicals

8.1 Evaluating Roots

567

568

8.6 Solving Equations with Radicals

8.2 Multiplying, Dividing, and Simplifying Radicals 579 8.3 Adding and Subtracting Radicals 8.4 Rationalizing the Denominator

587

592

8.5 More Simplifying and Operations with Radicals 599 SUMMARY EXERCISES Applying Operations with

Radicals

606

607

Chapter 8 Summary 616 Chapter 8 Review Exercises 619 Chapter 8 Mixed Review Exercises Chapter 8 Test 622 Chapters R-8 Cumulative Review Exercises 623

621

v

vi

Contents

9

Quadratic Equations

625

9.1 Solving Quadratic Equations by the Square Root Property 626

9.4 Graphing Quadratic Equations

9.2 Solving Quadratic Equations by Completing the Square 632 9.3 Solving Quadratic Equations by the Quadratic Formula 641 SUMMARY EXERCISES Applying Methods for Solving

Quadratic Equations

Answers to Selected Exercises Index

648

A-1

1-1

Additional topics avai lable in Mylab™ Math : Using Rational Numbers as Exponents; Complex Numbers; Sets; Introduction to Calculators; and Glossary

649

Chapter 9 Summary 656 Chapter 9 Review Exercises 659 Chapter 9 Mixed Review Exercises Chapter 9 Test 660 Chapters R-9 Cumulative Review Exercises 661

660

PREFACE WELCOME TO THE 13TH EDITION The first edition of Marge Lial's Beginning Algebra was published in 1969, and now we are pleased to present the 13th edition-with the same successful, well-rounded framework that was established 50 years ago and updated to meet the needs of today's students and professors. The names Lia! and Miller, two faculty members from American River College in Sacramento, California, have become synonymous with excellence in Developmental Mathematics, Precalculus, Finite Mathematics, and Applications-Based Calculus. With Chuck Miller's passing in 1986, Marge Lia! was joined by a team of carefully selected coauthors who partnered with her. John Hornsby (University of New Orleans) joined Marge in this capacity in 1992, and in 1999, Terry McGinnis became part of this developmental author team. Since Marge's passing in 20 12, John and Terry have dedicated themselves to carrying on the Lial/Miller legacy. In the preface to the first edition for Intermediate Algebra, Marge Lia! wrote " ... the strongest theme ... is a combination of readability and suitability for the book's intended audience: students who are not completely selfconfident in mathematics as they come to the course, but who must be self-confident and proficient . .. by the end of the course." Today's Lia! author team upholds these same standards. With the publication of the 13th edition of Beginning Algebra, we proudly present a complete course program for students who need developmental algebra. Revisions to the core text, working in concert with such innovations in the MyLab Math course as Skill Builder and Learning Catalytics, combine to provide superior learning opportunities appropriate for all types of courses (traditional, hybrid, online). We hope you enjoy using it as much as we have enjoyed writing it. We welcome any feedback that you have as you review and use this text.

WHAT'S NEW IN THIS EDITION? We are pleased to offer the following new features and resources in the text and MyLab.

IMPROVED STUDY SKILLS These special activities are now grouped together at the front of the text, prior to Chapter R. Study Skills Reminders that refer students to specific Study Skills are found liberally throughout the text. Many Study Skills

now include a Now Try This section to help students implement the specific skill.

REVISED EXPOSITION With each edition of the text, we continue to polish and improve discussions and presentations of topics to increase readability and student understanding. This edition is no exception. NEW FIGURES AND DIAGRAMS For visual learners, we have included more than 50 new mathematical figures, graphs, and diagrams, including several new "hand drawn" style graphs. These are meant to suggest what a student who is graphing with paper and pencil should obtain. We use this style when introducing a particular type of graph for the first time. ENHANCED USE OF PEDAGOGICAL COLOR We have thoroughly reviewed the use of pedagogical color in discussions and examples and have increased its use whenever doing so would enhance concept development, emphasize important steps, or highlight key procedures. INCREASED Concept Check AND WHAT WENT WRONG? EXERCISES The number of Concept Check, exercises, which facilitate students' mathematical thinking and conceptual understanding, and which begin each exercise set, has been increased. We have also more than doubled the number of WHAT WENT WRONG? exercises that highlight common student errors. INCREASED RELATING CONCEPTS EXERCISES We have doubled the number of these flexible groups of exercises, which are located at the end of many exercise sets. These sets of problems were specifically written to help students tie concepts together, compare and contrast ideas, identify and describe patterns, and extend concepts to new situations. They may be used by individual students or by pairs or small groups working collaboratively. All answers to these exercises appear in the student answer section. ENHANCED MYLAB MATH RESOURCES MyLab exercise coverage in the revision has been expanded, and video coverage has also been expanded and updated to a modern format for today's students. WHAT WENT WRONG? problems and all RELATING CONCEPTS exercise sets (both even- and oddnumbered problems) are now assignable in MyLab Math. SKILL BUILDER These exercises offer just-in-time additional adaptive practice in MyLab Math. The adaptive engine tracks student performance and delivers, to each individual, questions that adapt to his or her level of understanding. This new feature enables instructors to assign fewer questions for

vii

viii

Preface

homework, allowing students to complete as many or as few questions as they need.

LEARNING CATALYTIC$ This new student response tool uses students' own devices to engage them in the learning process. Problems that draw on prerequisite skills are included at the beginning of each section to gauge student readiness for the section. Accessible through MyLab Math and customizable to instructors' specific needs, these problems can be used to generate class discussion, promote peer-to-peer learning, and provide real-time feedback to instructors. More information can be found via the Learning Catalytics Link in MyLab Math. Specific exercises notated in the text can be found by searching LiaLBeginning# where the# is the chapter number.

Order of operations involving absolute value expressions (Section 1.5) Solving linear equations in one variable (Sections 2.1 and 2.2) Solving problems involving proportions and percent (Section 2. 7) Writing an equation of a line from a graph (Section 3.4) Solving systems of equations using the elimination method (Section 4.3) Adding, subtracting, and dividing polynomials (Sections 5.4 and 5.7) Finding reciprocals of rational expressions (Section 7 .2) Solving direct variation problems (Section 7 .8) Solving quadratic equations with no real solution (Sections 9. 1-9.3) Graphing quadratic equations (Section 9.4)

CONTENT CHANGES Specific content changes include the following: • Exercise sets have been scrutinized and updated with a renewed foc us on conceptual understanding and skill development. Even and odd pairing of the exercises, an important feature of the text, has been carefully reviewed. • Real-world data in all examples and exercises and in their accompanying graphs has been updated. • An increased emphasis on fractions, decimals, and percents appears throughout the text. We have expanded Chapter R to include new figures and revised explanations and examples on converting among fractions, decimals, and percents. And we have included an all-new set of Cumulative Review Exercises, many of which focus on fractions, decimals, and percents, at the end of Chapter 1. Sets of Cumulative Review Exercises in subsequent chapters now begin with new exercises that review skills related to these topics. • A new Section 2.4 provides expanded coverage of linear equations in one variable with fractional and decimal coefficients. Two new examples have been included, and the number of exercises has been doubled. • Solution sets of linear inequalities in Section 2.9 are now graphed first, before they are written using interval notation. • Expanded Mid-Chapter Summary Exercises in Chapter 2 continue our emphasis on the difference between simplifying an expression and solving an equation. New examples in the Summary Exercises in Chapters 4, 6, and 9 illustrate and distinguish between solution methods. • P resentations of the following topics have been enha nced and expanded, often including new examples and exercises.

UAL DEVELOPMENTAL HALLMARK FEATURES We have enhanced the following popular features, each of which is designed to increase ease of use by students and/ or instructors.

• Emphasis on Problem-Solving We introduce our sixstep problem-solving method in Chapter 2 and integrate it throughout the text. The six steps, Read, Assign a Variable, Write an Equation, Solve, State the Answer, and Check, are emphasized in boldface type and repeated in examples and exercises to reinforce the problem-solving process for students. We also provide students with boxes that feature helpful problem-solving tips and strategies. • Helpful Learning Objectives We begin each section with clearly stated, numbered objectives, and the included material is directly keyed to these objectives so that students and instructors know exactly what is covered in each section. • Cautions and Notes One of the most popular features of previous editions is our inclusion of information marked CAUTION and miIJ) to warn students about common errors and to emphasize important ideas throughout the exposition. The updated text design makes them easy to spot.

0

• Comprehensive Examples The new edition features a multitude of step-by-step, worked-out examples that include pedagogical color, helpful side comments, and special pointers. We give special attention to checking example solutions-more checks, designated using a special CHECK tag and ./ , are included than in past editions. • More Pointers There are more pointers in examples and discussions throughout this edition of the text. They provide students with important on-the-spot reminders, as well as warnings about common pitfalls.

Preface

• Numerous Now Try Problems These margin exercises, with answers immediately available at the bottom of the page, have been carefully written to correspond to every example in the text. This key feature allows students to immediately practice the material in preparation for the exercise sets. • Updated Figures, Photos, and Hand-Drawn Graphs Today's students are more visually oriented than ever. As a result, we provide detailed mathematical figures, diagrams, tables, and graphs, including a "hand-drawn" style of graphs, whenever possible. We have incorporated depictions of well-known mathematicians, as well as appealing photos to accompany applications in examples and exercises. • Relevant Real-Life Applications We include many new or updated applications from fields such as business, pop culture, sports, technology, and the health sciences that show the relevance of algebra to daily life. • Extensive and Varied Exercise Sets The text contains a wealth of exercises to provide students with opportunities to practice, apply, connect, review, and extend the skills they are learning. Numerous illustrations, tables, graphs, and photos help students visualize the problems they are solving. Problem types include skill building and writing exercises, as well as applications, matching, true/false, multiple-choice, and fill-in-theblank problems. Special types of exercises include Concept Check, WHAT WENT WRONG? , Extending Skills, and RELATING CONCEPTS • Special Summary Exercises We include a set of these popular in-chapter exercises in every chapter. They provide students with the all-important mixed review problems they need to master topics and often include summaries of solution methods and/or additional examples. • Extensive Review Opportunities We conclude each chapter with the following review components:

A Chapter Summary that features a helpful list of Key Terms organized by section, New Symbols, a Test Your Word Power vocabulary quiz (with answers immediately following), and a Quick Review of each section's main concepts, complete with additional examples. A comprehensive set of Chapter Review Exercises, keyed to individual sections for easy student reference. A set of Mixed Review Exercises that helps students further synthesize concepts and skills. A Chapter Test that students can take under test conditions to see how well they have mastered the chapter material.

ix

A set of Cumulative Review Exercises for ongoing review that covers material going back to Chapter R. • Comprehensive Glossary The online Glossary includes key terms and definitions (with section references) from throughout the text.

ACKNOWLEDGMENTS The comments, criticisms, and suggestions of users, nonusers, instructors, and students have positively shaped this text over the years, and we are most grateful for the many responses we have received. The feedback gathered for this edition was particularly helpful. We especially wish to thank the following individuals who provided invaluable suggestions. Barbara Aaker, Community College of Denver Kim Bennekin, Georgia Perimeter College Dixie Blackinton, Weber State University Eun Cha, College of Southern Nevada, Charleston Callie Daniels, St. Charles Community College Cheryl Davids, Central Carolina Technical College Robert Diaz, Fullerton College Chris Diorietes, Fayetteville Technical Community College Sylvia Dreyfus, Meridian Community College Sabine Eggleston, Edison State College LaTonya Ellis, Bishop State Community College Beverly Hall, Fayetteville Technical Community College Loretta Hart, NHTI, Concord's Community College Sandee House, Georgia Perimeter College Joe Howe, St. Charles Community College Lynette King, Gadsden State Community College Linda Kodama, Windward Community College Carlea McAvoy, South Puget Sound Community College James Metz, Kapi 'olani Community College Jean Millen, Georgia Perimeter College Molly Misko, Gadsden State Community College Charles Patterson, Louisiana Tech Jane Roads, Moberly Area Community College Melanie Smith, Bishop State Community College Erik Stubsten, Chattanooga State Technical Community College Tong Wagner, Greenville Technical College Rick Woodmansee, Sacramento City College Sessia Wyche, University of Texas at Brownsville Over the years, we have come to rely on an extensive team of experienced professionals. Our sincere thanks go to these dedicated individuals at Pearson who worked long and hard to make this revision a success. We would like to thank Michael Hirsch, Matthew Summers, Karen Montgomery, Alicia Frankel, Lauren Morse, Vicki Dreyfus, Stacey Miller, Eric Gregg, and all of the Pearson math team for helping with the revision of the text.

x

Preface

We are especially pleased to welcome Callie Daniels, who has taught from our texts for many years, to our team. Her assistance has been invaluable. She thoroughly reviewed all chapters and helped extensively with manuscript preparation. We are grateful to Carol Merrigan for her excellent production work. We appreciate her pos itive attitude, responsiveness, and expert skills. We would also like to thank SPi Global for their production work; Emily Keaton for her detailed help in updating real data applications; Connie Day for supplying her copyediting expertise; SPi Global for their photo research; and Lucie Haskins for

producing another accurate, useful index. Paul Lorczak and Hal Whipple did a thorough, timely job accuracychecking the page proofs and answers, and Sarah Sponholz checked the index. We particularly thank the many students and instructors who have used this text over the years. You are the reason we do what we do. It is our hope that we have positively impacted your mathematics journey. We would welcome any comments or suggestions you might have via email to [email protected]. John Hornsby Terry McGinnis

DEDICATION To Marin, Kasey, Erin, and Kellen Uncle Johnny

To Andrew and Tyler Mom

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Get the Most Out of Mylab Math for Beginning Algebra, Thirteenth Edition by Lial, Hornsby, McGinnis The Lial team has helped thousands of students learn algebra with an approachable, teacherly writing style and balance of skill and concept development. With this revision, the series retains the hallmarks that have helped students succeed in math, and includes new and updated digital tools in the Mylab Math course. Take advantage of the following resources to get the most out of your Mylab Math course.

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Get Students Engaged New! Learning Catalytics Learning Catalytics is an interactive student response tool that uses students' smartphones, tablets, or laptops to engage th em in more sophisticated tasks and thinking. In addition to a libra r y of deve lopmental math questions, Learning Catalytics questions created specifically for this text are pre-built to make it easy for instructors to begin using this tool! These questions, which cover prerequisite skills before each section, are noted in the margin of the Annotated Instructor's Edition, and can be found in Learning Catalytics by searching for "LialBeginning#" where# is the chapter number.

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Resources for Success Instructor Resources

Mylab

Student Resources

Annotated Instructor's Edition

Guided Notebook

Conta ins al l t he content found in t he student edition, plus answers to even and odd exerc ises on t he sa me text page, and Teac hing Tips and Classroom Exampl es t hroughout the t ext placed at key points.

This Guided Notebook helps students keep t heir work organized as they wor k through their course. The notebook includes:

The resources below are available through Pearson's Instructor Resource Center, or from Mylab Math.

Instructor's Resource Manual with Tests Incl udes m ini-lectures for each text section, seve ral forms of tests per cha pter-two diagnostic pretests, four free-response and two m ultipl e-choice test fo rms per chapter, and two fi nal exams.

Instructor's Solutions Manual Conta ins detailed, wor ked-out solutions to al l exe rcises in t he text.

• Guided Examples that are worked out for students, plus corresponding Now Try Th is exercises for each text objective. • Extra practice exercises fo r ever y section of the text, with ample space for students to show thei r work. • Learning objectives and key vocabulary terms for every text section, along with vocabulary practice problems.

Student Solutions Manual Provides comp letely worked -out solutions t o the odd-numbered section exercises and to all exercises in t he Now Trys, Relat ing Concepts, Chapter Reviews, Mixed Review, Chapter Tests, and Cumulative Rev iews.

TestGen® En ables instructors to build, edit, print, and administer tests using a comp uterized bank of questions developed to cover all the objectives of t he text. TestGen is algorithmica lly based, allow ing instructors to create multiple but eq u ivalent versions of the sa me question or test with the click of a butto n. Instr uctors can also modify test bank questions or add new questions.

PowerPoint Lecture Slides Ava ilable for down load only, these slides present key concepts and definitions from t he text. Access ible ve rsions of the PowerPo int sl ides are also avai lable for students who are vision impaired.

pearson.com/mylab/math

PHOTO CREDITS STUDY SKILLS p. S-1 Borchee/E+/Getty Images; p. S-2 Originoo stock/123RF; p. S-4 G-stockstudio/Shutterstock; p. S-6 Rawpixe l/Shutterstock; p. S-8 Studio DMM Photography, Designs & Art/Shutterstock; p. S-10 Doyeol (David) Ahn/Alamy Stock Photo CHAPTERR p. 11 Arek Malang/Shutterstock; p. 15 Ryan McVay/Stockbyte/Getty Images CHAPTERl p. 35 Cathy Yeulet/123RF; p. 42 David Pereiras/Shutterstock; p. 50 Kurt Kleemann/Shutterstock; p. 62 Terry McGinnis; p. 63 Africa Studio/Shutterstock; p. 65 GaryLHampton/iStock/Getty Images; p. 75 Belchonock/123RF; p. 80 Famveldman/1 23RF; p. 101 Sascha Burkard/Shutterstock; p. 102 Callie Danie ls CHAPTER2 p. 103 Callie Daniels; p. 134 PCN Photography/Alamy Stock Photo; p.142 Terry McGinnis; p. 143T Belchonock/123RF; p. 143B Brandon Alms/Shutterstock; p. 156 ZUMA Press Inc./Alamy Stock Photo; p. 158 Yulia Davidovich/ l 23RF; p. 161 Education & Exploration 3/Alamy Stock Photo; p. 163 Monkey Business Images/Shutterstock; p. 166T Jim West/Alamy Stock Photo; p. 166BL Chokkicx/iStock/360/Getty Images; p. 166BR Spass/Fotolia; p. 167 Tovovan/Shutterstock; p. 168 Sumire8/Fotolia; p. 169 Chris Cooper-Smith/Alamy Stock Photo; p. 170 Stuart Jenner/ Shutterstock; p. 174T Reiulf Gronnevik/Shutterstock; p. 174B Robynrg/Shutterstock; p. 178 Wavebreakmedia/Shutterstock; p.179 Linda Richards/Alamy Stock Photo; p. 181 Josh randall/ Shutterstock; p. 189 Kali9/Getty Images; p. 194 Sasi Ponchaisang/123RF; p. 195 Rob/Fotolia CHAPTER3 p. 207 Sergunt/Fotolia; p. 208 Blend Images/Alamy Stock Photo; p. 212 Georgios Kollidas/Alamy Stock Photo; p. 214 Maksym Topchii/ l 23RF; p. 218 Air Images/Shutters tock; p. 226 Wavebreakmedia/ Shutterstock; p. 253 Koksharov Dmitry/Shutterstock; p. 256 Matt Benoit/Shutterstock; p. 259 Rawpixel.com/Shutterstock; p. 261 Rebecca Fisher/Fotolia CHAPTER4 p. 287 Mmphotos/Fotolia; p. 304 Vadim Ponomarenko/Fotolia; p. 314 Dmitriy Shironosov/l 23RF; p. 315 Ed Rooney/Alamy Stock Photo; p. 320 Skycolors/Shutterstock; p. 321T Scott Rothstein/ Shutterstock; p. 321B Snvv/123RF; p. 323 Dwphotos/iStock/Getty Images; p. 335 Boleslaw Kubica/ Shutterstock CHAPTERS p. 341 Nikonomad/Fotolia; p. 350 Imagesbavaria/ 123RF; p. 364 Studio 72/Shutterstock; p. 367T Photoroad/l 23RF; p. 367B Albert Barr/Shutterstock; p. 368 Deepspace/Shutterstock; p. 379 Terry McGinnis; p. 409 NASA CHAPTER6 p. 413 Alexandre Solodov/Fotolia; p. 452 Nickolae/Fotolia; p. 460 Petr Jilek/Shutterstock; p. 463 Offscreen/Shutterstock; p. 465 Marmaduke St. John/Alamy Stock Photo; p. 470 Goran Bogicevic/123RF; p. 471 Elenabsl/123RF; p. 477 Stephen Barnes/Transport/Alamy Stock Photo CHAPTER7 p. 481 HodagMedia/Shutterstock; p. 514 Denis Tabler/Fotolia; p. 539 Tommaso Altamura/ l 23RF; p. 541 Johnny Habell/Shutterstock; p. 543TL Sakda saetiew/123RF; p. 543TR Sirtravelalot/Shutterstock; p. 543B Alexandr Shevchenko/Shutterstock; p. 544 Melinda Fawver/Fotolia; p. 546 Llandrea/Fotolia; p. 547 loana Davies (Drutu)/Shutterstock; p. 549 Germanskydiver/Shutterstock; p. 552 Richard Coencas/Shutterstock; p. 553 Olga Khoroshunova/123RF; p. 564 VJ Matthew/Shutterstock CHAPTERS p. 567 Terry McGinnis; p. 577L John Hornsby; p 577R AlessandroZocc/Shutterstock; p. 598 NASA; p. 605 Cristovao31/Fotolia; p. 614 S Curtis/Shutterstock; p. 615T SpaceKris/Shutterstock; p. 615B BetacamSP/Fotolia; p. 621 Nwdph/Shutterstock; p. 622 Thinkstock lmages/Stockbyte/Getty Images CHAPTER9 p. 625 Callie Danie ls; p. 630 Robert Ranson/123RF

xiv

STUDY SKILL 1 Using Your Math Text Your text is a valuable resource. You will learn more if you make full use of the features it offers.

NowTRYTHIS General Features of This Text Locate each feature, and complete any blanks. • Table of Contents This is located at the front of the text.

Find it and mark the chapters and sections you will cover, as noted on your course syllabus. • Answer Section This is located at the back of the text.

Tab this section so you can easily refer to it when doing homework or reviewing for tests. • List of Formulas This helpful list of geometric formulas, along with review information on triangles and angles, is found at the back of the text.

The formula for the volume of a cube is _ __

Specific Features of This Text Look through Chapter 1 or 2 and give the number of a page that includes an example of each of the following specific features. • Objectives The objectives are listed at the beginning of each section and again within the section as the corresponding material is presented. Once you finish a section, ask yourself if you have accomplished them. Seepage _ __ • Vocabulary List Important vocabulary is listed at the beginning of each section. You should be able to define these terms when you finish a section. See page _ __ • Now Try Exercises These margin exercises allow you to immediately practice the material covered in the examples and prepare you for the exercises. Check your results using the answers at the bottom of the page. See page _ __ • Pointers These small, shaded balloons provide on-the-spot warnings and reminders, point out key steps, and give other helpful tips. See page _ __ • Cautions These provide warnings about common errors that students often make or trouble spots to avoid. See page _ __ • Notes These provide additional explanations or emphasize other important ideas. Seepage _ __ • Problem-Solving Hints These boxes give helpful tips or strategies to use when you work applications. Look for them beginning in Chapter 2. See page _ __

S-1

I

STUDY SKILL 2 Reading Your Math Text Take time to read each section and its examples before doing your homework. You will learn more and be better prepared to work the exercises your instructor assigns.

Approaches to Reading Your Math Text Student A learns best by listening to her teacher explain things. She "gets it" w hen she sees the instructor work problems. She previews the section before the lecture, so she knows generally what to expect. Student A carefully reads the section in her text AFTER she hears the classroom lecture on the topic. Student B learns best by reading on his own . He reads the section and works through the examples before coming to class. That way, he knows what the teacher is going to talk about and what questions he wants to ask. Student B carefully reads the section in his text BEFORE he hears the classroom lecture on the topic. Which of these reading approaches works best for you-that of Student A or Student B?

Tips for Reading Your Math Text • Turn off your cell phone and the TV. You will be able to concentrate more fully on w hat you are reading. •

Survey the material. Glance over the assigned material to get an idea of the "big picture." Look at the list of objectives to see what you w ill be learning.

•

Read slowly. Read only one section-or even part of a section-at a sitting, w ith paper and pencil in hand.

•

Pay special attention to important information given in colored boxes or set in boldface type. Highlight any additional information you find helpful.

•

Study the examples carefully. Pay particular attention to the blue side comments and any pointer balloons.

•

Do the Now Try exercises in the margin on separate paper as you go. These problems mirror the examples and prepare you for the exercise set. Check your answers with those given at the bottom of the page.

•

Make study cards as you read. Make cards for new vocabulary, rules , procedures, formulas, and sample problems.

•

Mark anything you don 't understand. ASK QUESTIONS in class - everyone will benefit. Follow up with your instructor, as needed.

NowTRYTHIS Think through and answer each question.

1 . Which two or three reading tips given above w ill you try this week?

S-2

2. Did the tips you selected improve your ability to read and understand the material? Explain .

STUDY SKILL 3 Taking Lecture Notes }a.JUUtry 12

Come to class prepared. •

•

•

Exp?fU.ffi:s

ExpofU.ffi:s uoe.dto slwwrepede.dJUUi.t:~HA

Bring paper, pencils, notebook, text, completed homework, and any other materials you need. Arrive 10-15 minutes early if possible. Use the time before class to review your notes or study cards from the last class period.

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Work the Review Exercises. They are grouped by section . Answers are included at the back of the text. .,,. Pay attention to direction words, such as simplify, solve, and evaluate . .,,. Are your answers exact and complete? Did you include the correct labels , such as $, cm2 , ft, etc.? .,,. Make study cards for difficult problems.

•

Work the Mixed Review Exercises. They are in random order. Check your answers in the answer section at the back of the text.

•

Take the Chapter Test under test conditions . .,,. Time yourself . .,,. Use a calculator or notes only if your instructor permits them on tests . .,,. Take the test in one sitting . .,,. Show all your work . .,,. Check your answers in the answer section. Section references are provided.

Reviewing a chapter takes time. Avoid rushing through your review in one night. Use the suggestions over a few days or evenings to better understand and remember the material.

NowTRYTHIS Follow these reviewing techniques to prepare for your next test. Then answer each question. 1. How much ti me did you spend reviewing for your test? Was it enough?

3. Are you investing enough time and effort to really know the material and set yourself up for success? Explain.

2. Which reviewing techniques worked best for you?

4. What will you do differently when reviewing for your next test?

S-7

I

STUDY SKILL 8 Taking Math Tests Comments Come prepared with a pencil, eraser, paper, and calculator, if allowed.

Working in penc il lets you erase, keeping your work neat.

Scan the entire test, note the point values of different problems, and plan your time accordingly.

To do 20 problems in 50 minutes, allow 50 ..;- 20 = 2.5 minutes per problem. Spend less time on easier problems.

Do a "knowledge dump " when you get the test. Write important notes, such as formulas, in a corner of the test for reference.

Writing down tips and other special inform ation that you've learned at the beginning allows you to relax as you take the test.

Read directions carefully, and circle any significant words. When you finish a problem, reread the directions. Did you do what was asked?

Pay attention to any announcements written on the board or made by your instructor. Ask if you don't understand something.

Show all your work. Many teachers give partial credit if some steps are correct, even if the final answer is wrong. Write neatly.

If your teacher can't read your writing, you won't get credit for it. If you need more space to work, ask to use extra paper.

Write down anything that might help solve a problem: a formula, a diagram, etc. If necessary, circle the problem and come back to it later. Do not erase anything you wrote down.

If you know even a little bit about a problem, write it down. The answer may come to you as you work on it, or you may get partial credit. Don't spend too long on any one problem.

If you can 't solve a problem, make a guess. Do not change it unless you find an obvious mistake.

Have a good reason for changing an answer. Your first guess is usually your best bet.

Check that the answer to an application problem is reasonable and makes sense. Reread the problem to make sure you've answered the question.

Use common sense. Can the father really be seven years old? Would a month's rent be $32, 140? Remember to label your answer if needed: $, years, inches, etc.

Check for careless errors. Rework each problem without looking at your previous work. Then compare the two answers.

Reworking a problem from the beginning forces you to rethink it. If possible, use a different method to solve the problem.

NowTRYTHIS Think through and answer each question. 1 . What two or three tips will you try when you take your next math test?

3. What will you do differently w hen taking your next math test?

2. How did the tips you selected work for you when you took your math test?

4. Ask several classmates how they prepare for math tests. Did you learn any new preparation ideas?

S-8

STUDY SKILL 9 Analyzing Your Test Results An exam is a learning opportunity-learn from your mistakes. After a test is returned, do the following: •

Note what you got wrong and why you had points deducted.

•

Figure out how to solve the problems you missed. Check your text or notes, or ask your instructor. Rework the problems correctly.

•

Keep all quizzes and tests that are returned to you. Use them to study for future tests and the final exam.

Typical Reasons for Errors on Math Tests 1. You read the directions wrong. 2. You read the question wrong or skipped over something. 3. You made a computation error.

These are test-taking errors. They are easy to correct if you read carefully, show all your work, proofread, and doublecheck units and labels.

4. You made a careless error. (For example, you incorrectly copied a correct answer onto a separate answer sheet.) 5. Your answer was not complete. 6. You labeled your answer wrong . (For example, you labeled an answer "ft" instead of "ft 2 . ") 7. You didn't show your work. 8. You didn't understand a concept. 9. You were unable to set up the problem (in an application).

}

1 0. You were unable to apply a procedure.

These are test preparation errors. Be sure to practice all the kinds of problems that you will see on tests.

NowTRYTHIS Work through the following, answering any questions. 1. Use the sample charts at the right to track your test-taking progress. Refer to the tests you have taken so far in your course. For each test, check the appropriate box in the charts to indicate that you made an error in a particular category. 2. What test-taking erro rs did you make? Do you notice any patterns?

T Test-Taking Errors Test

4. What will you do to avoid these kinds of errors on your next test?

Read question wrong

Made computation error

Made careless error

Answer not complete

Answer labeled wrong

-

Didn 't show work

1

2

3

T Test Preparation Errors Test

3. What test preparation errors did you make? Do you notice any patterns?

Read directions wrong

Didn't understand concept

Didn 't set up problem correctly

Couldn't apply a procedure

1

2

3

S-9

STUDY SKILL 10 Preparing for Your Math Final Exam Your math final exam is likely to be a comprehensive exam, which means it will cover material from the entire term. One way to prepare for it now is by working a set of Cumulative Review Exercises each time your class finishes a chapter. This continual review will help you remember concepts and procedures as you progress through the course.

Final Exam Preparation Suggestions 1. Figure out the grade you need to earn on the final exam to get the course grade you want. Check your course syllabus for grading policies, or ask your instructor if you are not sure. 2. Create a fina l exam week plan. Set p riorities that allow you to spend ext ra time studying. This may mean making adj ustments, in advance , in your wo rk schedu le or enlisting extra help with fami ly responsibilities. 3. Use the following suggestions to guide your studying. •

Begin reviewing several days before the final exam. DON'T wait until the last minute.

•

Know exactly which chapters and sections will be covered on the exam.

•

Divide up the chapters. Decide how much you will review each day.

•

Keep returned quizzes and tests. Use them to review.

•

Practice all types of problems. Use the Cumulative Review Exercises at the end of each chapter in your text beginning in Chapter 1. All answers, with section references, are given in the answer section at the back of the text.

•

Review or rewrite your notes to create summaries of important information.

•

Make study cards for all types of problems. Carry the cards with you, and review them whenever you have a few minutes.

•

Take plenty of short breaks as you study to reduce physical and mental stress. Exercising, listening to music, and enjoying a favorite activity are effective stress busters.

Finally, DON'T stay up all night the night before an exam-get a good night's sleep.

NowTRYTHIS Think through and answer each question.

1. How many points do you need to earn on your math final exam to get the grade you want in your course? 2. What adjustments to your usual routine or schedule do you need to make for final exam week? List two or three . 3. Which of the suggestions for studying will you use as you prepare for your math final exam? List two or three. 4. Analyze your final exam results. How will you prepare differently next time?

S-10

PREALGEBRA REVIEW R.1 Fractions R.2 Decimals and Percents

Fractions OBJECTIVES 1 Write numbers in

The numbers used most often in everyday life are the natural (counting) numbers,

1, 2, 3, 4, . .. '

factored form.

2 Write fractions in lowest terms.

0, 1, 2, 3, 4, ... '

3 Convert between improper fractions and mixed numbers.

The three dots, or ellipsis points, indicate that each list of numbers continues in the same way indefinitely.

the whole numbers,

and fractions, such as

4 Multiply and d ivide fractions.

5 Add and subtract

1

2

2'

3'

11 12

The parts of a fraction are named as shown.

fractions. Fraction

6 Solve applied problems

bar ~

that involve fractions.

7 Interpret data in a circle

and

3 .,,.___ Numerator 8 .,,.___ Denominator

The fraction bar represents division ( ~

=a

+ b).

graph.

miID

Fractions are a way to represent parts of a whole. In a fraction, the numerator gives the number of parts being represented. The denominator gives the total number of equal parts in the whole. See FIGURE 1.

The shaded region represents ~ of the circle.

VOCABULARY natural (counting) numbers whole numbers fractions numerator denominator proper fraction improper fraction factors product prime number (continued)

D D D D D D D D D D

FIGURE 1

A fraction is classified as being either a proper fraction or an improper fraction. Proper fractions

Improper fractions

2 5' 7'

9 10'

23 25

3 2'

11 7'

28

5 5'

4

Numerator is less than denominator. Value is less than 1. Numerator is greater than or equal to denominator. Value is greater than or equal to 1.

1

2

CHAPTER R Prealgebra Review

VOCABULARY (continued) D D D D D D D D D

composite number lowest terms mixed number reciprocals quotient dividend divisor sum least common denominator (LCD) D difference D circle graph (pie chart)

OBJECTIVE 1 Write numbers in factored form.

In the statement 3 X 6 = 18, the numbers 3 and 6 are factors of 18. Other factors of 18 include 1, 2, 9, and 18. The result of the multiplication, 18, is the product. We can represent the product of two numbers, such as 3 and 6, in several ways. 3

x 6, 3 . 6, (3)(6) , (3)6, 3(6)

Products

We factor a number by writing it as the product of two or more numbers. Multiplication

3 . 6

t

t

Factoring

= 18

t

Factors Product

Factoring is the reverse of multiplying two numbers to get the product.

18 = 3 . 6

t

t

t

Product Factors

l1IlID

In algebra, a raised dot • is often used instead of the multiplication because x may be confused with the letter x .

x

symbol to indicate

A natural number greater than 1 is prime if it has only itself and 1 as factors. "Factors" are understood here to mean natural number factors. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37

First dozen prime numbers

A natural number greater than 1 that is not prime is a composite number. 4,~8 , 9 ,

10, 12, 14, 15, 16, 18, 20, 21

First dozen composite numbers

The number 1 is considered to be neither prime nor composite. /:.NOW TRY

~EXERCISE 1 Identify the number 60 as prime, composite, or neither. If the number is composite, write it as a product of prime factors.

IM4MQlll

Writing Numbers in Prime Factored Form

Identify each number as prime, composite, or neither. If the number is composite, write it as a product of prime factors. (a) 43

There are no natural numbers other than 1and43 itself that divide evenly into 43, so the number 43 is prime. (b) 35

The number 35 is composite and can be written as the product of the prime factors 5 and 7. 35 = 5 . 7 (c) 24 The number 24 is composite. We show a factor tree on the right, with prime factors circled. 24

/\

Divide by the least prime factor of 24, which is 2.

24 = 2 . 12

Divide 12 by 2 to find two factors of 12.

24 = 2 · 2 ·6

@ . 12

/\ @ ·6

/\

Now factor 6 as 2 · 3. NOW TRY ANSWER 1. composite; 2 · 2 · 3 · 5

24 = 2·2·2·3

@·®

All factors are prime. NOW TRY

g

Fractions

SECTION R.1

3

m!JD

No matter which prime factor we start with when factoring , we will always obtain the same prime factorization. We verify this in Example 1(c) by starting with 3 instead of 2. 24

/\

@ ·8

Divide 24 by 3.

24 = 3 · 8

Divide 8 by 2.

24 = 3 · 2 · 4

Divide4by2 .

24= 3 · 2 · 2 · 2

/\

@ ·4

/\

® ·@

The same prime factors result.

OBJECTIVE 2 Write fractions in lowest terms. The following properties are useful when writing a fraction in lowest terms. Properties of 1

Any nonzero number divided by itself is equal to 1. Any number multiplied by 1 remains the same.

~3 = 1

Example:

Example:

2 5

1=~ 5

A fraction is in lowest terms when the numerator and denominator have no factors in common (other than 1). Writing a Fraction in Lowest Terms

Step 1

Write the numerator and denominator in factored form.

Step 2

Replace each pair of factors common to the numerator and denominator with 1.

Step 3

Multiply the remaining factors in the numerator and in the denominator.

(This procedure is sometimes called "simplifying the fraction.")

li'4JMQlfj Writing Fractions in Lowest Terms Write each fraction in lowest terms. 10 2· 5 2 5 2 2 (a) - = = - • - = - · 1= -

15

(b)

3.5

3

5

3

3

Use the fi rst property of 1 to replace ~ with 1.

~

45 By inspection, the greatest common factor of 15 and 45 is 15. 15 45

15 3 . 15

3. 1

1 3

Remember to write 1 in the numerator.

If the greatest common factor is not obvious, factor the numerator and denominator into prime factors .

15 45

3.5 3.3.5

1. 1 3. 1. 1

1 3

The same answer results.

4

CHAPTER R Prealgebra Review

~NOW TRY

50 is the greatest common factor of 150 and 200.

"7 EXERCISE 2 Write each fraction in lowest terms.

30

10

(a) 42

(b) 70

Another strategy is to choose any common factor and work in stages.

72

150 200

(c) 120

15 . 10 20 . 10

3 . 5 . 10 4 . 5 . 10

3 1. 1=4

3 4

The same answer results. NOWTRY

~

OBJECTIVE 3 Convert between improper fractions and mixed numbers. A mixed number is a single number that represents the sum of a natural number and a proper fraction. The mixed number 2 ~ is illustrated in FIGURE 2. Mixed number ~

3 24

3 4

=2+-

l

2

4

EBEBEB

The mixed number 2~ is equivalent to the improper fraction ¥ ·

FIGURE2

~NOWTRY

1$f!1MQl#I Converting an Improper Fraction to a Mixed Number

. 92 "d num ber. W nte 5 as a m1xe

Wnte 8 as a rruxed number.

"7 EXERCISE 3

.

59

.

Because the fraction bar represents division(~

=a

-:-

b, or b)a), divide the numer-

ator of the improper fraction by the denominator. Denominator of fraction ~

7

4. 2

1!£MMQlll Using Inequality Symbols Determine whether each statement is true or false.

+1

(a) 6

~

(b) 5

+3
u

(e) 12

~

12

Because 12

(d)

1

7

(f) 0.5 (g)

1

2

2

15

3

$

1.

19 is true because 8 < 19.

20 · 2 is true because 15

= 30 are fal se, so 25

~

< 40.

30 is false.

= 12, this statement is true.

Because 0.5 = ~, this statement is false.

-

Both 15

10

6

10

6

~

2 .

> 15 and 15 = 15 are false, so 15 ~ 3 1s false.

NOW TRY ~

OBJECTIVE 5 Translate word statements to symbols. ~NOWTRY

lif:MMQl4j Translating from Words to Symbols

Write each word statement in symbols.

Write each word statement in symbols.

"7 EXERCISE 5

(b) Fifty is greater than fifteen. (c) Eleven is less than or equal to twenty.

(b) Nine is less than ten.

(a) Twelve equals ten plus two.

(a) Ten is not equal to eight minus two.

12

= 10 + 2

9 < 10

(c) Fifteen is not equal to eighteen. 15

~

(d) Seven is greater than four.

7> 4

18

(e) Thirteen is less than or equal to forty. 13

$

(f) Eleven is greater than or equal to

eleven.

40

11 ~ 11

NOW TRY~

OBJECTIVE 6 Write statements that change the direction of inequality symbols. Any statement involving < can be converted to one with >, and any statement involving > can be converted to one with < . We do this by reversing the order of

the numbers and the direction of the symbol. Interchange numbers.

6 < 10

becomes Reverse symbol.

lia;iMHii NOW TRY ANSWERS 4. (a) true (c) true 5. (a) 10 ¥ (c) II :s

(b) false (d) false 8 - 2 (b) 50 > 15 20

Converting between Inequality Symbols

Write each statement as another true statement with the inequality symbol reversed. (a) 5 > 2

is equivalent to

2 < 5.

(b)

1

3

2 4 $

is equivalent to

3

1

->-

4- 2

NOW TRY~

Exponents, Order of Operations, and Inequality

SECTION 1.1

33

T Summary of Equality and Inequality Symbols

/:.NOW TRY

~EXERCISES Write the statement as another true statement with the inequality symbol reversed.

Symbol

8

Is greater t han

15

:s

Is less t han or equal to

4 s 8 means 4 is less than or equal to 8.

2:

Is greater than or equal to

1

< 10 means 6 is less than 10. > 14 means 15 is greater than 14. 2:

0 means 1 is greater than or equal to 0.

0

CAUTION Equality and inequality symbols are used to write mathematical sentences. Operation symbols (+, - , · , and + ) are used to write mathematical expressions.

NOW TRY ANSWER 6. 9 > 8

1 .1 Exercises 0

Video solutions for select problems available in Mywb Math

Sentence:

4 < 10 .,.___ Gives the relationship between 4 and 10

Expression:

4 + 10 *"-- Tells how to operate on 4 and 10 to get 14

0

FOR

~~

Concept Check 2

1. 3

=

Mylab Math Decide whether each statement is true or false. If it is false, explain why. 2. 13

6

=

3. 3 1 = 1

3

4. The expression 62 means that 2 is used as a factor 6 times.

STUDY SKILLS REMINDER You will increase your chance of success in this course if you fully utilize your text. Review Study Skill 1, Using Your Math Text.

+ 3(8

5. When evaluated, 4

- 2) is equal to 42.

6. When evaluated, 12 + 2 · 3 is equal to 2.

Concept Check For each expression, label the order in which the operations should be performed. Do not actually perform them. 7. 18 - 2

+3

8. 28 - 6

00

9. 2 . 8 - 6

2

-7

00

10.40 +6(3 -l)

11.3 ·5- 2(4+2)

0 OOO

OOO

12.

-7

3

OOO 9 - 23 + 3 . 4 OOO 0

Find the value of each exponential expression. See Example 1. 13. 7 2

14. 82

15. 122

16. 142

17. 4 3

18. 5 3

19. 103

20. 113

21. 34

22. 64

23. 4 5

24. 35

25.

(~)

2

26.

29. (0.6)2

(~ )

2

30. (0.9)2

27.

(~)4

31. (0.4) 3

28.

(~)3

32. (0.5 ) 4

33. Concept Check The value of an expres- 34. Concept Check The value of an expression was found incorrectly as follows. sion was found incorrectly as follows. 8

+2 =

. 3 10. 3

=

30

WHAT WENT WRONG? Find the correct value of the expression.

16 - 23 + 5 = 16 - 6 + 5 = 10 + 5 = 15

WHAT WENT WRONG? Find the correct value of the expression.

34

CHAPTER 1

The Real Number System

Find the value of each expression. See Example 2 and 3.

35. 64-:- 4. 2 38. 11

+ 7. 6

41. 9. 4 - 8. 3

56. 5

+3

+ 4 [ 1 + 7 ( 3) l

40. 12.4 - 9.3 -:- 3.1

+ 10 .

2

2

11

+5

46. 18 - 7 . 2

+6

49. 18 - 2(3

+ 4)

51. 3(4 + 2) + 8. 3

52. 9( 1 + 7)

+2

54. 22 - 23 + 9

55. 2 + 3[5 + 4(2) ]

+ 4(22 ) ]

60. 42 [ ( 13

2(82 - 4)+8 29 - 33

1

43. - . -+-. 4 3 5 3

3

48. 12 + 64 -:- 8 - 4

57. 5[3

59. 32 [ ( 11+ 3) - 4]

62.

39. 25.2 - 12.6 -:- 4.2

45. 20 - 4 . 3

50. 30 - 3(4 + 2) 53. 18 - 4 2

37. 13

42. 11 . 4

9 2 4 5 44. - . - + - . 4 3 5 3 47. 10 + 40 -:- 5 . 2

+ 9. 5

36. 250-:- 5. 2

63.

58. 6 [2

•5

+ 8(3 3 ) l

6(3 2 - 1) + 8 61. 8 - 22

+ 4) - 8 ]

4(6 + 2) + 8(8 - 3) 6(4-2)-2 2

6( 5 + 1) - 9( 1 + 1) 64. - - - - - 5(8 - 6) - 2 3

Concept Check Insert one pair of parentheses in each expression so that the given value results when the operations are performed. 65. 3. 6 + 4. 2 = 60

66. 2. 8 - 1 . 3 = 42

67. 10 - 7 - 3

68. 8 + 2 2

=6

=

100

First simplify both sides ofeach inequality. Then determine whether the given statement is true or false. See Examples 2-4.

70. 6 . 5 - 12 s 18

69.9· 3 -11 S l6 71. 5 · 1 I 73. 0

~

75. 45

72. 9 . 3 + 4 . 5 ~ 48

74. 10 s 13 . 2 - 15 . 1

12 . 3 - 6 . 6

~

76. 55 ~ 3[4 + 3(4 + l) ]

2 [2 + 3(2 + 5) ]

77. [ 3 . 4 79.

+ 2 · 3 s 60

+ 5(2) l . 3 >

78. 2 . [ 7 . 5 - 3(2) l s 58

72

3+5(4-1) 2. 4 + 1 ~ 3

81. 3

~

80.

2(5 + 1) - 3(1+1) ------5(8 - 6) - 4 . 2

7(3+ 1)-2 3 + 5. 2 s 2

3(8 - 3)+2(4 - l) 82. 7 s - - - - - - - 9(6 - 2) - 11(5 - 2)

Write each statement in words, and determine whether it is true or false. See Examples 4 and 5.

83. 5 < 17 87. 7 91.

~

14

1

3

3

10

- = -

84. 8 < 12 88. 6 92.

~

10

-

6

12 3

= -

2

85. 5 =P 8

86. 6 =P 9

89. 15 s 15

90. 21 s 21

93. 2.5 > 2.50

94. 1.80 > 1.8

Write each word statement in symbols. See Example 5.

95. Fifteen is equal to five plus ten.

96. Twelve is equal to twenty minus eight.

97. Nine is greater than five minus four.

98. Ten is greater than six plus one.

99. Sixteen is not equal to nineteen. 101. One-half is less than or equal to two-fourths. 102. One-third is less than or equal to three-ninths.

100. Three is not equal to four.

SECTION 1.1

35

Exponents, Order of Operations, and Inequality

Write each statement as another true statement with the inequality symbol reversed.

See Example 6. 103. 5 < 20 106.

5 4

3 2

- 9

4 3 105. - > 5 4

107. 2.5 ;;::: 1.3

108. 4.1 :s: 5 .3

One way to measure a person's cardiofitness is to calculate how many METs, or metabolic units, he or she can reach at peak exertion. One MET is the amount ofenergy used when sitting quietly. To calculate ideal METs, we can use the following expressions.

14.7 - age · 0. 13

For women

14.7 - age· 0.11

For men

(Data from New England Journal of Medicine.) 109. A 40-yr-old woman wishes to calculate her ideal MET. (a) Write the expression, using her age. (b) Calculate her ideal MET.

(c) Researchers recommend that a person reach approximately 85% of his or her MET when exercising. Calculate 85% of the ideal MET from part (b). Then refer to the following table. What activity listed in the table can the woman do that is approximately this value?

Activity

Activity

METs

METs

Golf (with cart)

2.5

Skiing (water or downhill)

Walking (3 mph)

3 .3

Swimming

6.8 7.0

Mowing lawn (power mower)

4.5

Walking (5 mph)

8.0

Ballroom or square dancing

5.5

Jogging

10.2

Cycling

5.7

Skipping rope

12.0

Data from Harvard School of Public Health.

(d) Repeat parts (a)-(c) for a 55-yr-old man.

110. Repeat parts (a)-(c) of Exercise 109 using your age and gender.

The table shows the number of pupils per teacher in U.S. public schools in selected states.

111. Which states had a number greater than 12.6? 112. Which states had a number that was at most 15.2? 113. Which states had a number not less than 12.6? 114. Which states had a number less than 13.0?

State

Pupils per Teacher

Alaska

16.4

Texas

15.2

California

22.5

Virginia

12.6

Maine

12.4

Idaho

19.7

Missouri

12.1

Data from National Center for Education Statistics.

36

CHAPTER 1

The Real Number System

Variables, Expressions, and Equations OBJECTIVES 1 Evaluate algebraic expressions, given values for the variables. 2 Translate word phrases to algebraic expressions.

3 Identify solutions of

A constant is a fixed, unchanging number. A variable is a symbol, usually a letter, used to represent an unknown number.

Examples:

3 1 , 8 2, 10.8 4

5,

a,

Constants

X,

y,

Z

Variables

An algebraic expression is a sequence of constants, variables, operation symbols, and/or grouping symbols formed according to the rules of algebra.

equations.

4 Identify solutions of

Examples: x

+ 5, 2m -

9,

8p2

+ 6(p -

2)

Algebraic expressions

equations from a set of numbers.

5 Distinguish between expressions and equations.

OBJECTIVE 1 Evaluate algebraic expressions, given values for the variables. To evaluate an expression means to find its value. An algebraic expression can have different numerical values for different values of the variables.

lf:f4MQlll

Evaluating Algebraic Expressions

Evaluate each expression for x = 5. (a) 8

VOCABULARY D D D D D O D

constant variable algebraic expression equation solution set element

(b) 2 x - 9

=8 + 5

Let x = 5.

=2· 5 -9

Let x = 5.

=

Add.

= 10 =1

Multiply.

13

Subtract.

(d)-7

= 3 · x2_A s2= 5· 5J

= 3 · 52 = 3. 25 = 75

~EXERCISE 1 Evaluate each expression for x= 6.

9

4x - 2

(c) 3x 2

~NOWTRY

(a)

+x

Let x

=

4. 5 - 2

5.

Let x = 5.

7

Square 5.

20 - 2 Multiply.

Multiply.

7 18

9x- 5 (b) 4x2

7'

or

4

2-

7 NOW TRY

0

CAUTION

3x 2

means

3 · x 2 , not 3x · 3x.

g

See Example 1(c).

Unless parentheses are used, the exponent refers only to the variable or number just before it. We would need to use parentheses to write 3x · 3x with exponents. ( 3x) 2 NOW TRY ANSWERS

1.

(a)

49 {b)

144

means 3x · 3x.

SECTION 1.2

/:.NOW TRY

~EXERCISE2 Evaluate each expression for x = 4 and y = 7. 6x- 2y (a) 3x

+ 4y

(c) 4x2

-

(b) - 2y - 9

y2

lf)UMQlfj

2x

+ 7y

= 5 and y = 3. We could use parentheses and write 2( 5) + 7( 3) .

+7

= 2 • 5

Follow the rules for order of operations.

37

Evaluating Algebraic Expressions

Evaluate each expression for x (a)

Variables, Expressions, and Equations

•3

Let x = 5 and y = 3. Multiply.

= 31

Add.

9x - Sy

(b) - 2x - y

9· 5 - 8· 3 2. 5 - 3

Let x

45 - 24 10 - 3

Multiply.

21

5 andy = 3.

Subtract.

7

=3 (c)

=

Divide.

x2 - 2y 2

= 52 ~ = 25 -

2. 32

Let x = 5 and y = 3.

2. 9

Apply the exponents.

= 25 - 18

Multiply.

=7

Subtract. NOWTRY

~

OBJECTIVE 2 Translate word phrases to algebraic expressions.

i:if!f!1MQlll

Using Variables to Write Word Phrases as Algebraic Expressions

Write each word phrase as an algebraic expression, using x as the variable. (a) The sum of a number and 9

x

+ 9,

or

9

+x

"Sum" is the answer to an addition problem.

(b) 7 minus a number

7- x

"Minus" indicates subtraction.

The expression x - 7 is incorrect. We cannot subtract in either order and obtain the same result. (c) A number subtracted from 12

12 -

X

-
I ~ I·

Subtract the fract ions.

6

Write in lowest terms.

2

+ ( -4.6) = + (8.1 - 4.6)

(e) 8.1

L

~

= and = ; Subtract the lesser absolute value from the greater.

(

IB.1 1 = 8.1 and l-4.61 = 4.6; Subtract the lesser absolute value from the greater.

1s.1 1>1-4.6 1

= 3.5

+ 16 =O

(f) -16

(g) 42

l- 161= 16 and 1161= 16; The difference of the absolute values is 0, which is neither positive nor negative.

+ (-42) =0

When additive inverses are added, the sum is 0.

NOW TRY@)

OBJECTIVE 3 Use the definition of subtraction. Recall that the answer to a subtraction problem is a difference. In the subtraction x - y, x is the minuend and y is the subtrahend.

t

Minuend

t

Subtrahend

~NOW TRY

IUM!Qllj

Use a number line to find the difference.

Use a number line to find the difference 7 - 4.

"7 EXERCISE 5 6- 2

z

y

x

t

Difference

Subtracting Numbers on a Number Line

Step 1

Start at 0 and draw an arrow 7 units to the right. See

Step 2

From the right end of the first arrow, draw a second arrow 4 units to the left to represent the subtraction.

The number below the end of the second arrow is 3, so 7 - 4

FIGURE 13.

= 3.

-4 7

NOW TRY ANSWERS 4.(a)3 (c) - 2

5. 4

13

4

(b) -9, or - 19 (d) 0

- I

0

2

3 4 7- 4 =3 FIGURE 13

5

6

7

NOW TRY@)

SECTION 1.4

Adding and Subtracting Real Numbers

55

The procedure used in Example 5 to find the difference 7 - 4 is exactly the same procedure that would be used to find the sum 7 + ( - 4 ). 7- 4

is equal to

7

+ ( -4) .

This suggests that subtracting a positive number from a greater positive number is the same as adding the additive inverse of the lesser number to the greater.

Definition of Subtraction

For any real numbers x and y, the following holds true.

x-y=x+(-y) To subtract y from x, add the additive inverse (or opposite) of y to x. That is, change the subtrahend to its opposite and add.

Example:

4- 9

= 4+( - 9) = - 5

Subtracting Signed Numbers

Step 1

Change the subtraction symbol to an addition symbol, and change the sign of the subtrahend.

Step 2

Add the signed numbers.

l!f$JMQ11il

Subtracting Signed Numbers

Find each difference. (a)

12 - 3 =

12

No change __j

-r Change + (-3) L

5-7

Additive inverse of 3

-r Change -

= 5 + (-7) L

No change _____1'

= -2 (c)

-8 - 15

+.

12 has the greater absolute value, so the sum is positive.

=9 (b)

to

- 7 has the greater absolute value, so the sum is negative.

= -8 + (-15) L = -23

+.

Additive inverse of 7

-r Change -

No change ___j

to

to

+.

Additive inverse of 15

The sum of two negative numbers is negative.

56

CHAPTER 1

The Real Number System

- 3 - ( - 5)

~NOW TRY

(d)

Find each difference. (a) - 5 - ( - 11)

=- 3+5 No change _1' L

"7 EXERCISE 6

W

(b) 4 - 15

5

(c)

- 7- 3

(d) 5.25 - ( - 3.24)

(e)

Additive inverse of - 5

5 has the greater absolute value, so the sum is positive.

=2

1

Change - to + .

%- (-~) =~~ - (- !~) 15 40

32 40

=- +-

47 7 or 140 ' 40

Write equivalent fractions using the LCD, 40.

Definition of subtraction

Add the fractions. Write as a mixed number.

(f) - 8.75 - ( - 2.41 ) =

- 8.75

+ 2.41

= -6.34

Definition of subtraction Add the decimals.

NOW

TRY~

Uses of the Symbol -

We use the symbol - for three purposes. 1. It can represent subtraction, as in 9 - 5 = 4. 2. It can represent negative numbers, such as -10, -2, and -3. 3. It can represent the additive inverse (or opposite) of a number, as in "the additive inverse (or opposite) of 8 is -8." We may see more than one use of - in the same expression, such as -6 - (-9), where -9 is subtracted from -6. The meaning of the symbol - depends on its position in the algebraic expression.

OBJECTIVE 4 Use the rules for order of operations when adding and subtracting signed numbers.

1!£f¥MIQlll

Using the Rules for Order of Operations

Perform each indicated operation. (a) -6- [2- (8+3) ]

NOW TRY ANSWERS 6. (a) 6

(b) - 11 22

I

(c) -21, or - 121

(d) 8.49

Work from the inside out.

= - 6 - [2 - 11 J

Add inside the parentheses.

= -6 - [ 2 + ( -11) J = -6 - [ -9] = -6 + 9

Definition of subtraction Add inside the brackets.

= 3

Add.

Definition of subtraction

SECTION 1.4

(b) 5

/:.NOW TRY

~EXERCISE 7 Perform each indicated operation. (a)

I4 -

- 1) ]

57

Work within each set of parentheses inside the brackets.

=5+ [ (-3 +(-2)) - 3]

= 5 + [ (-5) - 3]

8 - [ (-3 + 7) - (3 - 9) ]

(b) 3 I6 - 9 I -

+ [ (-3 - 2) - (4

Adding and Subtracting Real Numbers

=

12 I

5 + [(-5) + (-3) ]

Show all steps to avoid sign errors.

=5+ [ -8 ]

= -3

Write equivalent fractions using the LCD, 12.

Work inside the brackets.

8 12

-

-

4 12

-

Add inside the brackets.

4 12

Subtract.

1 3

Write in lowest terms.

-

(d)

14-7 1 +2 16 - 3 1

Be careful. Multiply first.

The absolute value bars serve as grouping symbols.

= l-3 1 +2 13 1

Work with in the absolute value bars.

=3+2·3

Find each absolute value.

=3+6

Multiply.

=9

Add.

NOW TRY~

OBJECTIVE 5 Translate words and phrases involving addition and subtraction. T Words and Phrases That Indicate Addition Word or Phrase

l:f!f!!MQll:I

Example

Sum of

The sum of - 3 and 4

Added to

5 added to - 8

More than

12 more than - 5

Increased by

- 6 increased by 13

Plus

3 plus 14

Numerical Expression and Simplification

+ 4, which equals 1 + 5, which equals - 3 - 5 + 12, which equals 7 - 6 + 13, which equals 7 3 + 14 , which equals 17 - 3

- 8

Translating Words and Phrases (Addition)

Write a numerical expression for each phrase, and simplify the expression. NOW TRY ANSWERS 7. (a) - 2

(b) I

(a) The sum of -8 and 4 and 6 (Add inorder from lefttoright.>

- 8 + 4 + 6 simplifies to

- 4 + 6, which equals

2.

58

CHAPTER 1

The Real Number System

~NOW TRY

(b) 3 more than - 5, increased by 12

"7 EXERCISE 8 Write a numerical expression for the phrase, and simplify the expression.

(-5 + 3) + 12 simplifies to

-2

+ 12, which equals 10. NOW TRY~

Here we simplified each expression by performing the operations. T Words and Phrases That Indicate Subtraction

The sum of -3 and 7, increased by 10

Word, Phrase, or Sentence

Numerical Expression and Simplification

Example The difference of - 3 and - 8

Difference of

- 3 - ( - 8) simplifies to

- 3 + 8, which equals 5 Subtracted from*

12 subtracted from 18

18 - 12, which equals 6

From . ... subtract . . . .

From 12, subtract 8.

12 - 8 simplifies to 12 + (- 8), which equals 4

Less

6 /ess 5

6 - 5, which equals 1

Less than*

6 /ess than 5

5 - 6 simplifies to 5 + ( - 6), which equals - 1

Decreased by

9 decreased by -4

9 - ( - 4) simplifies to 9 + 4,

Minus

8 minus 5

8 - 5 , which equals 3

which equals 13

* Be careful with order when translating.

0

CAUTION In general ,

When subtracting two numbers, be careful to write them in the correct order.

x - y

'* y -

x.

For example, 5 - 3 # 3 - 5 .

Think carefully before interpreting an expression involving subtraction.

~NOWTRY

1£f!1MQl#I Translating Words and Phrases (Subtraction)

Write a numerical expression for each phrase, and simplify the expression.

Write a numerical expression for each phrase, and simplify the expression.

"7 EXERCISE 9

(a) The difference of 5 and - 8, decreased by 4 (b) 7 less than -2

(a) The difference of - 8 and 5

When "difference of" is used, write the numbers in the order given. - 8 - 5 simplifies to

- 8 + (-5),

which equals

- 13.

(b) 4 subtracted from the sum of 8 and -3 Here the operation of addition is also used, as indicated by the words sum of First, add 8 and -3. Next, subtract 4 from this sum. [8

+ ( -3) ] - 4 simplifies to 5 - 4, which equals 1.

(c) 4 less than -6 Here, 4 must be subtracted from -6, so write -6 first. Be careful with order.

-6 - 4

simplifies to

-6

+ (-4 ), which equals -10.

Notice that "4 less than -6" differs from "4 is less than -6." The second of these is symbolized 4 < -6 (which is a false statement). NOW TRY ANSWERS 8. (-3 + 7) + IO; 14

9. (a) [5 - (-8) ] - 4; 9

(d) 8, decreased by 5 less than 12 First, write "5 less than 12" as 12 - 5. Next, subtract 12 - 5 from 8.

(b) - 2 - 7; -9

8 - ( 12 - 5)

simplifies to

8 - 7, which equals

1. NOW TRY~

SECTION 1.4

/:.NOW TRY

~EXERCISE 10 Find the difference between a gain of 226 yd on the football field by the Chesterfield Bears and a loss of 7 yd by the New London Wildcats.

Adding and Subtracting Real Numbers

59

1!£f!1MQ11m Solving an Application Involving Subtraction The record-high temperature in the United States is 134°F, recorded at Death Valley, California, in 1913. The record low is - 80°F, at Prospect Creek, Alaska, in 1971. See FIGURE 14. What is the difference between these highest and lowest temperatures? (Data from National Climatic Data Center.) 134° We must subtract the lowest temperature from the highest temperature. Difference is 134° - (-80°).

134 - ( - 80)

Order of numbers matters in subtraction.

= 134 + 80 =

Definition of subtraction

214

Add.

-80°

The difference between the two temperatures is 214°F .

FIGURE 14

NOWTRY

~

OBJECTIVE 6 Use signed numbers to interpret data. /:.NOW TRY

~EXERCISE 11 Refer to FIGURE 15 and use a signed number to represent the change in enrollment from 1985 to 1990.

U£f4MQllll Using a Signed Number to Interpret Data The bar graph in FIGURE 15 shows public high school (grades 9-12) enrollment in millions of students for selected years from 1985 to 2015.

I

Public High School Enrollment

14.91 14.86 15.01 'iii" c

~ .E

c = ~ c

Q)

11

"O

....

:I

en

1985 1990

1995 2000 2005

2010

2015

Year Data from U.S. National Center for Education Statistics. FIGURE 15

(a) Use a signed number to represent the change in enrollment in millions from 2000 to 2005. Start with the number for 2005. Subtract from it the number for 2000. 2005

2000

14.91

13.52

t

t

= + 1.39 million students

A positive number indicates an The bar for 2005 is "higher" than the bar for 2000.

-E- increase.

(b) Use a signed number to represent the change in enrollment in millions from 2005 to 2010. Start with the number for 2010. Subtract from it the number for 2005. 2010 NOW TRY ANSWERS 10. 233 yd

2005

t

t

14.86

14.91

=-

0.05 million students

-E-

A negative number indicates a decrease. The bar for 201 O is "lower" than the bar for 2005.

11. - 1.05 mill ion stude nts

NOWTRY

~

60

CHAPTER 1

The Real Number System

1.4 Exercises O Video solutions f or select problems available in Mylab Math STUDY SKILLS REMINDER How are you doing on your homework? Review Study Skill 4, Completing Your Homework.

FOR

EXTRA

HELP

0

MyLab Math

Concept Check

Complete each of the following .

1. The sum of two negative numbers will always be a (positive I negative) number. Give a number-line illustration using the sum - 2 + ( - 3). 2. The sum of a number and its opposite will always be _ _ __ 3. When adding a positive number and a negative number, where the negative number has the greater absolute value, the sum will be a (positive I negative) number. Give a number-line illustration using the sum - 4 + 2.

4. To simplify the expression 8 + [ -2 + ( - 3 + 5) J, one should begin by adding _ _ __ and

, according to the rules for order of operations.

5. By the definition of subtraction, in order to perform the subtraction -6 - (-8), we must add the opposite of to to obtain _ _ __

6. "The difference of 7 and 12" translates as _ _ _ _ , while "the difference of 12 and 7" translates as _ _ __

Concept Check Suppose that x represents a positive number and y represents a negative number. Determine whether the given expression represents a positive or a negative number. 7. x- y

8. y - x

9. y-

lx l

10. x +

IYI

Find each sum. See Examples 1-7.

11. -6 + (-2)

12. -9 + ( - 2)

13. -5 + (- 7)

14.-1 1 +(-5)

15. 6 + ( -4)

16. 11 + (-8 )

17. 12 + (-15)

18. 3 + ( - 7)

19. -16 + 7

20. - 13 + 6

21. 6 + ( - 6)

22. - 11 + 11

2-)

23. - _!_ + ( 3 15

24. - _!_ + ( 4 12

1 2 25. - - + 6 3

6 19 26. - - + 25 20

5 ( 27. 3+

9 ( 28. 10+

_±_)

2~ + ( -3±)

17) - 12

~

11)

-3

30. 1 + ( - 2 ±)

31. - 3.5 + 12.4

32. -12.5 + 21.3

33. -2.34 + (- 3.67)

34. -1.25 + ( -6.88)

35. 4 + [ 13 + ( -5 ) ]

36. 6 + [ I 2 + ( - 3) J

37. 8 + [ -2 + (-1 ) J

38. 12 + [ -3+(-4) ]

39. -2 + [5 + ( -1 )]

40. -8 + [9 + (-2) ]

41. -6 + [6 + (-9) ]

42. - 3 + [ 3 + ( - 8) J

43. [ (-9) + ( - 3) ] + 12

44. [(-8) + ( -6) ] + 14

45. -6. 1 + [3.2 + ( -4.8) ]

46. -9.4 + [5.8 + (- 7.9)]

29.

47. [ - 3 + ( - 4)] + [5 + (- 6) ]

48. [ - 8 + (- 3) ] + [4 + (- 6) ]

49. [ - 4 + ( - 3)] + [8 + (- 1) ]

50. [ - 5 + (- 9) ] + [ 16 + (- 2) ]

51. [ - 4 + ( - 6)] + [ - 3 + ( - 8)] + [ 12 + (- 11) ] 52. [-2 + (-11) ] + [- 12 + (-2) ] + [ 18 + (- 6) ]

61

Add ing and Su btracting Real Numbers

SECTION 1.4

Find each difference. See Examples 1-7. 53. 4 - 7

54. 8 - 13

55. 5 - 9

56. 6 - 11

57. - 7 -

58. - 9 - 4

59. - 8 - 6

60. - 9 - 5

61. 7 - ( - 3)

62. 9 - ( - 2)

63. - 6 - ( - 2)

64. - 7 - ( - 5)

65. 2 - (3 - 5)

66. - 3 - (4 - 11 )

67.

3 5 69. - - - 4 8

5 1 70. - - - 6 2

lli)

68.

~-

71.

%- (- ~ - ~)

(-

72.

74. 5.7 - ( - 11.6)

1- 10

l

(-!-]__) 8 10

~ - (- ±)

73. 3.4 - (-8.2) 76. - 4.4 - 8.6

75. - 6.4 - 3.5

Perform each indicated operation. See Ex amples 1-7.

+4

77. (4 - 6)+ 12

78. (3 - 7)

80. (9 - 3) - 15

81. 6 - ( - 8+3)

83. 2

+ (- 4 - 8)

86. I - 4

84. 6

82. 8 - ( - 9

21- I - 9 - 3 I 90. ( -

91. ( -

~ + 0.25) -

(-

~ + 0.75)

93. - 9 + [ (3 - 2) - (- 4 95. -3

+ 2) ]

92.

88. I - 4 -

97. -9.1 2 + [ (-4.8 - 3.25) + 11.279 ]

+ 21

21- I- 8 - 11

~ - %) - ( - ~ - 1)

(-%-

0.75 ) - ( 0.5 -

94. - 8 - [ (- 4 - 1)

+ [ (-5 - 8) - (-6 + 2) ]

+ 5)

85. I- 5 - 6 I + I9

+ ( - 9 - 2)

87. I - 8 -

+ 8 I + I6 - 1 I

79. (8 - 1) - 12

~)

+ (9 - 2) ]

96. -4 + [ ( - 12 + 1) - (- 1 - 9) ] 98. - 7.62 - [ (-3.99 + 1.427 ) - ( - 2.8) ]

Write a numerical expression for each phrase, and simplify the expression. See Examples 8 and9. 99. The sum of - 5 and 12 and 6

100. The sum of - 3 and 5 and -1 2

101. 14 added to the sum of - 19 and - 4

102. - 2 added to the sum of - 18 and 11

103. The sum of - 4 and - 10, increased by 12 104. The sum of - 7 and -13, increased by 14 2

5

9

105. 7 more than the sum of 7 and - 7

106. 1.85 more than the sum of - 1.25 and - 4.75

107. The difference of 4 and - 8

108. The difference of7 and - 14

109. 8 less than - 2

110. 9 less than -1 3

111. The sum of9 and - 4, decreased by 7

112. The sum of 12 and - 7, decreased by 14

113. 12 less than the difference of 8 and - 5

114. 19 less than the difference of 9 and - 2

62

CHAPTER 1

The Real Number System

The table gives scores (above or below par-that is, above or below the score "standard") for selected golfers during the 2017 Blue Bay LPGA Tournament. Write a signed number that represents the total score (above or below par) for the four rounds for each golfer. Round 1

Round2

Round3

Round4

115.

Shanshan Feng

Golfer

-3

-5

+1·

-2

116.

Jessica Korda

-1

0

-2

-1

117.

Pernilla Lindberg

-5

+7

-1

+2

118.

Brittany Lang

+3

+5

+2

-4

*Golf scoring commonly includes a Data from LPGA.

+ symbol with a score over par.

Solve each problem. See Example 10. 119. Based on 2020 population projections, Illinois will probably Jose 2 seats in the U.S. House of Representatives, Minnesota will lose 1 seat, and New York will lose 1. Write a signed number that represents the total number of seats these three states are projected to lose. (Data from Election Data Services.) 120. Both Alabama and Ohio are projected to Jose 1 seat in the U.S. House of Representatives in 2020. The states projected to gain the most seats are Texas with 4 and Florida with 2. Write a signed number that represents the algebraic sum of these changes. (Data from Election Data Services.) 121. The surface, or rim, of a canyon is at altitude 0. On a hike down into the canyon, a party of hikers stops for a rest at 130 m below the surface. The hikers then descend another 54 m. Write the new altitude as a signed number.

J30m

122. A pilot announces to the passengers that the current altitude of their plane is 34,000 ft. Because of turbulence, the pilot is forced to descend 2100 ft. Write the new altitude as a signed number.

34,000 ft 2100 ft

SECTION 1.4

Adding and Subtracting Real Numbers

123. The lowest temperature ever recorded in Arkansas was -29°F. The highest temperature ever recorded there was 149°F more than the lowest. What was this highest temperature? (Data from National Climatic Data Center.)

63

124. On January 23, 1943, the temperature rose 49°F in two minutes in Spearfish, South Dakota. If the starting temperature was -4°F, what was the temperature two minutes later? (Data from Guinness World Records.)

125. The lowest temperature ever recorded in Illinois was -36°F on January 5, 1999. The lowest temperature ever recorded in Utah was on February 1, 1985, and was 33°F lower than Illinois 's record low. What is the record low temperature for Utah? (Data from National Climatic Data Center.) 126. The lowest temperature ever recorded in South Carolina was - 19°F. The lowest temperature ever recorded in Wisconsin was 36° lower than South Carolina's record low. What is the record low temperature for Wisconsin? (Data from National Climatic Data Center.) 127. Nadine enjoys playing Triominoes every Wednesday night. Last Wednesday, on four successive turns, her scores were

-19, 28,

-5, and

13.

What was her final score for the four turns? 128. Bruce enjoys playing Triominoes. On five successive turns, his scores were

- 13,

15,

- 12,

24,

and

14.

What was his total score for the five turns? 129. In 2005, Americans saved -0.5 % of their after-tax incomes. In September 2017, they saved 3.1 %. (Data from U.S. Bureau of Economic Analysis.)

(a) Express the difference between these amounts as a positive number. (b) How could Americans have a negative personal

savings rate in 2005?

130. In 2000, the U.S. federal budget had a surplus of $236 billion. In 2016, the federal budget had a deficit of $616 billion. Express the difference between these amounts as a positive number. (Data from U.S. Office of Management and Budget.) 131. In 2006, bachelor's degree recipients from private four-year institutions had an average of $ 18,800 in total student debt. This average increased $1400 by 2011 and then dropped $300 by 2016. What was the average amount of total student debt in 2016? (Data from The College Board.)

132. The average annual spending per U.S. household on apparel and apparel services was $ 1786 in 2014. This amount increased $60 by 2015 and then decreased $43 by 20 16. What was the average household expenditure for apparel and apparel services in 2016? (Data from U.S. Bureau of Labor Statistics.)

64

CHAPTER 1

The Real Number System

133. In August, Susan began with a checking account balance of $904.89. She made the following withdrawals and deposits.

134. In September, Jeffery began with a checking account balance of $537.12. He made the following withdrawals and deposits.

Withdrawals

Deposits

Withdrawals

$35.84

$85.00

$41.29

Deposits $80.59

$26.1 4

$ 120.76

$13.66

$276.13

$84.40

$3.12

Assuming no other transactions, what was her new account balance?

Assuming no other transactions, what was his new account balance?

135. Linda owes $870.00 on her MasterCard account. She returns two items costing $35 .90 and $ 150.00 and receives credit for these on the account. Next, she makes a purchase of $82.50 and then two more purchases of $10.00 each. She makes a payment of $500.00. She then incurs a finance charge of $37.23. How much does she still owe? 136. Marcial owes $679.00 on his Visa account. He returns three items costing $36.89, $29.40, and $ 113.55 and receives credit for these on the account. Next, he makes purchases of $135.78 and $4 12.88 and two purchases of $20.00 each. He makes a payment of $400. He then incurs a finance charge of $24.57. How much does he still owe?

The graph shows annual returns in percent for Class A shares of the Invesco S&P 500 Index Fund from 2012 to 2016. Use a signed number to represent the change in percent return for each period. See Example 11. ( S&P 500 Index Fund Annual Returns 31.69 30

....c: Q) ()

...

&.

25 20 15 10 5 0

2012

2013

2014

2015

2016

Year Data from Invesco.

137. 2012 to 2013

138. 2013 to 2014

139. 2014 to 2015

140. 2012 to 20 16

The two tables show the heights of some selected mountains and the depths of some selected trenches. Use the information given to answer each question. Mountain

Height (in feet, as a positive number)

Trench

Depth (in feet, as a negative number)

Foraker

17,400

Philippine

- 32,995

Wilson

14,246

Cayman

- 24,721

Pikes Peak

14,110

Java

- 23,376

Data from The World Almanac and Book of Facts.

141. What is the difference between the height of Mt. Foraker and the depth of the Philippine Trench?

SECTION 1.5

Mu ltiplying and Dividing Real Numbers

65

142. What is the difference between the height of Pikes Peak and the depth of the Java Trench? 143. How much deeper is the Cayman Trench than the Java Trench? 144. How much deeper is the Philippine Trench than the Cayman Trench? 145. How much higher is Mt. Wilson than Pikes Peak? 146. If Mt. Wilson and Pikes Peak were stacked one on top of the other, how much higher would they be than Mt. Foraker?

Multiplying and Dividing Real Numbers OBJECTIVES

1 Find the product of a positive number and a negative number.

2 Find the product of two negative numbers.

The result of multiplication is a product. We know that the product of two positive numbers is positive. We also know that the product of 0 and any positive number is 0, so we extend that property to all real numbers. Multiplication Property of 0

For any real number x, the following hold true.

3 Identify factors of

x · 0 = 0 and 0 · x

integers.

=0

4 Use the reciprocal of a number to apply the definition of division.

5 Use the rules for order of operations when multiplying and dividing signed numbers.

OBJECTIVE 1 Find the product of a positive number and a negative number.

Observe the following pattern.

6 Evaluate algebraic

=

15

3.4

=

12

=9 3. 2 = 6 3. 3

expressions given values for the variables.

7 Translate words and

The products decrease by 3.

3. 1 = 3

phrases involving multiplication and division.

3. 0

=

0

3· (- 1) = ?

8 Translate simple sentences into equations.

3. 5

What should 3 · ( - 1) equal? The product 3 · ( -1 ) represents the sum

-1 + (-1) + (-1),

which equals

-3,

so the product should be -3. Also, 3 · (-2) and 3 · (-3) represent the sums

and

-2

+ ( -2) + ( -2), which equals

-6

-3

+ (-3) + (-3),

-9.

which equals

These results maintain the pattern in the list and suggest the following rules.

66

CHAPTER 1

The Real Number System

VOCABULARY D D D D D D

product factor multiplicative inverse (reciprocal) quotient dividend divisor

i:.NOWTRY EXERCISE 1

\:7

Find each product. (a) - 11(9)

(b) 3.1( - 2.5)

Multiplying Signed Numbers (Different Signs)

For any positive real numbers x and y , the following hold true.

x(-y)= -(xy) and (-x)y= -(xy) That is, the product of two numbers with different signs is negative. Examples:

li!f4MQlll

6(-3) = -18

and

( -6)3 = -18

Multiplying Signed Numbers (Different Signs)

Find each product. (a) 8( - 5)

(b)

= - (8·5)

= - 40

-9(1) =-(9·l)

The product of two numbers with different signs is negative.

(c) -6.2(4.1 )

= -(6.2. 4.1) = -25.42 NOW TRY

= -3

g

OBJECTIVE 2 Find the product of two negative numbers. Look at another pattern.

= -20

-5 (4)

-5 (3) = -15 -5 (2)

=

-10

- 5(1)

=

-5

- 5(0)

=0

The products increase by 5.

-5 (- 1) = ? The numbers in color on the left side of the equality symbols decrease by 1 for each step down the list. The products on the right increase by 5 for each step down the list. To maintain this pattern, -5 ( -1) should be 5 more than -5 ( 0), or 5 more than 0, so - 5 ( - 1) = 5. The pattern continues with -5 ( -2 ) = 10 -5 (-3) = 15

= 20

-5 ( -4 )

- 5(-5 ) = 25,

and so on.

These results suggest the next rule. Multiplying Two Negative Numbers

For any positive real numbers x and y , the following holds true.

-x(-y) = xy That is, the product of two negative numbers is positive. NOW TRY ANSWERS 1. (a) - 99 (b) - 7.75

Example:

- 5(-4)

= 20

li:tlJMQlfJ Multiplying Two Negative Numbers

i:.NOWTRY

~EXERCISE2 Find each product. (a) -8 ( -11) (b)

67

Multiplying and Dividing Real Numbers

SECTION 1.5

Find each product. (a) - 9( -2)

-~( -%)

(b)

-~( -%)

= 18

=

(c) - 0.5 (- 1.25)

= 0.625

1

The product of two numbers with the same sign is positive.

NOWTRY

~

The following box summarizes multiplying signed numbers. Multiplying Signed Numbers

The product of two numbers with the same sign is positive. The product of two numbers with different signs is negative. Examples:

7 (3) = 21,

-7(-3 ) = 21 ,

-7(3) = -21, and 7 (- 3) = -21

OBJECTIVE 3 Identify factors of integers. The definition of factor can be extended to integers. If the product of two integers is a third integer, then each of the two integers is a factor of the third. T Integer Factors Integer

Pairs of Factors

18

20

15

7

1

1, 18

1, 20

1, 15

1, 7

1, 1

- 1, - 7

- 1, - 1

2, 9

2, 10

3, 6

4, 5

- 1, - 18

- 1, - 20

- 2, -9

- 2,- 10

-3, - 6

- 4,-5

3,5 - 1, - 15 - 3 ,-5

T Reciprocals Reciprocal (Multiplicative Inverse)

Number

The definition of division depends on the idea of a reciprocal, or multiplicative inverse, of a number.

1

4

4 3

10

0.3, or 10

3

-5

1

1

- s •or - 5

5

8 5

-3

A number and its reciprocal have a product of 1. For example, 4 ·

Reciprocals, or Multiplicative Inverses

Pairs of numbers whose product is 1 are reciprocals, or multiplicative inverses, of each other.

±= 1.or I.

0 has no reciprocal because the product of 0 and any number is 0, not I.

NOW TRY ANSWERS 2. (a) 88

OBJECTIVE 4 Use the reciprocal of a number to apply the definition of division.

5

(b) 14

The table in the margin gives examples of reciprocals. Recall that the answer to a division problem is a quotient. For example, we can write the quotient of 15 and 3 as 15 -:- 3, which equals 5. We obtain the same answer if we multiply 15 · ~ , the reciprocal of 3. This discussion suggests the next definition.

68

CHAPTER 1

The Real Number System

Definition of Division

For any real numbers x and y, where y # 0, the following holds true. 1 y

x+y=x·That is, to divide two numbers, multiply the first number (the dividend) by the reciprocal, or multiplicative inverse, of the second number (the divisor). 15 -:- 3

Example:

= 15

1 . - =5 3

Recall that an equivalent form of x -:- y is ~, where the fraction bar represents division. In algebra, quotients are usually repr~sented with a fraction bar. 15 -:- 3

EmD

is equivalent to

15 3

The following all represent division of x by y, where y ¥- 0.

x -o- y , ~. y Example:

15 -o- 3,

~5 ,

15 / 3,

and

x / y, 3)15

and

y}x

are equivalent forms of 15 divided by 3.

Because division is defined in terms of multiplication, the rules for multiplying signed numbers also apply to dividing them. Dividing Signed Numbers

The quotient of two numbers with the same sign is positive. The quotient of two numbers with different signs is negative. Examples:

I::. NOW TRY

~EXERCISE 3 Find each quotient.

- 10 (a) - 53

(c)

- 1.44 (b) - 0.12

1£f;1MQl#I

15

- 15

Dividing Signed Numbers

8 2

(a) _

(d)

=

-4

- 100

(b) - - = - 20 5

- 4.5

(c) - - = 50 - 0.09

-i -:- (- ~)

=-i·(-1) NOW TRY ANSWERS 15 3. (a) - 2 (b) 12 (c) -23

- 15

Find each quotient.

7

-8-o--10

15

= 5 - = - 5 and - = - 5 -=5 3 , --3 -3 , 3

1

6

Multiply by the reciprocal of the divisor. Multiply the fractions. Write in lowest terms.

NOW TRY

g

Multiplying and Dividing Real Numbers

SECTION 1.5

69

. C ons1"der the quotient 312 ·

12 = 4 3

-

4 .3

because

Multiply to check a division problem.

= 12.

Using this relationship between multiplication and division, we investigate division by 0. Consider the quotient~.

0.3

because Now consider

= 0.

5. 3

-= ?

0

.

We need to find a number that when multiplied by 0 will equal 3, that is, ? · 0 = 3. No real number satisfies this equation because the product of any real number and 0 must be 0. Thus,

~ is not a number, and division by 0 is undefined. If a division problem involves division by O, write "undefined."

Division Involving 0

For any real number x, where x ¥= 0, the following hold true.

0

=0

~

Examples:

_O_=O _ 10

x.

0 1s undefined.

and

--10 - is undefined.

and

0

From the definitions of multiplication and division of real numbers,

-40 - = -5 8

and

40 -8

-=

-5 , so

-40

40

8

-8

- - = -.

Based on this example, the quotient of a positive number and a negative number can be expressed in different, yet equivalent, forms. Also,

-40 -= 5

-8

and

40

8= 5,

so

-40

-8

40 8

Equivalent Forms

For any positive real numbers x and y, the following are equivalent.

-x

x

y

-y '

-x -y

x

and and

y

Step2

16 - 12x = 32 - 8x - 16

Distributive property

16 - 12x = 16 - 8x

Combine like terms.

16 - 12x + 12x = 16 - 8x + 12x Adding 12x keeps a positive coefficient on x.

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

Step 3

Add 12x.

Combine like terms. Subtract 16.

0 = 4x

Combine like terms.

0 4

Divide by 4.

4x

4

O =x

Step 4

CHECK

4(4 - 3x )=32-8(x +2) 4 [ 4- 3(0 ) ] ~ 32 - 8(0 + 2) 4 (4 - 0) ~ 32 - 8(2)

4(4)~32-16 16 = 16

./

Original equation Let x = 0. Multiply and add . Subtract and multiply. True

A true statement results, so the solution set is { 0} .

m!lDI

NOW TRY

g

It is perfectly acceptable for an equation to have solution set { 0}.

OBJECTIVE 2 Solve equations with no solution or infinitely many solutions. Each equation so far has had exactly one solution. An equation with exactly one solution is a conditional equation because it is true only under certain conditions. Some equations may have no solution or infinitely many solutions.

lf:fi!MQllil

Solving an Equation That Has Infinitely Many Solutions

Solve 5x - 15

= 5(x -

3). 5x - 15

= 5(x -

3)

5x - 15 = 5x - 15 5x - 15 - 5x = 5x - 15 - 5x Notice that the variable "disappeared."

-

= -15 15 + 15 = -15 + 15 -15

O=O NOW TRY ANSWER 5. {O}

Solution set:

Distributive property Subtract 5x. Combine like terms. Add 15.

True

{all real numbers}

122

CHAPTER 2

Linear Equations and Inequalities in One Variable

~NOWTRY

Because the last statement (0 = 0) in Example 6 is true, any real number is a solution. We could have predicted this from the second line in the solution.

~EXERCISES Sol ve.

-3(x - 7)

=

5x - 15 = 5x - 15

-3x + 21

a }

I

{x lx2a }

/:.NOW TRY EXERCISE 2

[a, oo)

L

a

"7

Solve the inequality, and graph the solution set. 5

(a, oo)

'a

( - oo, oo)

{x lx is a real number }

+ 5x 2: 4x + 3

1!£MMQl#j Using the Addition Property of Inequality Solve 7

+ 3x :::::: 2x - 5, and graph the solution set.

+ 3x:::::: 2x -

7

7

+ 3x 7

As with equations, our goal is to isolate x.

5

2x :::::: 2x - 5 - 2x

+ x:::::: -5

Combine like terms.

7 + x - 7 :::::: -5 - 7

Subtract 7.

x :::::: - 12

NOW TRY ANSWER 2· I I [ I I) - 4 -3 - 2 - 1 0

[-2, oo)

I

2

- 13 - 12 - 11 - 10

-9

-8

Subtract 2.x.

-7

Combine like terms. -6

-5

-4

-3

-2

-1

0

FIGURE24

The solution set is graphed in

FIGURE 24

and written [ - 12, oo).

NOW TRY~

SECTION 2.9

Solving Linear Inequalities

185

miID

Because an inequality has many solutions, we cannot c heck all of them by substitution as we did with the single solution of an equation. To check the solutions in the interval [ - 12, oo) in Example 2, we first substitute - 12 for x in the related equation .

CHECK

7+3x = 2x - 5 7

+ 3( -

Related equation

12) J, 2( - 12) - 5

7 - 36 J, -24 - 5 -29 = -29

./

Letx = - 12. Multiply. True

A true statement results, so - 12 is indeed the "boundary" point. Next we test a number other than - 12 from the interval [ - 12, oo). We choose 0.

CHECK

7 + 3x ,.----.,7 + 3(0 ) Ois easy to substitute.

2! ?

~

2x - 5

Original inequality

2( 0 ) - 5

Letx = 0.

?

7 + 0 ?! 0 - 5 7

2!

Multiply.

-5 ./

True

Again, a true statement results. The checks confirm that solutions to the inequality are in the interval [ - 12, oo). Any number "outside" the interval [ - 12, oo ) - that is, any number in ( -oo, - 12)-will give a false statement w hen tested. (Try this with - 13 . A false statement, - 32 2! -31, results.)

OBJECTIVE 3 Use the multiplication property of inequality. Consider the true inequality 3 < 7. Multiply each side by the positive number 2.

3< 7

2(3) < 2(7) 6

< 14

Multiply each side by 2. True

The result is a true statement. Now multiply each side of 3 < 7 by the negative number - 5.

3< 7 - 5(3) < - 5(7) -15 < -35

Multiply each side by -5. False

To obtain a true statement when multiplying each side by -5, we must reverse the direction of the inequality symbol.

3

-5 (7)

-15 > - 35

Multiply by -5. Reverse the direction of the symbol. True

miID

The above illustrations began with the inequality 3 < 7, a true statement involving two positive numbers. Similar results occur when one or both of the numbers are negative. Verify this by multiplying each of the following inequalities first by 2 and then by -5. -3< 7, 3>-7,

and

-7 b, and a

b.

2:

As with the multiplication property of equality, the same nonzero number may be divided into each side of an inequality.

Note the following differences for positive and negative numbers.

1. When each side of an inequality is multiplied or divided by a positive number, the direction of the inequality symbol does not change. 2. When each side of an inequality is multiplied or divided by a negative number, reverse the direction of the inequality symbol.

li:f4MQl#I Using the Multiplication Property of Inequality Solve each inequality, and graph the solution set. (a) 3x < -18 We divide each side by 3, a positive number, so the direction of the inequality symbol does not change. (It does not matter that the number on the right side of the inequality is negative.)

3x < -18

- 18 3

3x

3 is positivie. Do NOT reverse the direction of the symbol.

- < -3

Divide by 3.

x < -6

-7

-6

-5

-4

-3

-2

- 1

0

FIGURE 25

The solution set is graphed in

FIGURE 25

and written (- oo, -6).

(b) -4x:::::: 8

Here, each side of the inequality must be divided by -4, a negative number, which does require changing the direction of the inequality symbol. -4x:::::: 8 -4 is negative. Change ;,,, to ,,; .

-4x 8 -4 - -4

-- < x::::;

- 2

To avoid errors, show the division as a separate step.

Divide by - 4. Reverse the symbol.

SECTION 2.9

Solving Linear Inequalities

187

~NOW TRY

~EXERCISE3 Solve the inequality, and graph the solution set.

-Sk

2'.:

-6

-4

-5

-2

-3

0

- I

2

FIGURE 26

The solution set is graphed in FIGURE 26 and written ( -

15

oo,

- 2 J.

NOWTRY

~

OBJECTIVE 4 Solve linear inequalities using both properties of inequality. Solving a Linear Inequality in One Variable

Step 1

Simplify each side separately. Use the distributive property as needed. • Clear any parentheses. • Clear any fractions or decimals. • Combine like terms.

Step 2

Isolate the variable terms on one side. Use the addition property of inequality so that all terms with variables are on one side of the inequality and all constants (numbers) are on the other side.

Step 3

Isolate the variable. Use the multiplication property of inequality to obtain an inequality in one of the following forms, where k is a constant (number). variable < k,

variable

:5

variable > k,

k,

variable ;::: k

or

Remember: Reverse the direction of the inequality symbol only when multiplying or dividing each side of an inequality by a negative number.

li!f!1MQlll Solving a Linear Inequality

~NOW TRY

~EXERCISE4 Solve the inequality, and graph the solution set. 6 - 2x

+ 5x < 8x -

Solve 3x + 2 - 5 > -x + 7 + 2x, and graph the solution set. Step 1

Simplify by combining like terms.

4

3x + 2 - 5 > - x + 7 + 2x 3x - 3 > x + 7 Step 2

Isolate the variable term using the addition property of inequality.

3x - 3 - x > x + 7 - x 2x-3 > 7

Combine like terms.

2x-3 + 3 >7 +3

Add3.

2x > 10 Step 3

2x

10

2

2

- >-

Divide by 2.

x>S

NOW TRY ANSWERS EI I I ]

Combine like terms.

Isolate the variable using the multiplication property of inequality. Because 2 is positive, keep the symbol >.

3·

Subtract x.

I I I•

- 6 -5 - 4 -3 -2 - 1 0

(-oo, - 3] 4.

0

I I I ( I -1

0

(2, oo)

1 2

I)

2

3

4

5

6

7

8

9

10

FIGURE 27

3 4

The solution set is graphed in FIGURE 21 and written (5,

oo ).

NOW TRY~

188

CHAPTER 2

Linear Equations and Inequalities in One Variable

l!:m. NOW TRY

~EXERCISE 5 Solve the inequality, and graph the solution set.

2t - 3(t - 6) :::; 4(t

+ 7)

1!£MMQl4j Solving a Linear Inequality Solve 5(z - 3) - 7z

4(z - 3)

2:

~ 5(z - 3) - 7z

Step 1

15 - 7z - 2z - 15 -2z - 15 - 4z -6z - 15 -6z - 15 + 15 5z -

Step2

-6z Step3

-7

2:

~ 4(z - 3)

2:

4z -

2:

4z - 3

Combine like terms.

2:

4z - 3 - 4z

Subtract 4z.

2:

-3

Combine like terms.

2:

-3

12 12 -

2:

Start by clearing parentheses.

+9 +

12 9

Distributive property

15

+

Add 15. Combine like terms.

-6z -6 - -6

--
6(2x + 8)

l:J:MMQllil Solving a Linear Inequality (Fractional Coefficients) Solve ~(x - 6)

< ~(5x + 1), and graph the solution set.

l (x _ 6 ) < .?_ ( 5x + l )

S teP 1

4

3 3 9 10 -x - - < - x 4 2 3

Clear the parentheses first. Then clear the fractions.

2 3

+-

Distributive property

12(~x - ~) < 12(13°x + ~)

Multiply each side by the LCD, 12.

12(~x) + 12(-~) < 12(13°x) + 12(~) < 40x + 8 9x - 54 - 40x < 40x + 8 -3lx - 54 < 8 - 3 lx - 54 + 54 < 8 + 54 -3lx < 62 9x - 54

Step2

5·

-3lx - 31

I I EI I I I) -4 -3 - 2 -1

0

I

Multiply.

40x

Subtract 40x. Combine like terms. Add 54. Combine like terms.

62 - 31

Divide by - 31. Reverse the symbol.

-- > -

Step3

NOW TRY ANSWERS

Distributive property

x> -2

2

[-2, oo) 6·

EI I )

I I I I•

-5

-4

-3

-2

- l

0

2

3

4

5

-6 -5 -4 -3 -2 - 1 0

(-oo, - 4)

FlGURE29

The solution set is graphed in FIGURE 29 and written ( -2, oo) .

NOW TRY~

SECTION 2.9

Solving Linear Ineq ualities

189

OBJECTIVE 5 Solve applied problems using inequalities. T Words and Phrases That Indicate Inequality Word or Phrase

0

CAUTION

Example

Inequality

Is more than

A number is more than 4

x >4

Is less than

A number is less than -1 2

x < -1 2

Exceeds

A number exceeds 3.5

x > 3.5

Is at least

A number is at least 6

Is at most

A number is at most 8

x "" 6 x :S 8

Do not confuse a statement such as "5 is more than a number," expressed as

5 > x , with the phrase "5 more than a number," expressed as x + 5 or 5 + x .

The next example uses the idea of finding the average of a number of scores. To find the average of n numbers, add the numbers and divide by n. We continue to use the six problem-solving steps, changing Step 3 to "Write an inequality." ~NOWTRY

lif!fl1MQlll

Will has grades of 98 and 85 on his first two tests in algebra. If he wants an average of at least 90 after his third test, what possible scores may he make on that test?

John has grades of 86, 88, and 78 on his first three tests in geometry. If he wants an average of at least 80 after his fourth test, what possible scores may he make on that test?

~EXERCISE 7

Finding an Average Test Score

Step 1

Read the problem again.

Step 2

Assign a variable. Let x

Step 3

Write an inequality.

= John's score on his fourth test.

Average

is at least 80.

i

i i

+ 88 + 78 + x ~------::::: 80 86

To find his average after four tests, add the test scores and divide by 4.

4

Step 4

252 + x ---::::: 80 4

Solve.

4(

252 + 4 252

252

+x

Add in the numerator.

x) ::::: 4(80)

+ x::::: 320

- 252 ::::: 320 - 252 x:::=::: 68

NOW TRY ANSWER 7. 87 or more

Multiply by 4.

Subtract 252. Combine li ke terms.

Step 5

State the answer. He must score 68 or more on the fourth test to have an average of at least 80.

Step 6

Check.

86

+ 88 + 78 + 68

320

------ = -

4

4

=

80

To complete the check, also show that any number greater than 68 (but less than or equal to 100) makes the average greater than 80.

NOW TRY~

190

CHAPTER 2

Linear Equations and Inequalities in One Variable

0

CAUTION

In applied problems, remember that

and

is at least

translates as

is greater than or equal to

is at most

translates as

is less than or equal to.

OBJECTIVE 6 Solve linear inequalities with three parts. An inequality that says that one number is between two other numbers is a three-part inequality. For example,

- 3 < 5 < 7 says that 5 is between - 3 and 7.

/:m.NOWTRY

~EXERCISES Graph the inequality, and write it using interval notation.

O:s x< 2

lj!f4MQlj:i

Graphing a Three-Part Inequality

Graph the inequality - 1 :::; x < 3, and write it using interval notation. The statement is read "- 1 is less than or equal to x and x is less than 3." We want the set of numbers between -1 and 3, with -1 included and 3 excluded. We use a square bracket at -1 because -1 is part of the graph and a parenthesis at 3 because 3 is not part of the graph. See FIGURE 30.

-3

- I is included.

3 is excluded.

t

t

-2

2

0

- I

3

4

5

Graph of the interval [-1, 3) FIGURE30

NOW TRY~

The interval is written [ - 1, 3 ).

The three-part inequality 3 < x + 2 < 8 says that x + 2 is between 3 and 8. We solve this inequality by working with all three parts at the same time.

3 -2 < x+2 -2 < 8 -2 1
5

15. - - :s x 2

3 16. - - :s x 4

18. 1 ;::::: x

19. - 2 < x

20. - 1 < x

13. t

17. 0;::::: x

1

12. r < -1 1

Solve each inequality. Graph the solution set, and write it using interval notation. See Example2. 21. z - 8;::::: -7 24. 3x + 7 ;::::: 2x

22. p - 3 ;::::: -1 1

+ 11

25. 3n

+5
49

38. -k < 0

3 39. - -4 r < - 15

41. - 0.02x s 0.06

42. - 0.03v

~

~

- 0.12

Solve each inequality. Graph the solution set, and write it using interval notation. See Examples 4-6.

43. 8x

+ 9 s - 15

44. 6x + 7 s - 17

45. - 4x - 3 < 1

46. - 5x - 4 < 6

47. Sr + 1 ~ 3r - 9

48. 6t + 3 < 3t

49. 5x - 2 s - x 51. - 7x

+ 10

+ 4 > - 3x -

50. 3x - 9

~

+ 12 - 2x + 6

52. - 8x + 1 < - 4x

2

+ 11

53. 6x

+ 3 + x < 2 + 4x + 4 55. - x + 4 + 7x s - 2 + 3x + 6

54. - 4w + 12

57. 5 ( t - 1) > 3( t - 2)

58. 7 ( m - 2) < 4(m - 4)

59. 5 (x

+ 9w ~ w + 9 + w 56. 14x - 6 + 7x > 4 + lOx - 10

+ 3) - 6x s 3(2x + 1) - 4x

1

2

61.

3( 5x -

63.

3( p + 3) > 6( p -

4) ~ 5 (x

2

4 1 65. -x - - (x

5

2

67. 4x - (6x 69. 5 (2k

4)

3 10

+ 3) s -

+ 1) s 8x + 2(x - 3)

7

64.

9( x -

66.

6x + 3(x -

1

+ 4(z + 6)

~

2(3z

4) s

1

68. 2z - (4z

+ 3) - 2(k - 8) > 3(2k + 4) + k - 2

70. 2(3z - 5)

+ 3x > 4(x -

6)

+

1

5 5 62. U(5x - 7) < r;(x - 5)

+ 3)

5

60. 2(x - 5)

+ 2) + 3z - 15

4

3( x + 5) 1) >

1

2

+ 3) > 6z + 3(z + 4)

194

CHAPTER 2

Linear Equations and Inequalities in One Variable

Concept Check

Translate each statement into an inequality. Use x as the variable.

71. You must be at least 18 yr old to vote. 72. Less than 1 in. of rain fell.

73. Chicago received more than 5 in. of snow. 74. A full-time student must take at least 12 credits. 75. Tracy could spend at most $20 on a gift. 76. The car's speed exceeded 60 mph.

Solve each problem. See Example 7. 77. Christy has scores of 76 and 81 on her first two algebra tests. If she wants an average of at least 80 after her third test, what possible scores may she make on that test? 78. Joseph has scores of 96 and 86 on his first two geometry tests. What possible scores may he make on his third test so that his average is at least 90?

79. A student has scores of 87, 84, 95, and 79 on four quizzes. What scores may she make on the fifth quiz to have an average of at least 85?

80. Another student has scores of 82, 93, 94, and 86 on four quizzes. What scores may he make on the fifth quiz to have an average of at least 90? 81. The average monthly precipitation in Houston, Texas, for October, November, and December is 4.6 in. If 5.7 in. falls in October and 4.3 in. falls in November, how many inches must fal l in December so that the average monthly precipitation for these months exceeds 4.6 in.? (Data from National Climatic Data Center.)

82. The average monthly precipitation in New Orleans, Louisiana, for June, July, and August is 6.7 in. If 8.1 in. fall s in June and 5.7 in. falls in July, how many inches must fall in August so that the average monthly precipitation for these months exceeds 6.7 in.? (Data from National Climatic Data Center.) 83. When 2 is added to the difference of six times a number and 5, the result is greater than 13 added to five times the number. Find all such numbers. 84. When 8 is subtracted from the sum of three times a number and 6, the result is less than 4 more than the number. Find all such numbers.

85. The formula for converting Fahrenheit temperature to Celsius is C

=

5

- (F - 32). 9

If the Celsius temperature on a certain winter day in Minneapolis is never less than -25°,

how would we describe the corresponding Fahrenheit temperatures? (Data from National Climatic Data Center.)

86. The formula for converting Celius temperature to Fahrenheit is F =

9

5C+ 32.

The Fahrenheit temperature of Phoenix has never exceeded 122°. How would we describe this using Celsius temperature? (Data from National Climatic Data Center.)

SECTION 2.9

87. For what values of x would the rectangle have a perimeter of at least 400?

Solving Linear Ineq ualities

195

88. For what values of x would the triangle have a perimeter of at least 72?

~ 2x + 5

89. A phone call costs $2.00, plus $0.30 per minute or fractional part of a minute. If x represents the number of minutes of the length of the call, then 2 + 0.30x represents the cost of the call. If Alan has $5.60 to spend on a call , what is the maximum total time he can use the phone?

90. At the Speedy Gas ' n Go, a car wash costs $3.00 and gasoline is selling for $3.60 per gallon. Carla has $48.00 to spend, and her car is so dirty that she must have it washed. What is the maximum number of gallons of gasoline that she can purchase?

A company that produces DVDs has found that revenue from sales of DVDs is $5 per DVD, less sales costs of $100. Production costs are $125, plus $4 per DVD. Profit (P) is given by revenue (R) less cost (C), so the company must find the production level x that makes P

> 0, that is, R - C > 0.

P = R - C

91. Write an expression for revenue R, letting x represent the production level (number of DVDs to be produced).

92. Write an expression for production costs C in terms of x. 93. Write an expression for profit P, and then solve the inequality P > 0. 94. Describe the solution in terms of the problem. Concept Check Write a three-part inequality using the variable x that corresponds to each graph of solutions on a number line. - I

0

2

-I

0

2

0

2

[

I

- I

0

J2

- I

98.

Graph each inequality, and write it using interval notation. See Example 8.

99. 8 ::; x::; 10 102. - 3 ::; x < 0

100. 3 ::; x ::; 5

101. 0 < y ::; 10

103. 4 > x > - 3

104. 6 2: x

2:

-4

Solve each inequality. Graph the solution set, and write it using interval notation. See Example 9.

105. - 8 < 4x ::; 4

106. - 3::; 3x < 12

107. - 5 ::; 2x - 3 ::; 9

108. - 7::; 3x - 4 ::; 8

109. 10 < 7p + 3 < 24

110. - 8 ::; 3r - 1 ::; -1

111. 6::; 3(x - 1) < 18

112. - 4 < 2(x+ 1) :S6

I

113. - 12 ::; - z 2

+ 1 ::; 4

1 114. -6::; - x 3

+ 3 ::; 5

196

CHAPTER 2

Linear Equations and Inequalities in One Variable

115. 1 :s: 3

2

+-p 3

3 4

116. 2 < 6 + - x < 12

:s: 7

5 117. - 7 :s: - r - 1 :s: - 1

3 118. - 12 :s: - x

4

7

+ 2 :s:

- 4

Extending Skills Solve each inequality. Graph the solution set, and write it using interval notation. 119. -4 < - 2x < 12

120. 9 < - 3x < 15

121. 5 < I - 6m < I 2

122. - I :s: I - Sq :s: I 6

RELATING CONCEPTS For Individual or Group Work (Exercises 123-126)

Work Exercises 123-126 in order, to see the connection between the solution of an equation and the solutions of the corresponding inequalities. 123. Solve the following equation, and graph the solution set on a number line. 3x + 2

=

14

124. Solve the following inequality, and graph the solution set on a number line. 3x

+2>

14

125. Solve the following inequality, and graph the solution set on a number line. 3x

+2
,or 2: .

ANSWERS 1. C; Example: {8} is the solution set of 2x + 5

=

21.

2. B; Example: Angles with measures 35° and 55° are complementary angles.

3. C; Example: Angles with measures 112° and 68° are supplementary angles.

6. D; Examples: x < 5, 7

7 in.

7

2

8

4. A; Example: J2j;;:, or 12 5. B; Example: 3 = 12

+ 2y 2: 11, - 5 < 2z - 1 s 3

Quick Review CONCEPTS

EXAMPLES

DI The Addition Property of Equality The same number may be added to (or subtracted from) each side of an equation without changing the solution set.

x- 6

Solve.

x - 6

=

12

+ 6 = 12 + 6 x = 18

Add 6. Combine like terms.

Solution set: { 18}

DI The Multiplication Property of Equality Each side of an equation may be multiplied (or divided) by the same nonzero number without changing the solution set.

3 -x = - 9 4

Solve.

4

3

4

3 . 4x = 3 . (- 9 ) x

=

-12

Solution set: { -12}

Multiply

by ~. the reciprocal of ~-

198

CHAPTER 2

Linear Equations and Inequalities in One Variable

CONCEPTS

f!I Solving Linear Equations Using Both

EXAMPLES Solve.

Properties of Equality

2x + 2(x + I )

=

14

+x

2x + 2x + 2

=

14

+x

4x + 2

=

14

HI Clearing Fractions and Decimals When Solving Linear Equations Solving a Linear Equation in One Variable Step 1

4x + 2 - x - 2 =

Simplify each side separately.

• Combine like terms.

Step 3

Isolate the variable.

Step 4

Check.

x- 2

Subtract x. Subtract 2. Combine like terms.

12

3x 3

• Clear any fractions or decimals .

Isolate the variable terms on one side.

Combine like terms.

3x = 12

• Clear any parentheses.

Step 2

+x 14 + x -

Dist ribut ive property

Divide by 3.

3

x =4 CHECK

2(4)

+ 2(4 + 1) 1=

14 + 4

18 = 18

Let

.I

x = 4.

True

Solution set: { 4}

fD Applications of Linear Equations Solving an Applied Problem

One number is five more than another. Their sum is 21. What are the numbers? Let Then

x = the lesser number. = the greater number.

Step 1

Read.

Step2

Assign a variable.

x + (x + 5)

Step3

Write an equation.

2x

Step4

Solve the equation.

Steps

State the answer.

Step 6

Check.

x

+5

=

21

+5=

21

2x = 16

Subtract 5.

x =8

Divide by 2.

+5

The numbers are 8 and 8

13 is five more than 8, and 8

f!lll Formulas and Additional Applications

Combine like terms.

=

13.

+ 13 =

2 1. The answer checks.

Find L if .sl/. = LW, given that .sl/. = 24 and W = 3.

from Geometry To find the value of one of the variables in a formula, given values for the others, substitute the known values into the formula.

.sl/.

=

LW

24

=

L · 3

24

L · 3

3

3

.sl/. = 24, W = 3 Divide by 3.

8=L Solving a Formula for a Specified Variable (Solving a Literal Equation) To solve a formula for one of the variables, isolate that variable by treating the other variables as constants (numbers) and using the steps for solving equations.

Solve P = 2a

+ 2b for b.

P=2a+2b P-2a=2a+2b - 2a

Subtract 2a.

P-2a=2b

Combine li ke terms.

P -2a 2

Divide by 2.

2b 2

P- 2a - 2= b, or

P-2a b = --2

CONCEPTS

199

Summary

CHAPTER 2

EXAMPLES

lfll Ratio, Proportion, and Percent To write a ratio, express quantities using the same units.

4 ft to 8 in. can be written 48 in. to 8 in., which is the ratio 48

8 To solve a proportion, use the method of cross products.

6

, or

Solve.

x

35

12

60

60x

=

12 · 35

Cross products

60x

=

420

Multiply.

x=?

Divide by 60.

Solution set: {7 }

Percent, Fraction, and Decimal Equivalents 1 10 100 1 % = 100 = 0.01, 10% = 100 = 0.10, 100% = 100 = 1

5 5% = 100 = 0.05,

To solve a percent problem, use the percent equation.

percent (as a decimal) · base

35 35% = 100 = 0.35,

What percent

of

t

t

= amount

325

t

200 200% = 100 = 2 is

65?

t t

325

p

65 65 p = 325

p = 0 .2,

or p

=

20%

Thus, 20% of 325 is 65.

lfll Further Applications of Linear Equations Solving an Applied Problem Step 1

Read.

Step2

Assign a variable . Make a sketch, diagram, or table, as needed.

Two cars leave from the same point, traveling in opposite directions. One travels at 45 mph and the other at 60 mph. How long will it take them to be 210 mi apart? Lett= time it takes for the two cars to be 210 mi apart.

There are three forms of the formula relating distance, rate, and time. d =rt,

r

Step3

Write an equation.

Step 4

Solve the equation.

Step 5

State the answer.

Step 6

Check.

d = -, t

t

= d-r Rate

Time

Distance

Slower Car

45

I

451

Faster Car

60

I

601

45t

The sum of the distance is 210 mi.

+ 60t = 210 105t = 210 t= 2

Combine like terms. Divide by 105.

It will take the cars 2 hr to be 210 mi apart.

200

CHAPTER 2

Linear Equations a nd Inequa lities in One Variable

CONCEPTS

EXAMPLES

flil Solving Linear Inequalities

Solve the inequality, and graph the solution set.

3( 1 - x)+5 - 2x > 9 - 6

Solving a Linear Inequality in One Variable

Step 1

3 - 3x

Simplify each side separately.

+5-

• Clear any parentheses. • Clear any fractions or decimals.

8-

• Combine like terms.

Step 2

Isolate the variable terms on one side.

Step 3

Isolate the variable.

>98 - 5x > 3 5x - 8 > 3 - 5x > - 5 2x

- 5x

Subtract 8.

8

Combine like terms.

- 5

Divide by - 5. Change > to < .

x< I

(-oo, 1)

Solution set: -2 - I

To solve a three-part inequality such as

0

2

Solve the inequality, and graph the solution set.

< 2x + 6 :::; 8,

4
.... 0

"'c:

150 . 100

~ 50 ~

0 '-''~20_1_0_ _2_0_1_3_~2-0-16~x

(b) Use the ordered pairs from part (a) to find the slope of the line.

Year Data from !DC.

(c) Interpret the meaning of the slope in the context of this problem. 80. Locate the line on the graph that represents PCs. Repeat parts (a)-(c) of Exercise 79.

RELATING CONCEPTS For Individual or Group Work (Exercises 81-86) FIGURE A gives the percent offreshmen at 4-year colleges and universities who planned to major in the Biological Sciences. FIGURE B shows the percent of the same group of students who planned to major in Business. Work Exercises 81-86 in order.

I

I

Biological Sciences Majors y

....c: Q)

...u

Business Majors y

12

Q)

a. 10 8 6 0 2000

2008

2016

Year Data from Higher Education Research Institute.

FIGURE A

x

0 ~-------~~ x 2000 2008 2016 Year Data from Higher Education Research Institute.

FIGURE B

81. Find the slope of the line in FIGURE A to the nearest hundredth. 82. The slope of the line in FIGURE A is (positive I negative). This means that during the period represented, the percent of freshmen planning to major in the Biological Sciences (increased/ decreased). 83. The slope of a line represents the rate of change. Based on FIGURE A, what was the increase in the percent of freshmen per year who planned to major in the Biological Sciences during the period shown? 84. Find the slope of the line in FIGURE B to the nearest hundredth. 85. The slope of the line in FIGURE B is (positive/negative) . This means that during the period represented, the percent of freshmen planning to major in Business (increased/ decreased) . 86. Based on FIGURE B, what was the decrease in the percent of freshmen per year who planned to major in Business?

SECTION 3.4

Slope-Intercept Form of a Li near Equation

245

Slope-Intercept Form of a Linear Equation OBJECTIVES

OBJECTIVE 1 Use slope-intercept form of the equation of a line.

1 Use slope-intercept form of the equation of a line.

Recall that we can find the slope (steepness) of a line by solving the equation of the line for y. In that form, the slope is the coefficient of x. For example, the line with equation

2 Graph a line using its slope and a point on the line. 3 Write an equation of a line using its slope and any point on the line.

+ 3 has slope 2.

y = 2x

What does the number 3 represent? To find out, suppose a line has slope m and y-intercept (0, b).We can find an equation of this line by choosing another point (x, y) on the line, as shown in FIGURE 32. Then we apply the slope formula.

4 Graph and write equations of horizontal and vertical lines.

y - b .,,.__ Change in y-values m=-x - 0 .,,.__ Change in x-values

y-b m= - x

Subtract in the denominator.

mx=y-b

Multiply by x.

mx+b=y

Add b .

y = mx

y

+b

0 FIGURE 32

Rewrite.

This result is the slope-intercept form of the equation of a line. Both the slope and the y-intercept of the line can be read directly from this form. For the line with equation y = 2x + 3, the number 3 gives the y-intercept (0, 3). Slope-Intercept Form

The slope-intercept form of the equation of a line with slope m and y-intercept (0, b) is

= mx +b.

y

Slope _J

L

(0, b) is they-intercept.

The intercept given is they-intercept. l:i.NOW TRY

~EXERCISE 1 Identify the slope and y- intercept of the line with each equation.

3

(a) y =

- 5x - 9

(b) y =

- 3+3

x

7

lif$JMQlll

Identifying Slopes and y-lntercepts

Identify the slope and y-intercept of the line with each equation. y = - 4x

(a)

Slope _J

L

y-intercept (0, 1)

y = lx

+ (- 8).

Slope _J =

6x

can be written as

y

=

6x

(d) y = ~-~

can be written as

L

y-intercept (0, - 8)

+ 0.

Slope _J

NOW TRY ANSWERS

(b) slope: -

1

(b) y = x - 8 can be written as

(c) y

1. (a) slope: - ~; y-intercept: ( 0, - 9)

+

L

y-intercept (0, 0)

y= ~x+( -~).

~ ; y-intercept: ( 0, ~ ) Slope

_J

L

y-intercept ( 0, -

~)

NOW TRY~

246

CHAPTER 3

Linear Equations and Inequalities in Two Variables; Functions

lmii)

Slope-intercept form is an especially useful form for a linear equation because of the information we can determine from it. It is also the form used by graphing calculators and the one that describes a linear function.

OBJECTIVE 2 Graph a line using its slope and a point on the line. We can use the slope and the point represented by the y-intercept to graph a line. Graphing a Line Using the Slope and y-lntercept

= mx + b, if necessary, by

Step 1

Write the equation in slope-intercept form y solving for y.

Step 2

Identify they-intercept. Plot the point ( 0, b).

Step 3

Identify the slope m of the line. Use the geometric interpretation of slope ("rise over run") to find another point on the graph by counting from they-intercept.

Step 4

Join the two points with a line to obtain the graph. (If desired, obtain a third point, such as the x-intercept, as a check.)

lif!MMQl#j Graphing Lines Using Slopes and y-lntercepts Graph the equation of each line using the slope and y-intercept. (a) y

2 3

=-x- 1

Step 1

The equation is in slope-intercept form.

y

=

2 -x- 1 3

t

t

Slope

Value of b in y-intercept (0, b)

Step2

They-intercept is (0, - 1). Plot this point. See FIGURE 33.

Step 3

The slope is~- By definition, slope m

=

change in y (rise)

.

change m x (run )

2

=-. 3

From the y-intercept, count up 2 units and to the right 3 units to the point (3, 1 ).

Step 4 (b) 3x

Draw the line through the points ( 0, - 1 ) and (3, 1) to obtain the graph in FIGURE 33.

FIGURE33

+ 4y = 8

Step 1

Solve for y to write the equation in slope-intercept form.

3x + 4y

=8

~-~

Isolate y on one side.

4y

Slope-intercept form __,. y

=

-3x 3

+8

= - 4:-X + 2

Subtract 3x . Divide by 4.

SECTION 3.4

/:.NOW TRY

~EXERCISE2 Graph 3x + 2y = 8 using the slope and y-intercept.

Slope-Intercept Form of a Li near Equation

Step 2

They-intercept in y = - ~x + 2 is (0, 2 ). Plot this point. See FIGURE 34.

Step 3

~ , which can be written as -3 either 4 or _ 4 . We use 4 here.

24 7

y

The slope is .

-3

m=

3

change in y (rise) change in x (run)

- 3

=-

4

From they-intercept, count down 3 units (because of the negative sign) and to the right 4 units to the point ( 4, -1).

Step 4

Draw the line through the two points ( 0, 2) and ( 4, - 1) to obtain the graph in

NOW TRY~

FIGURE34.

III!I]) In Step 3 of Example 2(b), we could use

3

for the slope. From they-intercept 4 (0, 2) in FIGURE 34, count up 3 units and to the left 4 units (because of the negative sign) to the point (-4, 5). Verify that this produces the same line.

/:.NOW TRY ~EXERCISE3

Graph the line passing through the point ( -3, - 4 ), with slope

~·

l:f:t!JMQl#I Graphing a Line Using Its Slope and a Point Graph the line passing through the point (-2, 3) , with slope -4. First, plot the point (- 2, 3). See FIGURE 35. Then write the slope -4 as

-4 change in y (rise) m = change in x (run) = - 1- .

'

y

c.:.2 · 3l

Locate another point on the line by counting down 4 units from ( -2, 3) and then to the right 1 unit. Finally, draw the line through this new point ( -1 , - l) and the given point ( - 2, 3). See FIGURE 35. We could have written the slope as ~ 1 instead. In this case, we would move up 4 units from ( -2, 3) and then to the left 1 unit. Verify that this produces the same line.

FIGURE35

NOW TRY~

OBJECTIVE 3 Write an equation of a line using its slope and any point on the line. We can use the slope-intercept form to write the equation of a line if we know the slope and any point on the line. NOW TRY ANSWERS 2. ft+t+f-118.t-'.cl-t+i

l@JMQlll

,.

Write an equation in slope-intercept form of the line passing through the given point and having the given slope.

Using Slope-Intercept Form to Write Equations

2

(a) (O, - l ), m=3

Because the point ( 0, - 1) is the y-intercept, b = - 1. We can substitute this value for band the given slope m = ~ directly into slope-intercept form y = mx an equation.

+ b to write

248

Linear Equations and Inequalities in Two Variables; Functions

CHAPTER 3

Slope -ii

~NOW TRY

"7 EXERCISE 4

y = mx

Write an equation in slopeintercept form of the line passing through the given point and having the given slope.

+b

2

y

2

=-x-

Substitute form and b.

1

3

is (0, b ) .

Slope-intercept form

y = 3 x+( - l )

(a) (0, 2), m = -4

(b) ( -2, 1), m = 3

~ y- intercept

Definition of subtraction

(b) (2, 5), m = 4 This line passes through the point (2, 5) , which is not they-intercept because the x-coordinate is 2, not 0. We cannot substitute directly as in part (a). We can find the y-intercept by substituting x = 2 and y = 5 from the given point and the given slope m = 4 into y = mx + b and solving for b.

y = mx (0, b) is the y-intercept. Don't stop here.

+b

Slope-intercept form

= 5, m = 4, and x = 2.

5 = 4(2)+ b

Let y

5=8 + b

Multiply.

- 3

=b

Subtract 8.

Now substitute the values of m and b into slope-intercept form. y= mx+ b y

Write an equation in slopeintercept form of the line graphed. y tttt" ti

J ( 3, oi,. '£1 O-

- 3

NOW TRY~

Let m = 4 and b = - 3.

li:f4MQl4j Writing an Equation of a Line from a Graph

~NOWTRY

"7 EXERCISE 5

~

= 4x

Slope-intercept form

Write an equation in slope-intercept form of the line graphed in FIGURE 36. From the graph in FIGURE 36, identify the y-intercept ( 0, 4). Thus, b = 4. To find the slope, count grid squares down 4 units from the y-intercept and to the right 2 units to the x-intercept (2, 0 ).

!I

\01 ~~l+I

slope m =

-4

2 '

or

y

- 2.

We now have the information needed to write the equation of the line. Slope -ii

~ y-intercept

= mx + b y = - 2x + 4 y

is (0, b ) .

Slope- intercept form Let m = - 2 and b = 4.

FIGURE36

NOWTRY~

OBJECTIVE 4 Graph and write equations of horizontal and vertical lines.

1!£MMQl#ill Graphing Horizontal and Vertical Lines Using Slope and a Point Graph each line passing through the given point and having the given slope. NOW TRY ANSWERS 4. (a) y = - 4x + 2 (b) y = 3x + 7 2

5. y = 3x - 2

(a) (4,-2) , m=O

Recall that a horizontal line has slope 0. To graph this line, plot the point ( 4, -2) and draw the horizontal line through it. See FIGURE 37 on the next page.

SECTION 3.4

Slope-Intercept Form of a Linear Equation

y

/:.NOW TRY

249

y

~EXERCISES Graph each line passing through the given point and having the given slope.

x

(a) ( - 3, 3 ), undefined slope (b) (3, -3), slope 0 FIGURE 37

FIGURE 38

(b) (2, -4) , undefined slope Recall that a vertical line has undefined slope. To graph this line, plot the point (2, - 4) and draw the vertical line through it. See FIGURE 38. NOW TRY~

/:.NOW TRY

li!f!1MQlll Writing Equations of Horizontal and Vertical Lines

~EXERCISE 7 Write an equation of the line passing through the point ( - 1, 1) and having the given slope.

Write an equation of the line passing through the point (2, -2) and having the given slope.

(a) Undefined slope (b) m = 0 NOW TRY ANSWERS 6. (a) (b)

(·- ·i 3 \ r1t>!

7. (a) x = - 1

y

(a) Slope 0

II

(b) y = I

3.4 Exercises O Video solutions to select problems available in MyLab Math

STUDY SKILLS REMINDER Reread your class notes before working t he assigned exercises. Review Study Skill 3, Taking Lecture Notes.

This line is horizontal because it has slope 0. Recall that a horizontal line through the point (a, b) has equation y = b. The y-coordinate of the point ( 2, - 2) is - 2, so the equation is y = - 2. See FIGURE 39.

x

(b) Undefined slope This line is vertical because it has undefined slope. A vertical line through the point (a , b) has equation x = a. The x-coordinate of (2, -2) is 2, so the equation is x = 2. See FIGURE 39.

0

FIGURE 39

NOW TRY~

Mylab Math

Concept Check

Fill in each blank with the correct response.

1. In slope-intercept form y y-intercept i s - - ·

=

2. The line with equation y

= -

mx + b of the equation of a line, the slope is _ _ and the ~ - 3 has slope _ _ and y-intercept _ _ .

3. Concept Check Match each equation in parts (a) - (d) with the graph in A- D that would most closely resemble its graph. (a) y = x

+3

(b) y = - x

+3

(c) y = x - 3

A.+' B. 'i C .j O

x

~x

~x

(d) y = -x - 3

D.,J

~x

250

CHAPTER 3

Linear Equations and Inequalities in Two Variables; Functions

4. Concept Check

Match the description in Column I with the correct equation in Column II. I

II

A. y

(a) Slope = 2, passes through (0, 4)

=

B. y

4x I

(b) Slope= -2, y-intercept (0, I )

E. y

=

-4x I

C. y = 4x

(c) Passes through (0, 0) and (4, 1)

=

D. y = -4x

- 2x + 1

F. y

=

2x + 4

(d) Passes through (0, 0) and (I , 4)

Concept Check Answer each question. 5. What is the common name given to a vertical line whose x-intercept is the origin? 6. What is the common name given to a line with slope 0 whose y-intercept is the origin? Identify the slope and y -intercept of the line with each equation. See Example I.

s

7 3

8. y = - x - 6

7. y = -x - 4 2

9. y = -x

x s 12. y= - - 7 14

x 3 11. y = - - s 10

10. y = x + 1

+9

Graph the equation of each line using the slope and y-intercept. See Example 2. 1

13. y = 3x + 2

14. y = 4x - 4

15. y = -3x

I 16. y = - -x 2

+2

17. y = 2x

18. y = -3x

19. 2x + y =

-s

20. 3x + y = -2

21. 4x - Sy = 20

22. 6x - Sy = 30

23. x

+ 2y = 0

+4

24. x - 3y = 0

Graph each line passing through the given point and having the given slope. See Examples 3 and 6.

1

2

25.(0, l) , m=4

26. (0, -3), m =

27. (0,4),m = -4

28. (0, -S), m = -2

2 29. (1, -S), m = - -

2 31. (- 1,4),m=5

32. (-2, 2), m =

34. (0, 0), m = -3

35. (-2, 3), m = 0

36. (3, 2), m = 0

37. (2, 4 ), undefined slope

38. (3, -2), undefined slope

39. (S, -S) , slope 0

40. ( -4, 4 ), slope 0

41. (-2, 2), undefined slope

42. ( - 1, - 3), undefined slope

1 30. ( 2, - I) , m = - 3

s

3

33. (0, 0), m = -2

2

Write an equation in slope-intercept form of each line graphed. See Example 5.

43.

44.

y

x

y

x

SECTION 3.4

45.

y

46.

y

47.

y

48.

y

.. ""'

;

""'

2)

l

,

.... ....

~

-

•n

251

Slope-Intercept Form of a Li near Equation

x

n J

,

~

x

, 0, -. I) ... ;

Write an equation in slope-intercept form (if possible) of the line passing through the given point and having the given slope. See Examples 4 and 7. 49. slope 4, y-intercept ( 0, - 3)

50. slope - 5, y-intercept (0, 6)

51. (0, - 7), m

=

52. (0, - 9), m

=

53. ( 4, 1), m

2

54. (2, 7), m

3

=

- 1

=

1

55. (-1,3),m= -4

56. (-3, l ), m= -2

57. (9, 3), m

58. (8, 4), m

=

1

=

1

m =

25

3 59. (-4, l ),m=4

60. (2, 1),

61. (0, 3), m

62. (0, -4), m

=

0

=

0

63. (2, -6), undefined slope

64. ( - 1, 7), undefined slope

65. (0, - 2), undefined slope

66. (0, 5), undefined slope

67. (6, -6), slope 0

68. (- 3, 3), slope 0

Each table of values gives several points that lie on a line. (a) Use any two of the ordered pairs to.find the slope of the line. (b) Identify the y-intercept of the line. (c) Use the slope and y-intercept from parts (a) and (b) to write an equation of the line in slope-intercept form. (d) Graph the equation.

69. * Y 70. * Y 0

4

71. ~ Y1 -9

0

-1

3

5

2

2

-6

0

5

9

4

0

0

-2

72. * Y - 10

- 1

0

3

5

5

252

CHAPTER 3

Linear Equations and Inequalities in Two Variables; Functions

Extending Skills

Solve each problem.

73. Andrew earns 5% commission on his sales, plus a base salary of $2000 per month. This is illustrated in the graph and can be modeled by the linear equation

y

0.05x

=

+ 2000,

where y is his monthly salary in dollars and x is his sales, also in dollars. Monthly Salary y

4500 4000

~0

3500

3

3000

"O

~ 2500

" 2000 Cl)

10,000 20,000 30,000 40,000 50,000 Sales (in dollars)

0

x

(a) What is the slope? With what does the slope correspond in the problem? (b) What is the y-intercept? With what does the y-value of the y-intercept correspond in the problem? (c) Use the equation to determine Andrew's monthly salary if his sales are $10,000. Confirm this using the graph. (d) Use the graph to determine his sales if he wants to earn a monthly salary of $3500. Confirm this using the equation. 74. The cost to rent a moving van is $0.50 per mile, plus a flat fee of $100. This is illustrated in the graph and can be modeled by the linear equation

y

=

0.50x

+ 100,

where y is the total rental cost in dollars and x is the number of miles driven. Rental Van Charge y

600 500

~

0

400

"O

5 "aeo ..c

300 200

u

100 0

200

400 600 Miles

800 1000

(a) What is the slope? With what does the slope correspond in the problem? (b) What is the y-intercept? With what does the y-value of the y-intercept correspond in the problem? (c) Use the equation to determine the total charge if 400 mi are driven. Confirm this using the graph. (d) Use the graph to determine the number of miles driven if the charge is $500. Confirm this using the equation.

Poi nt-Slope Form of a Linear Equation and Modeling

SECTION 3.5

253

Extending Skills The cost y of p roducing x items is, in some cases, expressed in the form y = mx + b. The value of b gives the fixed cost (the cost that is the same no matter how many items are produced), and the value of m is the variable cost (the cost of producing an additional item). Use this information to work each problem. 75. It costs $400 to start a business selling campaign buttons. Each button costs $0.25 to make. (a) What is the fi xed cost?

(b) What is the variable cost?

(c) Write the cost equation. (d) Find the cost of producing 100 buttons. (e) How many campaign buttons will be produced if the total

cost is $775? 76. It costs $2000 to purchase a copier, and each copy costs $0.02 to make. (a) What is the fi xed cost?

(b) What is the variable cost?

(c) Write the cost equation. (d) Find the cost of producing 10,000 copies. (e) How many copies will be produced if the total cost is $2600?

RELATING CONCEPTS For Individual or Group Work (Exercises 77-80) A line with equation written in slope-intercept form y = mx + b has slope m and y-intercept ( 0, b ). The standard form of a linear equation in two variables is

Ax+ By= C,

Standard form

where A, B, and C are real numbers and A and B are not both 0. Work Exercises 77-80 in order.

77. Write the standard form of a linear equation in slope-intercept form- that is, solved for y- to show that, in general, the slope is given by - ~ (where B 716 0) . 78. Use the fact that m (a) 2x

+ 3y =

= -

18

~ to find the slope of the line with each equation. (b) 4x - 2y = - 1

(c) 3x - 7y = 21

79. Refer to the slope-intercept form found in Exercise 77. What is the y-intercept? 80. Use the result of Exercise 79 to find the y-intercept of each line in Exercise 78.

Point-Slope Form of a Linear Equation and Modeling OBJECTIVES

OBJECTIVE 1 Use point-slope form to write an equation of a line.

1 Use point-slope form to write an equation of a line.

There is another form that can be used to write an equation of a line. To develop this form , let m represent the slope of a line and let (x 1, y 1 ) represent a given point on the line. We let (x, y) represent any other point on the line. See FIGURE 40.

2 Write an equation of a line using two points on the line. 3 Write an equation of a line that fits a data set.

y - y, m = --x - x1 m (x - x 1) = y - y 1 y - y 1 = m (x - x 1)

y

Definition of slope

1

Multiply each side by x - x1 .

°'""

A"Y point (x, y)

Slope is 111.

Interchange sides.

This result is the point-slope form of the equation of a line.

x

0 FIGURE40

254

CHAPTER 3

Linear Equations and Inequalities in Two Variables; Functions

Point-Slope Form

The point-slope form of the equation of a line with slope m passing through the point (xi, Yi) is ~Sl ope

Y - Y1 = m(x - x 1).

L

I:. NOW TRY

~EXERCISE 1 Write an equation of the line passing through (3, -1 ),

with slope-~. Give the final answer in slope-intercept form.

Given _j point

li:f4MQlll Using Point-Slope Form to Write Equations Write an equation of each line. Give the final answer in slope-intercept form. (a) The line passing through (-2, 4 ), with slope -3 Here Xi = - 2, Yi = 4, and m = -3. Substitute into the point-slope form. Y - Yi = m( x - Xi )

Only y,, m, and x,

are~e~~~~with

= - 3 [x - (- 2) J 4 = -3 (x + 2) 4 = -3x - 6

Distributive property

y = -3x - 2

Add 4.

y- 4

Let y1 = 4, m = - 3, X1 = -2.

y-

Definition of subtraction

y-

The answer is in

Point-slope form

y= mx+ bform as specified.

(b) The line passing through ( 4, 2) , with slope ~ y- yi = m(x- x 1 )

3

y- 2

= 5 (x -

y- 2

=

Oo not clear fractions here because we want the answer in slope-intercept form-that is, solved for y.

Point-slope form 3

4)

Let y 1 = 2, m = 5, x1 = 4.

3 12 -x - 5 5

Distributive property

3 12 10 y=-x - - + 5 5 5 y

=

3 2 -x - 5 5

Add 2 =

~ to each side.

Combine like terms.

NOW TRY~

OBJECTIVE 2 Write an equation of a line using two points on the line. Many of the linear equations we have worked with were given in standard form

Ax+ By= C,

Standard form

where A, B, and C are real numbers and A and B are not both 0. In most cases, A, B, and C are rational numbers. For consistency in this text, we give answers so that A, B, and Care integers with greatest common factor 1 and A ;;:::: 0. (If A = 0, then we give B > 0.)

Ell.DJ

The definition of standard form is not the same in all texts. A linear equation can be written in many different, yet equally correct, ways. For example, 3x

NOW TRY ANSWER 2

I

1. y =- 5x+5

+ 4y =

12,

6x + 8y = 24, and

-9x - 12y = -36

all represent the same set of ordered pairs. When answers are given in standard form, the form 3x + 4y = 12 is preferable to the other forms because the greatest common factor of 3, 4, and 12 is 1 and A ~ 0.

Poi nt-Slope Form of a Linear Equation and Modeling

SECTION 3.5

/:.NOW TRY

~EXERCISE2 Write an equation of the line passing through the points (4, 1) and (6, - 2). Give the final answer in

255

lf)UMQlfj Writing an Equation of a Line Using Two Points Write an equation of the line passing through the points ( - 2, S) and (3, 4 ) . Give the final answer in slope-intercept form and then in standard form. First, find the slope of the line using the given points.

(a) slope-intercept form and

slope m

(b) standard form.

4-S = 3 _ ( - 2)

Subtract y-values. Subtract x-values in the same order.

- 1 Simplify the fraction.

s 1

-a

a

b

s

-;;

= - ~and (- 2, S) or (3, 4) as (x 1> y 1)

Use m

in point-slope form. We choose ( 3, 4 ) .

y- y 1 = m(x- x 1)

Point-slope form

1

1

y - 4 = -5(x - 3)

Let y1 = 4, m = - 5, x1 = 3.

1 3 y -4=- - x+ S S 1 3 20 y =- -x + - + S S S 1 23 Slope-intercept form ~ Y = - Sx + 5

Distributive property Add 4 = ~ to each side.

Combine like terms.

Sy = -x + 23 Standard form ~ x

Multiply by 5 to clear fractions.

+ Sy = 23

Add x .

NOW TRY

m!IDI

~

There is often more than one way to w rite an equation of a line. In Example 2, the same equation w ill result using the point ( - 2, 5) for (x1 , y 1 ) in point -slope form . We cou ld also use slope-intercept form y = mx + b and substitute the slope and either given point, solving for b.

T Summary of the Forms of Linear Equations Equation y=mx+b

Description Slope-intercept form

Example 3

y = :zx - 6

Slope ism . y-intercept is (0, b).

Y -

Y1

= m(x - x1 )

Point-slope form

3

y+3=:z(x-2)

Slope ism . Line passes through (x1 , y,) .

Ax+ By= C (where A, 8 , and Care real numbers and A and 8 are not both 0)

Standard form Slope 1s . - A (8 # 0). 8 x-intercept

is(~. 0)

y-intercept is ( 0,

x=a

3x - 2y = 12

(A# 0) .

~) (8 # 0).

Vertical line

x=3

Slope is undefined. x-intercept is (a, 0).

NOW TRY ANSWERS 3

2. (a) y=-2x+7 (b) 3x

+ 2y =

14

y=b

Horizontal line Slope is 0. y-intercept is (0, b).

y=3

256

CHAPTER 3

Linear Equations and Inequalities in Two Variables; Functions

OBJECTIVE 3 Write an equation of a line that fits a data set. If a given set of data fits a linear pattern-that is, if its graph consists of points lying close to a straight line-we can write a linear equation that models the data.

l:&f!1MQl#I Writing an Equation of a Line That Models Data The table lists average annual cost y (in dollars) of tuition and fees for instate students at public 4-year colleges and universities for selected years, where x = 1 represents 2011, x = 2 represents 2012, and so on. Year

x

2011

1

8820

2012

2

9080

2013

3

9150

Cost y (in dollars)

2014

4

9240

2015

5

9500

2016

6

9650

Data from The College Board.

(a) Plot the data and write an equation that approximates it. We plot the data as shown in FIGURE 41 . Average Annual Costs at Public 4-Year Colleges )'

10,000 9500 ..

•••

9000 8500 ..

8"'

8000 0 ~~~2-~3-4~~5-~6... x

Year FIGURE41

The points appear to lie approximately in a straight line. To find an equation of the line, we choose two ordered pairs ( 1, 8820) and ( 6, 9650) from the table and determine the slope of the line through these points. Y2 -

Y1

m= - - = X1

X2 -

9650 - 8820 6- 1

= 166

Let (6, 9650) = (x2 , Y2) and ( 1, 8820)

=

(x1 , y 1 ) .

The slope, 166, is positive, indicating that tuition and fees increased $166 each year. Now substitute this slope and the point ( 1, 8820) in the point-slope form to find an equation of the line . Y-

Y1

= m (x -

x 1)

y - 8820

= 166(x -

1)

y - 8820

= 166x -

166

Let (x1 , y1 )

= (1, 8820 ), m = 166.

Distributive property

= 166x + 8654 Add 8820. = 166x + 8654 can be used to model the data. y

Thus, the equation y

Point-slope form

Point-Slope Form of a Linear Equation and Modeling

SECTION 3.5

/:.NOW TRY

~EXERCISE3 Use the points (3, 9150) and ( 5, 9500) to write an equation in slope-intercept form that approximates the data of Example 3. How well does this equation approximate the cost in 2016?

(b) Use the equation found in part (a) to determine the cost of tuition and fees in 2015. We let x = 5 (for 2015) in the equation from part (a), and solve for y.

y y y

= 166x + 8654 = 166(5) + 8654 = 9484

Equation of the line Substitute 5 for x. Multiply, and then add.

Using the equation, tuition and fees in 2015 were $9484. The corresponding value in the table for x = 15 is 9500, so the equation approximates the data reasonably well. NOW TRY

NOW TRY ANSWER 3. y = I75x + 8625; The equation gives y = 9675 when x = 6, which approximates the data reasonably well.

3.5 Exercises O

Video solutions for select problems available in MyLab Math

257

miJD

g

In Example 3, if we had chosen two different data points, we would have obtained

a slightly different equation. See Now Try Exercise 3. Also, we could have used slope-intercept form y = mx + b (instead of point-slope form) to write an equation that models the data.

~M

HELP

0

Mylab Math

Concept Check

Fill in each blank with the correct response.

1. In point-slope form y - y 1 = m(x - x 1) of the equation of a line, the slope is _ _ and the Line passes through the point _ _ . 2. The line with equation y - 4

Concept Check

=

~(x + 2) has slope _ _ and passes through the point

Work each problem.

3. Match each form or description in Column I with the corresponding equation in Column II. II

I (a) Point-slope form

A. x=a

(b) Horizontal line

B. y

=

mx

+b

(c) Slope-intercept form

c.

(d) Standard form

D. y - y 1 = m(x - x 1)

(e) Vertical line

E. Ax +By=C

4. Write the equation y

+ 1 = -2(x -

y= b

5 ) in slope-intercept form and then in standard form.

S. Which equations are equivalent to 2x - 3y

=

6?

2 A . y=-x-2 3

B. -2x

3 C. y= --x+3 2

D. y - 2

+ 3y = -6 =

2

-(x 3 - 6)

6. In the summary box following Example 2, we give the equations

y

3 x- 6 2

= -

and

y

+3 =

3 2

-(x

- 2)

as examples of equations in slope-intercept form and point-slope form, respectively. Write each of these equations in standard form. What do you notice?

258

CHAPTER 3

Linear Equations and Inequalities in Two Variables; Funct ions

Write an equation of the line passing through the given point and having the given slope. Give the final answer in slope-intercept form. See Example 1.

7. (1 , 7), m = 5

8. (2, 9), m = 6

9. (6, -3) , m = 1

10. ( - 4, 4 ), m = 1

11. (1 , -7), m = -3

12. ( 1, - 5) , m = - 7

13. (3, - 2),m = - 1

14. ( - 5, 4 ), m = - 1

2 15. ( - 2 , 5) , m = -3

4 17. (6 - 3) m = - ' ' 5

7 18. (7' -2) ' m =- -2

16. (4, 2), m =

- 31

19. Concept Check When asked to write the equation y = 2x + 4 in standard form , a student wrote the fo llowing equation, which did not agree with the answer in the answer section. -2x + y = 4 WHAT WENT WRONG?

Write the equation in standard form as defined in this text.

20. Concept Check When asked to write the equation 6y = 12 - 3x in standard form, a student wrote the fo llowing equation , which did not agree with the answer in the answer section. 3x + 6y = 12 WHAT WENT WRONG?

Write the equation in standard form as defined in this text.

Write an equation of the line passing through the given pair of points. Give the final answer in (a) slope-intercept form and (b) standard form. See Example 2.

21. (4, 10) and (6, 12)

22. (8, 5) and (9, 6)

23. ( - 4, 0) and (0, 2)

24. (0, - 2) and ( - 3, 0)

25. ( - 2, -1 ) and (3, -4)

26. (-1 , - 7) and (- 8, - 2)

27. (5 , - 7) and ( - 3, 2)

28. ( - 4, 6) and (9, -1 )

29. (

30.

(I_4 ' 2_) and ( ~ - I_) 4 4' 4

31.

(I_2' ~)and (_I_4' ~) 2 4

32.

- ~3'3~)and U-3'3~)

(~

3'

_I_) and (_I_ I_) 3 3' 3

Write an equation of the given line through the given points. Give the final answer in (a) slope-intercept form and (b) standard form.

33.

y

34.

y

35.

y

36.

y

37.

y

38.

y

x

SECTION 3.5

Point-Slope Form of a Linear Equation and Modeling

259

Extending Skills Write an equation of the line satisfying the given conditions. Give the fina l answer in slope-intercept form. (Hint: Recall the relationships among slopes of parallel and perpendicular lines.) 39. Parallel to 5x - y = 10; y-intercept (0, - 2)

40. Parallel to 3x + y y-intercept (0, 4)

41. Perpendicular to x - 2y = 7; y-intercept (0, -3 )

42. Perpendicular to x + 4y y-intercept (0, -1 )

43. Passes through (2, 3); parallel to 4x - y = -2

44. Passes through (2, -3 ); parallel to 3x - 4y = 5

45. Passes through ( 4, 2) ; perpendicular to x - 3y = 7

46. Passes through ( - 1, 4); perpendicular to 2x + 3y

=

7; =

=

-5;

8

Solve each problem. See Example 3.

47. The table lists the average annual cost y (in dollars) of tuition and fees at 2-year colleges for selected years x, where year 1 represents 2012, year 2 represents 2013, and so on. x 1

Cost (in dollars)

2013

2

3340

2014

3

3370

2015

4

3460

2016

5

3520

Year

2012

3310

Data from The College Board.

(a) Write five ordered pairs (x, y) for the data. (b) Plot the ordered pairs (x, y) . Do the points lie approximately in a straight line?

(c) Use the ordered pairs (2, 3340) and (4, 3460) to write an equation of a line that approximates the data. Give the final equation in slope-intercept form. (d) Use the equation from part (c) to estimate the average annual cost at 2-year colleges in 2017 to the nearest dollar. (Hint: What is the value of x for 2017?)

48. The table gives the worldwide number y of monthly active Twitter users in millions for selected years, where x = 1 represents 2014, x = 2 represents 2015, and so on. Year

x

Twitter Users (in millions)

2014

1

271

2015

2

304

2016

3

313

2017

4

326

Data from Twitter.

(a) Write four ordered pairs (x, y) for the data. (b) Plot the ordered pairs

(x, y) . Do the points lie approximately in a straight line?

(c) Use the ordered pairs ( 1, 271 ) and ( 3, 313) to write an equation of a line that approx imates the data. Give the final equation in slope-intercept form. (d) Use the equation from part (c) to estimate the worldwide number of monthly active Twitter users in 2018. (Hint: What is the value of x for 2018?)

260

CHAPTER 3

Linear Equations and Inequalities in Two Variables; Funct ions

The points on the graph show the annual average number y ofpaid music streaming subscriptions (in millions) in the United States for recent years x. The graph of a linear equation that models the data is also shown.

( The Sound of Music y V)

12

0 +:l

10

c:

Q.

·.:::::() V) V) c:

8

.0 0

=

j

CJ) :=

6

-0 c: E ~ :.:.. .0

4

E j z

3

2

0

4

5

Year

Data from RIAA.

Here x = 1 represents 2011, x = 2 represents 2012, and so on. 49. Use the ordered pairs shown on the graph to write an equation of the line that models the data. Give the final equation in slope-intercept form. SO. Use the equation from Exercise 49 to do the following.

(a) Estimate the annual average number of paid music streaming subscriptions in the United States in 2016. (Hint: What is the value of x for 2016?) (b) How does the answer in part (a) compare to the actual value of 22.6 million subscrip-

tions in 2016? The points on the graph indicate years of life expected at birth y in the United States for selected years x. The graph of a linear equation that models the data is also shown.

( Life Expectancy y

~ 111 Cl)

>-

:§. 60 Cl) Cl

x does not contain points in quadrant IV.

0.

SECTION 3.6

Concept Check

Determine whether the g iven ordered pair is a solution of the given inequality.

12. 2x + Sy < 10

8

~

11. x - 4y

267

Graphing Linear Inequalities in Two Variables

(a) (0, 0)

(b) (0, 2)

(a) (0, 0)

(b) (S, 0)

(c) (4, -1)

(d) ( - 4, 1)

(c) ( -S, 2)

(d) (-2, -3 )

13. Concept Check When graphing the linear inequality

3x + 2y

~

12,

a student incorrectly used a dashed line when graphing the boundary line. WHAT WENT WRONG? Explain the student's error.

14. Concept Check

When graphing the linear inequality

+ 4y >

x

0,

a student used (0, 0) as the test point to determine which region to shade. WHAT WENT WRONG? Explain the student's error. For each inequality, the straight-line boundary has been drawn. Complete each g raph by shading the correct reg ion. See Examples 1- 4.

15. x

+y

~

+y s

16. x

4

2

y

y

l'I.

''

l

,.... ,.... '···· '···· :o,

' '

'

. . .... .... ~

1',

" ,··· ~,

~

19. - 3x

S

+ 4y >

It

12

20. 4x - Sy

< 20

l \

:o

y

"

, ~·

\

•.... •....

x .....

y

y

..... .

7

~

' ' '

'

+y

~

y

"'

'···· '···· '····'····

x

;

18. 2x

+ 2y

l'I. ~

'····'···· '···· '····

17. x

x

'

'

·'·:· ''

"'-

' '

,

'···· l

"

.. .............

~

'

21. x > 4

···· ·· ···· ·· JI ······ ·· ·· • ·· -;--;--;.--;--o-+--;-;--;-+-;~ x

'·····'············'·····' vl'l' .1... ~ .... ;.... ;........

,

,··· .... .•..~ 1, 'c

'

23. y < 0

22. x s 0 y

l

"

' ' '

;

\

'····'···· ' ····

x

y

'···· '····•···· •···· l •·····o····o·····o·····o·····o ~-;--;.--;-~ f1-+--'---;--;.--;-• x

···,················

y

x

268

CHAPTER 3

Linear Equations and Inequalities in Two Variables; Functions

24. y < -1

25. x ::::; - y

26. x ::::; 3y

y

y

y

Graph each linear inequality. See Examples 1-4.

27. x

+ y::::; 5

28. x+y2=3

29. x

30. x

+ 3y > 6

31. 2x + 3y > -6

32. 3x + 4y < 12

33. y

2:

2x + 1

34. y < -3x

+

1

+ 2y < 4

35. x < -2

36. x > 1

37. y::::; 5

38. y::::; -3

39. x

40. y

41. y 2: 4x

2:

0

42. y::::; 2x

2:

0

43. x < -2y

44. x > - Sy

Extending Skills For the given information, (a) graph the inequality (here, x 2: 0 and y 2: 0, so graph only the part of the inequality in quadrant I) and (b) g ive two ordered pairs that satisfy the inequality.

45. A company will ship x units of merchandise to outlet I and y units of merchandise to outlet II. The company must ship a total of at least 500 units to these two outlets. The preceding information can be expressed using the linear inequality x

+ y 2: 500.

46. A toy manufacturer makes stuffed bears and geese. It takes 20 min to sew a bear and 30 min to sew a goose. There is a total of 480 min of sewing time available to make x bears and y geese. These restrictions lead to the l.inear inequality 20x

+ 30y ::s 480.

Introduction to Functions OBJECTIVES

OBJECTIVE 1 Define a relation.

1 Define a relation.

Suppose gasoline costs $3.00 per gal. If we purchase 1 gal, then we must pay $3.00( 1) = $3.00. If we purchase 2 gal, then the cost is $3.00(2) = $6.00. For 3 gal, the cost is $3.00(3) = $9.00, and so on. Generalizing, if x represents the number of gallons, then the cost y is $3.00x. The equation

2 Define a function.

3 Determine whether a graph or an equation represents a function.

4 Find domains and ranges.

5 Use function notation. 6 Apply the function concept in an application.

y

= 3.00x

relates the number of gallons, x, to the cost in dollars, y. The set of ordered pairs (x, y ) that satisfy this equation forms a relation. In an ordered pair (x, y ), x and y are the components of the ordered pair.

SECTION 3.7

Introduction to Functions

269

Relation

Any set of ordered pairs is a relation. • The set of all first components in the ordered pairs of a relation is the domain of the relation.

VOCABULARY D D D D D

components relation domain range function

• The set of all second components in the ordered pairs is the range of the relation.

l!f$JMQlll

Identifying Domains and Ranges of Relations

Give the domain and range of each relation. (a) { (O, 1), (2, 5), (3, 8), (4, 2)}

Domain: { 0, 2 , 3, 4} .

Step 2

Step 3

Choose a test point not on the line, and substitute the coordinates of that point in the inequality. Shade the region that includes the test point if it satisfies the original inequality. Otherwise, shade the region on the other side of the boundary line.

2x+y=5 using the intercepts ( 0, 5) and ( ~, 0). Make it solid because the symbol :s: includes equality. y

Use (0, 0 ) as a test point.

2x + y < 5 ?

2(0 )

+ 0 0

Work each problem.

26. Determine whether each relation is a function. Give the domain and range. (a) {(2,3), (2,4), (2,5)}

(b) {(0,2), ( 1,2), (2, 2)}

27. Use the vertical line test to determine whether the graph is that of a function. y

28. If f(x)

Chapters R-3

=

3x + 7, find f(-2).

Cumulative Review Exercises 1. For the decimal number 135.264, round to the place value indicated. (a) hundredths

(b) tenths

(c) ones or units

(d) tens

Perform each indicated operation. 2.

2

5

9

6

-+-

5. 8.3

+ 2.09

8. 5 - (-4) + ( - 2)

5

I

3

I

4

8

3. 10-- 3 8 10

4.

6. 0.2 x 0.08

7. 530.26

9

-13(-5) · 41 - 62

10

-7-

(-3) 2 ·

7

100

-

(-4)(24 )

5(2) - (-2) 3

286

CHAPTER 3

Linear Equations and Inequalities in Two Variables; Funct ions

11. True or false?

4(3 - 9) 2- 6

;;::: 6

12. Find the value of .xz 3 - 5y 2 for x = - 2, y = - 3, and z = -1 . 13. What property does 3(-2 + x)

=

- 6 + 3x illustrate?

14. Simplify - 4p - 6 + 3p + 8 by combining like terms.

Solve. 15. V =

l

37rr2h for h

16. 6 - 3(1

+ x)

=

2(x

+ 5) - 2

x -2 2x + 1 18. - - = - 3 5

17. - (m - 3) = 5 - 2m

Solve each inequality. Graph the solution set, and write it using interval notation. 19. - 2.5x < 6.5

2 1 21. -x - -x:::::; - 2

20.4(x+3) - 5x < 12

3

6

Solve each problem. 22. Find the measure of each marked angle. (a)

4+

( 14x - 8)0

(b)

8) 0

)

In 2016, a total of 2,014,220 electric cars were on the road worldwide. The circle graph shows the distribution of these electric cars by country. 23. How many electric cars could be found in China in 20 16? Round to the nearest whole number. 24. How many electric cars could be found in the Netherlands in 2016? Round to the nearest whole number.

I

Global Electric Car Distribution

6% Norway 7%

Data from International Energy Agency.

Consider the linear equation - 3x + 4y 25. The x- and y-intercepts

STUDY SKILLS REMINDER It is not too soon to beg in preparing for your final exam.

Review Study Skill 10, Preparing for Your Math Final Exam.

=

12. Find the following.

26. The slope

28. Are the lines with equations x + 5y neither?

=

- 6 and y

27. The graph =

5x - 8 parallel, perpendicular, or

Write an equation of each line. Give the final answer in slope-intercept form. 29. The line passing through (2, - 5), slope 3 30. The line passing through ( 0, 4) and ( 2, 4)

SYSTEMS OF LINEAR EQUATIONS AND INEQUALITIES The point of intersection of two lines can be found using a system of linear equations, the subject of this chapter.

ID

Solving Systems of Linear Equations by Graphing

ID

Solving Systems of Linear Equations by Substitution

m

ID

Solving Systems of Linear Equations by Elimination

m

Applying Techniques for Solving Systems of Linear Equations

SUMMARY EXERCISES

Applications of Linear Systems Solving Systems of Linear Inequalities

287

288

CHAPTER 4

Systems of Linear Equations and Inequalities

Solving Systems of Linear Equations by Graphing OBJECTIVES 1 Determine whether a given ordered pair is a solution of a system. 2 Solve linear systems by graphing. 3 Solve special systems by graphing. 4 Identify special systems without graphing.

A system of linear equations, or linear system, consists of two or more linear equations with the same variables. 2x + 3y

=4 y = -S

x - y= l + 3y = 1 Systems of linear equations 3x - y = 4 - 2x y =3 In the system on the right, think of y = 3 as an equation in two variables by writing it as Ox + y = 3. Examples:

x

OBJECTIVE 1 Determine whether a given ordered pair is a solution of a system. A solution of a system of linear equations is an ordered pair that makes both equations true at the same time. A solution of an equation is said to satisfy the equation.

lf!f4MQlll

Determining Whether an Ordered Pair Is a Solution

Determine whether the ordered pair (4, - 3) is a solution of each system. (a)

VOCABULARY D system of linear equations (linear system) D solution of a system D solution set of a system D consistent system D inconsistent system D independent equations D dependent equations

l!:m. NOW TRY

+ 4y = -8 3x + 2y = 6 x

To decide whether ( 4, - 3) is a solution of the system, substitute 4 for x and - 3 for y in each equation. x + 4y = - 8

4 + 4( - 3) J, -8

+ ( -12) J, -8

4

-8 = -8 ./

3x Substitute.

+ 2y = 6

3(4 ) + 2( - 3) J, 6 12

Multiply.

+ ( -6) J, 6 6=6 ./

True

~EXERCISE 1

Because (4, - 3) satisfies both equations, it is a solution of the system.

Determine whether the ordered pair ( 5, 2) is a solution of each system.

(b) 2x +Sy= -7

(a) 3x - y = 13

2x + y

y= 7

Multiply. True

+ 4y = 2

Again, substitute 4 for x and - 3 for y in each equation.

= 12

(b) 2x+5y = 20

x -

3x

Substitute.

2x +Sy = -7

2(4 )

+ S( - 3) J, -7

8+(-lS)J,-7 - 7 = - 7 ./

3x + 4y = 2 Substitute.

+ 4( - 3) J, 2 12 + ( -1 2) J, 2

3(4)

Multiply. True

0=2

Substitute. Multiply. False

The ordered pair (4, - 3) is not a solution of this system because it does not satisfy the second equation. NOW TRY ~

OBJECTIVE 2 Solve linear systems by graphing. The set of all ordered pairs that are solutions of a system is its solution set. One way to find the solution set of a system of two linear equations is to graph both equations on the same axes. Any intersection point would be on both lines and would therefore be a solution of both equations. NOW TRY ANSWERS 1. (a) yes {b) no

Thus, the coordinates of any point at which the lines intersect give a solution of the system.

SECTION 4.1

Solving Systems of Linear Equations by Graphing

The graph in FIGURE 1 shows that the solution of the system in Example l(a) is the intersection point

289

y The point of intersection is the solut ion of

(4, - 3). The solution (point of intersection) is always written as an ordered pair. Because two different straight lines can intersect at no more than one point, there can never be more than one solution for such a system.

+ +

-4

FIGURE 1

l.:.NOW TRY

li$JMQ!fJ Solving a System by Graphing (One Solution)

Solve the system by graphing.

Solve the system by graphing.

"7 EXERCISE 2

+ 3y = 4 3x - y = - 5

2x

x - 2y = 4

2x+ y = 3

We graph these two lines by plotting several points for each line. The intercepts are often convenient choices. We show finding the intercepts for 2x + 3y = 4. To find the y-intercept, let x

+ 3y = 4 2(0) + 3y = 4 3y = 4

= 0.

To find the x-intercept, let y

2x

y

4

2x Let x = 0.

=3

y-intercept ( 0,

+ 3y = 4

+ 3(0) = 4 2x = 4

2x

= 0.

x=2

~)

Let y = 0.

x-intercept (2, 0)

The tables show these intercepts and a check point for each graph. 2x + 3y = 4 x o

4

3

--+-We recommend finding a third ordered pair as a check.

2

3x - y

=

O

--+-- 2 8 3

x

+ 3y

12,

we first graph

= 12

by finding and plotting a few ordered pairs that satisfy the equation. (The x- and y-intercepts are good choices.) Because the symbol > does not include equality, the points on the line do not satisfy the inequality and we graph it using a dashed line. To decide which region includes the points that are solutions, we choose a test point not on the line.

+ 3y > 12 0 + 3(0) :> 12 x

VOCABULARY D system of linear inequalities D solution set of a system of linear inequalities

Original inequality

?

(0, 0) is a convenient test point.

Let x = 0 and y = 0.

0 > 12

False

This false result indicates that the solutions are those points in the region that does not include (0, 0), as shown in FIGURE 14. y ,:-....

• (2, 5)

.. 4 ........

x+3y> 12 ..........

.... ....

Points that lie in the region above the boundary line satisfy the inequality.

•(10, 2)

....

12

0 (0, 0)

.... "'>

x

l

FIGURE 14

We use the same techniques to solve systems of linear inequalities. OBJECTIVE 1 Solve systems of linear inequalities by graphing. A system of linear inequalities consists of two or more linear inequalities. The solution set of a system of linear inequalities includes all ordered pairs that make all inequalities of the system true at the same time. Solving a System of Linear Inequalities

Step 1

Graph each linear inequality on the same axes.

Step 2

Choose the intersection. Indicate the solution set by shading the intersection of the graphs-that is, the region where the graphs overlap.

miDI

In Example 1, we graph the system of linear inequalities in stages, on multiple sets of axes. In the remaining examples, only one graph is shown. Be sure that the region of the final solution set is clearly indicated .

SECTION 4.5

/:.NOW TRY

~EXERCISE 1 Graph the solution set of the system.

10

Step 1

Graph 3x + 2y:::; 6 with the solid boundary line 3x + 2y = 6 using the intercepts (0, 3) and (0, 2) . Determine the region to shade. We are testing 3X + 2Y < 6 the region .

?

3(0) + 2(0) 10 with dashed boundary line 2x - Sy = 10 using the intercepts (0, - 2) and (S, 0). Determine the region to shade. 2x - Sy> 10 ?

2(0) - S(O) ::> 10

Use (0, 0) as a test point.

0 > 10

False

Shade the region that does not include (0, 0). See FIGURE 15(b). y

3x+ 2ys 6

y

y

3

(0,0)

!I I

+- + +- +

+ +

r

f

+- + +- + +-

r

Test

(0, 0) ~ point

x

-2

............ 14' ....

.... ........

x

5

2x-5y> 10

le.,...,..

.... .... ....

Test~•(0,-4)

point (a)

t

(b)

Solution set FIGURE 16

FIGURE 15

Step 2

The solution set of this system includes all points in the intersection-that is, the overlap-of the graphs of the two inequalities. As shown in FIGURE 16, this intersection is the purple shaded region and portions of the boundary line 3x + 2y = 6 that surrounds it.

To confirm the solution set in FIGURE 16, select a test point in the purple shaded region , such as (0, -4) , and substitute it into both inequalities to make sure that true statements result. (Using an ordered pair that has one coordinate 0 makes the substitution easier.) CHECK

1. _

NOW TRY ANSWER

3x

+ 2y < 6

2x - Sy > 10

?

3(0) + 2( - 4) 10 20 > 10

Test (0, - 4 ). True

We have shaded the correct region in FIGURE 16. Test points selected in the other three regions will satisfy only one of the inequalities or neither of them. NOW TRY~ (Verify this.) ./

328

CHAPTER 4

Systems of Linear Equations and Inequalities

1!£MMQl#j Solving a System of Linear Inequalities

~NOW TRY

"7 EXERCISE 2 Graph the solution set of the system.

Graph the solution set of the system.

x-y>O

2x +Sy> 10 x - 2y


2 shows the graphs of both x - y > 0 and 2x + y > 2. Dashed lines indicate that the graphs of the inequalities do not include their boundary lines. For x - y > 0, the dashed boundary line passes through the origin. To determine the region to shade, we must select a test point other than (0, 0 )-that is, a point off the line. We use (2, 0) , which gives the true statement 2 > 0, and so we shade the region that includes this point. For 2x + y > 2, we can use (0, 0) as the test point. The false statement 0 > 2 results, and so we shade the region that does not include this point. The solution set of the system is the region with purple shading. Neither boundary line is included. FIGURE 11

y

Solution set FIGURE 17

CHECK To confirm the solution set in FIGURE 11, select a test point in the purple shaded region, such as (3, 0), and substitute it into both inequalities. The resulting statements, 3 > 0 and 6 > 2, are true, so we have shaded the correct region as the NOW TRY ~ solution set. ./

~NOW TRY

1!£1!jtfilQl#I Solving a System of Three Linear Inequalities

Graph the solution set of the system.

Graph the solution set of the system.

"7 EXERCISE 3 x - y< 2 x

2:

- 2

y::::; 4

4x - 3y :s: 8 x;:::

2

y:s: 4 Graph the solid boundary line 4x - 3y

-H

= 8 through

(Because the y-intercept the intercepts ( 2, 0) and ( 0, does not have integer coordinates, we also use the point

NOW TRY ANSWERS 2. 2 ~

j

2x+ Sy> 10 x-2y< 0

y

Solution set

(-1, -4) to help draw an accurate line.) Recall that x = 2 is a vertical line through the point ( 2, 0) , and y = 4 is a horizontal line through the point (0, 4) . Use (0, 0) as a test point to determine the region to shade for each inequality. For 4x - 3y :s: 8, shade the region "above" the line. For x ;::: 2, shade the region to FIGURE 18 the "right" of the line. And for y :s: 4, shade the region "below" the line. The graph of the solution set is the shaded region in FIGURE 18, including all boundary lines.

3.

CHECK Use (3, 2) in the shaded region as a test point to confirm the solution set. ./ NOW TRY~

SECTION 4.5

4.5 Exercises 0 Video solutions for select problems available in Mylab Math

0

FOR

EXTRA

HELP

329

Solving Systems of Linear Ineq ualities

Mylab Math

Concept Check Match each system of inequalities with the correct graph from choices A-D. 1. x

~

A.

5

B.

y

y

y :s - 3

x

x

2. x :s 5 y~

-3

c.

D.

y

y

3. x > 5 y

< -3 x

x

- 3

4. x

Concept Check 5.

Describe the region that is the solution set of each system of inequalities.

0

6. x :s 0

7. x > 0

8. x < 0

y~O

y :s 0

yO

x~

Concept Check

Determine whether each ordered pair is a solution of the given system of inequalities. Then shade the solution set of each system. Boundary lines are already graphed.

9. x - 3y :s 6 x

~

10. x - 2y

(a) (-5, - 4)

~

4

x :s - 2

-4 (b) (0, - 4)

(c) (0, 0)

(a) (-3, 0)

(b) (0, 0)

y

11. x

+

y

5x - 3y

y

>4 < 15

(a) (0, 0)

(c) ( - 4, -5)

12. 3x - 2y > 12 4x + 3y

(b ) (3, 3) y

(c) (5, 0)

< 12

(a) (0, 0)

(b) (3, -3) y

(c) (6, 0)

330

CHAPTER 4

Systems of Linear Equations and Inequalities

Graph the solution set of each system of linear inequalities. See Examples 1 and 2.

13. x

+ y :S

14. x

6

+ 4y :S 8

2x 19.

x

+ y> + 2y


+ 6y >

18

+3

x + y> O

+4

2

27. y :S x

3x < y

y :S 4

x~

6x - 2y

24. x :S 4y

x+ y< O

I

30. x - 2y

32.

~

37. x

+y

+ y 9

34. 2x -

y

36. x - 2y

~

- 2

y> - 2

y~

-2

x> -4

x:S 3

~

- 3

38. x

+y
-4

y :S

3

y> - 1

39. 3x - 2y

~

6

x+ y:S 4 x ~ O y~

-4

2

x :S - 3

- S

Graph the solution set of each system of linear inequalities. See Example 3.

35. 4x +Sy< 8

4

x- y < - 1

x - 2y > 4

28. y

y ~

s

21. 4x + 3y < 6

2y

26. x

31. - 3x+

+ 2y
- 2x + 4

20. x

+y < 3 2x > y

25. x

18. x

x- y< S

4

+ Sy

x - 2y :S

17. 2x + 3y < 6

y ~ 4

y :S 2x - s x < 3y + 2

22. 3x

15. 4x

x -y~ 3

x -y~ l

16. x

+ y :S 2

40. 2x - 3y < 6 x+ y

>3

x< 4 y< 4

Summary

CHAPTER 4

331

----·----------

Chapter 4

Summary

STUDY SKILLS REMINDER Be prepared for your math test on this chapter. Review Study Skills 7 and 8, Reviewing a Chapter and Taking Math Tests.

Key Terms

Ill system of linear equations (linear system)

solution of a system solution set of a system consistent system

inconsistent system independent equations dependent equations

system of linear inequalities solution set of a system of linear inequalities

Test Your Word Power See how well you have learned the vocabulary in this chapter.

1. A system of linear equations consists of A. at least two linear equations with different variables B. two or more linear equations that have an infinite number of solutions c. two or more linear equations with the same variables D. two or more linear inequalities.

2. A consistent system is a system of equations A. with at least one solution B. with no solution c. with graphs that do not intersect D. with solution set 0.

C. with an infinite number of solutions D. that have the same graph.

4. Dependent equations

3. An inconsistent system is a system of equations A. with one solution B. with no solution

A. B. C. D.

have different graphs have no solution have one solution are different forms of the same equation.

ANSWERS 1. C; Example: 2x + y = 7, 3x - y = 3 2. A; Example: The system 2x + y = 7, 3x - y = 3 is consistent. The graphs of the equations intersect at exactly one point- in this case, the solution (2, 3 ). 3. B; Example: The equations of two parallel lines make up an inconsistent system. Their graphs never intersect, so there is no solution of the system. 4. D; Example: The equations 4x - y = 8 and 8x - 2y = 16 are dependent because their graphs are the same line.

Quick Review CONCEPTS

m

EXAMPLES

Solving Systems of Linear Equations by Graphing

An ordered pair is a solution of a system if it makes all equations of the system true at the same time.

Is ( 4, - I ) a solution of the following system?

x + y= 3 2.x-y = 9 Because 4 + ( - I ) = 3 and 2( 4) - ( -1 ) ordered pair ( 4, - 1) is a solution.

Solving a Linear System by Graphing

Step 1

Graph each equation of the system on the same axes.

Step 2

Find the coordinates of the point of intersection.

Step 3

Check. Write the solution set.

Solve the system by graphing.

x+y =S 2.x-y=4 The ordered pair (3, 2) satisfies both equations, so { (3, 2)} is the solution set.

=

9 are both true, the y

332

CHAPTER 4

Systems of Linear Equations and Inequalities

CONCEPTS

EXAMPLES

If the graphs of the equations do not intersect (that is, the lines are parallel), then the system has no solution and the solution set is 0 .

y

y

If the graphs of the equations are the same line, then the system has an infinite number of solutions. Use set-builder notation to write the solution set as

{(x,y) I

where a form of the equation is written on the blank.

m

'

},

Solving Systems of Linear Equations by Substitution

Solution set: {(x, y) I 3x + 2y = 6 }

Solutio n set: 0

Solve the system by substitution.

+ 2y = -5

(1)

y = - 2x - 1

(2)

x

Solving a Linear System by Substitution Step 1

Solve one equation for either variable.

Equation (2) is already solved for y.

Step 2

Substitute for that variable in the other equation to obtain an equation in one variable.

Substitute - 2x - 1 for y in equation ( 1).

Step 3

Solve the equation from Step 2.

Step 4

Find the other value by substituting the result from Step 3 into the equation from Step 1 and solving for the remaining variable.

Step 5

x

+ 2y

= - 5

(1) Let y = - 2x - 1.

x+2( - 2x - l ) = - 5 x - 4x - 2 = -5

- 3x - 2

Check. Write the solution set.

=

Distributive property

-5

Combine like terms. Add 2. Divide by - 3.

x= I

To find y, let x = 1 in equation (2).

y

=

-2x - I

(2)

y

=

- 2( 1) - 1

Let x = 1.

y

=

-3

Multiply, and then subtract.

A check confirms that { ( 1, - 3) } is the solution set.

m

Solving Systems of Linear Equations by Elimination

Solve the system by elimination.

Step 1

Write both equations in the form Ax+ By= C.

Step 2

Multiply to transform the equations so that the coefficients of o ne pair of variable terms are opposites.

Step 3

Add the equations to eliminate a variable.

Step 4

Solve the equation from Step 3.

Step 5

Find the other value by substituting the result from Step 4 into either of the original equations and solving for the remaining variable.

Step 6

Check. Write the solution set.

+ 3y =

x

Solving a Linear System by Elimination

3x -

y

=

7

(1)

1

(2)

Multiply equation (1) by - 3 to eliminate the x-terms .

-3x - 9y = - 21 3x -

1

y=

-lOy

=

Multiply equation (1) by - 3.

(2)

-20

Add.

y =2

Divide by - 10.

Substitute to find the value of x.

x

+ 3y

=

7

x+3( 2) =7 x =l

(1) Let y = 2. Mult iply. Subtract 6.

A check confirms that { ( 1, 2)} is the solution set.

CHAPTER 4

CONCEPTS

x - 2y - x

Step 1

+ 2y = - 2

Assign variables.

Step 3

Write two equations using both variables.

0=

Solve the system.

Step 5

State the answer.

Step 6

Check.

0

Solution set:

{ (x , r) Ix - 2y = 6 }

The sum of two numbers is 30. Their difference is 6. Find the numbers. Let x

one number, and let y

=

+y x- y x

= =

the other number.

=

30 6

(1 ) (2)

= 36

2x

Add.

x = 18

Step 4

6

-x + 2y = - 6

Solution set: 0

Read.

Step 2

x - 2y =

6

=

O= 4

If the result is a true statement, such as 0 = 0, the graphs are the same line, and an infinite number of ordered pairs are solutions.

Solving an Applied Problem Use the modified six-step method.

333

EXAMPLES

If the result of the addition step (Step 3) is a false statement, such as 0 = 4, the graphs are parallel lines and there is no solution.

m Applications of Linear Systems

Summary

Divide by 2.

To find y , let x = 18 in equation (1).

+y

=

30

(1)

18 + y

=

30

Let x = 18.

x

y = 12

Subtract 18.

The two numbers are 18 and 12. The sum is 18

+ 12 = 30,

and the difference is 18 - 12 = 6, as required, so the answer checks.

m

Solving Systems of Linear Inequalities

Graph the solution set of the system.

Solving a System of Linear Inequalities Step 1

Graph each inequality on the same axes.

Step 2

Graph the intersection. Indicate the solution set by shading the intersection of the graphs-that is, the regions where the graphs overlap.

2x + 4y

2:

5

x

2:

1

y

First graph the solid boundary lines 2x

+ 4y = 5 and x

= 1.

Then use a test point, such as (0, 0), to determine the region to shade for each inequality. Using (0, 0), two false statements result. For 2x

+ 4y 2: 5, shade the region "above" the line.

For x;::: 1, shade the region to the "right" of the line. The intersection, the purple shaded region, is the solution set of the system.

334

CHAPTER 4

Chapter 4

Systems of Linear Equations and Inequalities

Review Exercises

m

Determine whether the given ordered pair is a solution of the given system.

2. ( - S, 2)

1. (3 , 4)

4x - 2y Sx

x - 4y = - 13

4

=

+ y=

2x

19

+ 3y =

4

Solve each system by graphing.

3. x

+y =

4

m

5. 2x + 4 = 2y y- x = - 3

4. x - 2y = 4 2x + y = - 2

2x-y = S

6. x - 2 = 2y 2x - 4y = 4

Answer each question.

7. To solve the following system by substitution, which variable in which equation would be easiest to solve for in the first step? Sx - 3y = 7

-x + 2y

=

4

8. After solving a system of linear equations by the substitution method, a student obtained the false statement "O = S." What is the solution set of this system? Solve each system by the substitution method.

9. 3x + y

=

7

11. Sx

10. 2x - Sy= -1 9

x = 2y

y=x +2

x

+ l Sy = + 3y =

30

+ Sy +2=

12. 4x

6

x

=

44

2y

l!J Answer each question. 13. Which system does not require that we multiply one or both equations by a constant to solve the system by the elimination method?

+ 3y =

A. - 4x

7

B.

3x - 4y = 4

+ 8y = 12x + 24y = Sx

13

C. 2x

+ 3y = S

D.

x - 3y = 12

36

x

+ 2y =

3x -

y

=

9

6

14. Ifwe were to multiply equation (1) by - 3 in the system below, by what number would we have to multiply equation (2) in order to do the fo llowing?

+

12y = 7

(1)

3x + 4y = 1

(2)

2x

(a) Eliminate the x-terms when solving by the elimination method. (b) Eliminate the y-terms when solving by the elimination method.

Solve each system by the elimination method.

15. 2x - y

=

13

x+y=8

llltfl

16. - 4x

+ 3y

=

2S

6x - Sy= - 39

17. Sx - 3y

x - 2y

= 11

=4

=

3

- 4x - 2y

=

6

2x

Solve each system by any method.

19. 2x + 3y = - S

20. 6x - 9y = 0

3x + 4y = -8

2x - 3y = 0

+

y

18.

21. x - 2y = s y = x- 1

CHAPTER 4

22.

~+ ~=

7

3 1 23. -x - -y 4 3

-+-=8

1 2 -x+-y 2 3

x

2y

4

3

7 6 5 3

24. 0.3x

= -

=

Review Exercises

+ 0.2y =

335

- 0.5

0.4x - 0.5y = -2.2

llJ Solve each problem using a system of equations. 25. The two leading pizza chains in the United States are Pizza Hut and Domino's Pizza. In November 2016, Pizza Hut had 3063 more locations than Domino's, and together the two chains had 29,063 location s. How many locations did each chain have? (Data from Pizza Today.)

26. Two widely circulated magazines in the United States are Better Homes and Gardens and Game Informer. Together, the circulation for these two magazines during the first six months of 2017 was 14.0 million. The circulation for Game Informer was 1.2 million less than that of Better Homes and Gardens. What were the circulation figures for each magazine? (Data from Alliance for Audited Media.) 27. The perimeter of a rectangle is 90 m. Its length is l~ times its width. Find the length and width of the rectangle.

Perimeter= 90 m

28. Laura has 20 bills, all of which are $ 10 or $20 bills. The total value of the money is $330. How many of each denomination does she have? 29. Sharon has candy that sells for $ 1.30 per lb, to be mixed with candy selling for $0.90 per lb to obtain 100 lb of a mix that will sell for $1 per lb. How much of each type should she use? 30. A plane flying with the wind travels 540 mi in 2 hr. Flying against the same wind, the plane travels 690 mi in 3 hr. Find the rate of the plane in still air and the wind speed. 31. A 40% antifreeze solution is to be mixed with a 70% solution to obtain 90 L of a 50% solution. How many liters of the 40% and 70% solutions will be needed?

Liters of Solution

Percent (as a decimal)

x y

90

m 33. x

32. Nancy invested $ 18,000, part of it at 3% annual simple interest and the rest at 4%. Her interest income for the first year was $650. How much did she invest at each rate?

Liters of Pure Antifreeze

Amount Invested (in dollars)

Rate (as a decimal)

0.40

x

0.03

0.70

y

0.04

0.50

18,000

Graph the solution set of each system of linear inequalities.

+y

2::

2

x-y::=;4

34.

y 2:: 2x 2x

+ 3y::::; 6

35. 3x - y ::::; 3 x

2::

-1

y ::::; 2

Interest (in dollars)

336

CHAPTER 4

Chapter 4

Systems of Linear Equations and Inequalities

Mixed Review Exercises Solve each system by any method.

1. 3x

+ 4y = 6

2.

-3x

4x - Sy= 8

3. y = 4x - 4 3x 5.

+y =

=

0

+ 2y =

0

Sx - 4y

4. x

=

y

+6

2y - 2x = -12

-11

y 14 4 3 x y 8 -+- = 2 12 3

2x

-+ -

6. 0.4x - 0.9y = 0.7

3

0.3x + 0.2y

=

1.4

Solve each problem.

7. Why would it be easier to solve System B by the substitution method than System A? System A: - Sx

+ 6y =

System B: 2x + 9y

- 7

y

2x +Sy= -S

8. Patricia compared the monthly payments she would incur for two types of mortgages: fixed rate and variable rate. Her observations led to the graph shown.

l

=

=

13

3x - 2

Mortgage Shopping

(a) For which years would the monthly pay-

ment be more for the fixed rate mortgage than for the variable rate mortgage? (b) In what year would the payments be the

same, and what would those payments be?

4

8

16

12

Year

9. The perimeter of an isosceles triangle measures 29 in. One side of the triangle is S in. longer than each of the two equal sides. Find the lengths of the sides of the triangle.

y

y =x+5

10. In 2016, a total of 12.8 million people visited the Lincoln Memorial and the World War II Memorial, two popular tourist attractions. The World War II Memorial had 3.0 million fewer visitors than the Lincoln Memorial. How many visitors did each of these attractions have? (Data from National Park Service.) 11. Two cars leave from the same place and travel in opposite directions. One car travels 30 mph faster than the other. After 2~ hr, they are 26S mi apart. What are the rates of the cars?

265 mi

CHAPTER 4

12. Which system of linear inequalities is graphed in the figure? A. x :s: 3

y :s: 1

B. x :s: 3

c. x :2: 3

1

y :s: 1

y

:2:

337

Test

y

D . x::::: 3

y

:2:

1

Graph the solution set of each system of linear inequalities. 13. x

+y< s

y :s: 2x

14.

x - y:=:: 2

x

+ 2y > 4

~~~~~~~~~~~-

Chapter 4

Test

O View the complete solutions

FOR

Step-by-step test solutions are found on the Chapter Test Prep Videos available in MyLab Math.

EXTRA

HELP

1. Determine whether the ordered pair ( 1, - S) is a solution of the system.

to all Chapter Test exercises in MyLabMath.

+y

=

- 3

x- y

=

-9

2x

2. Solve the system by graphing. x + 2y - 2x + y

= =

6 - 7

3. Suppose that the graph of a system of two linear equations consists of lines that have the same slope but different y-intercepts. How many solutions does the system have?

Solve each system by the substitution method. 4. 2x + y

-4

=

5. 4x + 3y x+ y

x =y+ 7

= =

3S

6. y = 6x - 8

3x + y=- l l

0

Solve each system by the elimination method. 7. 2x - y = 4

3x + y 10.

=

8. 4x + 2y = 2 Sx

21

4x + Sy = 2 - 8x - lOy = 6

11.

+ 4y

=

9. 3x + 4y = 9

2x +Sy= 13

7

6x - Sy= 0

12. -2x

- 2x + 3y = 0

7x

+ Sy = + 6y =

14

- 2

Solve each system by any method.

+S - 8y + 10

13. 4y = - 3x

6x

=

14.

6 1 -x - -y

s

3

=

-20

15.

+ 0.3y = - 0.3x + O. ly = 0 .2x

1.0 1.8

2 1 - -x + -y = 11

3

6

Solve each problem using a system of equations. 16. The distance between Memphis and Atlanta is 782 mi less than the distance between Minneapolis and Houston. Together, the two distances total 1S70 mi. How far is it between Memphis and Atlanta? How far is it between Minneapolis and Houston?

338

CHAPTER 4

Systems of Linear Equations and Inequalities

17. In 2016, the two most popular amusement parks in the United States were Disneyland and the Magic Kingdom at Walt Disney World. Disneyland had 2.4 million fewer visitors than the Magic Kingdom, and together they had 38.4 million visitors. How many visitors did each park have? (Source: www.aecom.com) 18. A 15% solution of alcohol is to be mixed with a 40% solution to make 50 L of a final mixture that is 30% alcohol. How much of each of the original solutions should be used?

Lifers of Solution

Percent (as a decimal)

19. Chicago and Kansas City are 500 mi apart. A car leaves Chicago traveling toward Kansas City at the same time that a camper leaves Kansas City traveling toward Chicago. The camper travels 20 mph slower than the car. If they meet after 5 hr, find the rate of each vehicle.

Lifers of Pure Alcohol

Graph the solution set of each system of linear inequalities. 20. 2x + 1y ::::; 14

x-

y> 6 4y+l2;::=:-3x

21. 2x -

y ;::=: 1

22. Which system of linear inequalities is graphed in the figure? A. x>4

B. x3

y< 3

C. x > 4 y4

Cumulative Review Exercises Complete the table offraction, decimal, and percent equivalents. Fraction in Lowest Terms

Decimal

Percent

1 100

1.

2.

5%

3.

0.5

4.

75%

5. Multiply. (b) 1.5 x 100

(a) 1.5 x 10

(c) 1.5 x 1000

6. Divide. (a) 1.5

7-

10

(b) 1.5

7-

100

(c) 1.5

7-

1000

Find the value of each exponential expression.

7. 52

8. 103

10. (0.1) 2

CHAPTERS R-4

11. Find the value of the expression

+ 2y 2 lOy + 3

3x 2

Cumulative Review Exercises

for x

=

1 and y

=

339

5.

12. Evaluate -2 + 6 [ 3 - (4 - 9)] .

Solve each linear equation or inequality. Write solution sets of inequalities using interval notation. 13. 2-3(6x + 2) =4(x+ 1) + 18 5

16. - 8 < 2x + 3

15. - -x 6 < 15 kT 17. Solve the formula P = - for T.

v

Solve each p roblem. 18. A survey of 2000 American teens and young adults in February 2017 measured their participation in leading social media sites. Complete the results shown in the table. Percent Participation among Respondents

Social Media Site Snapchat

79%

Facebook

76%

Actual Number of Respondents Who Participate

1460

lnstagram

Twitter

800

Data from Edison Research, Triton Digital.

19. Two angles of a triangle have the same measure. The measure of the third angle is 4 ° less than twice the measure of each of the equal angles. Find the measures of the three angles.

Graph each linear equation. 21. 3x + y = 6

20. x - y = 4

Find the slope of each line. 22. Through (-5, 6) and ( 1, -2 )

23. Perpendicular to the line y = 4x - 3

Write an equation of each line. Give the final answer in slope-intercept form. I

24. Through ( -4, 1), slope 2

25. Through the points (1 , 3) and ( - 2, - 3)

26. Write an equation of the l.ine satisfying the given conditions. (a) Vertical, through (9, -2)

(b) Horizontal, through ( 4, -1 )

Solve each system by any method.

27. 2x x

y

=

+ 2y =

- 8 11

28. 4x + Sy = - 8 3x +4y =-7

29. 3x + 4y

=

2

6x +Sy= 1

340

CHAPTER 4

Systems of Linear Equations and Inequalities

Solve each problem using a system of equations.

30. Admission prices at a high school football game were $6 for adults and $2 for children. The total value of the tickets sold was $2528, and 454 tickets were sold. How many adult and how many child tickets were sold? Number of Tickets

Cost of Each (in dollars)

Total Value (in dollars)

x

6

6x

y

454

STUDY SKILLS REMINDER It is not too soon to begin preparing for your final exam. Review Study Skill 10, Preparing for Your Math Final Exam.

31. A chemist needs 12 L of a 40% alcohol solution. She must mix a 20% solution and a 50% solution. How many liters of each will be required to obtain what she needs? 32. Graph the solution set of the system of linear inequalities. x

+ 2y

::::; 12

2x- y::S 8

EXPONENTS AND POLYNOMIALS Exponents and scientific notation, two of the topics of this chapter, are often used to express very large or very small numbers. Using this notation, one light-year, which is about 6 trillion miles, is written

DI

The Product Rule and Power Rules for Exponents

DI

Adding, Subtracting, and Graphing Polynomials

Ill

Integer Exponents and the Quotient Rule

Ill Ill

Multiplying Polynomials

SUMMARY EXERCISES

Applying the

Rules for Exponents

m

m

Special Products Dividing Polynomials

Scientific Notation

6 x 1012•

341

342

CHAPTER 5

Exponents and Polynomials

The Product Rule and Power Rules for Exponents OBJECTIVES 1 Use exponents. 2 Use the product rule for exponents.

OBJECTIVE 1 Use exponents. Recall from earlier work that in the expression 52, the number 5 is the base and 2 is the exponent, or power. The expression 5 2 is an exponential expression. Although we do not usually write the exponent when it is 1, in general

3 Use the rule (am)n = amn. 4 Use the rule (ab) m = ambm. 5 Use the rule

(EJ

= : :.

6 Use combinations of the rules for exponents.

7 Apply the rules for

a1

= a,

for any quantity a.

1!£MMQlll Using Exponents Write 3 · 3 · 3 · 3 in exponential form and evaluate. Because 3 occurs as a factor four times, the base is 3 and the exponent is 4. The exponential expression is 34 , read "3 to the f ourth power" or simply "3 to the fo urth."

exponents in a geometry application.

3 · 3 ·3 · 3

means

34 ,

which equals

81. NOW TRY~

4 factors of 3

IJ:fdMQl#j Evaluating Exponential Expressions Identify the base and the exponent of each expression. Then evaluate.

VOCABULARY

Expression

D base D exponent (power) D exponential expression

(a) 5 4

(b) - 54 (c) ( - 5) 4

Base

Exponent

Value

5 . 5 . 5 . 5, which equals 625

5

4

5

4

- 1 . (5 . 5. 5 . 5),

-5

4

(-5)( - 5 )( - 5 )(-5), which equals 625

which equals

NOW TRY~

~NOWTRY

~EXERCISE 1 Write 4 · 4 • 4 in exponential form and evaluate.

~NOWTRY

~EXERCISE2 Identify the base and the exponent of each expression. Then evaluate. (a) 34

(b) - 34

(c)

(- 3) 4

0

CAUTION

Compare Examples 2(b) and 2(c). In -54 , the absence of parentheses

means that the exponent 4 applies only to the base 5, not -5. In (-5) 4 , the parentheses mean that the exponent 4 applies to the base - 5. In summary, -an and (-a )n are not necessarily the same. Expression

- an ( - a)n

Base

Exponent

Example

a

n

-a

n

= - (3 . 3 ) = - 9 2 ( - 3) = ( - 3 ) ( - 3 ) = 9 - 32

OBJECTIVE 2 Use the product rule for exponents. To develop the product rule, we use the definition of exponents. 24 . 23 4 factors

3 factors

~~~-~

NOW TRY ANSWERS 1. 43; 64 2. (a) 3; 4; 8 1 (b) 3; 4 ; -8 1 (c) -3 ; 4; 8 1

- 625

= (2 . 2. 2. 2)(2. 2. 2) =2·2·2·2·2·2·2 4 + 3 = 7 factors

= 21

SECTION 5.1

The Product Rule and Power Rules for Exponents

343

62. 63

Also,

= (6. 6)(6. 6. 6) =6·6·6·6·6

= 65. Generalizing from these examples, we have the following. 24

•

23

is equal to

62 · 63 is equal to

2 4 +3 ,

which equals

27 .

62+3, which equals

65 .

In each case, adding the exponents gives the exponent of the product, suggesting the

product rule for exponents. Product Rule for Exponents

= am+n.

For any positive integers m and n, a 111 ·a" (Keep the same base and add the exponents.)

62

Example:

•

6 5 = 6 2 +5 = 6 7

0

CAUTION When using the product rule, keep the same base and add the exponents. Do not multiply the bases. 62

/:.NOW TRY

~EXERCISE3 Use the product rule for exponents to simplify each expression, if possible. (a) Y2 • Y • y5

(d)

32

•

65

is equal to

67,

Use the product rule for exponents to simplify each expression, if possible. (a) 6 3

65

•

(b) x 2

= 63+5 =

68

+

(c) m 4

(d)

•

m3

•

m5

(e) 23

x

Product rule Add the exponents.

= x2+ 1 = x3

a = a1, for all a. Product rule Add the exponents.

This can be written as m'm3m5.

= m4+3+5

Product rule

= m12

Add the exponents.

Think: 23 means 2 . 2. 2.

•

= x2 . x l

53 33

not 36 7 .

l:f:t!JMQl#I Using the Product Rule

(b) (2x3 ) ( 4x6 ) (c) 24

•

23 . 32

Think: 32 means

3 . 3.

The product rule does not apply. The bases are different.

= 8. 9

Evaluate 23 and 32 .

= 72

Multiply.

+ 24 The product rule does not apply. This is a sum, not a product. = 8 + 16 Evaluate 23 and 24. = 24 Add .

NOW TRY ANSWERS 3. (a) y 8 (b) 8x9 (c) The product rule does not appl y; 2000 (d) The product rule does not apply; 36

(f) (2x3 )(3x7 )

-c::::'.2x 3 means 2 • x 3 and 3x 7 means3 • x 7.)

(2 · 3) · (x3 • x7 ) = 6x3+7

Multiply; product rule

= 6x10

Add the exponents.

=

Commutative and associative properties

NOW TRY~

344

CHAPTER 5

Exponents and Polynomials

0

CAUTION expressions.

Note the important difference between adding and multiplying exponential 8x 3

+ 5x 3

(8x 3 )(5x 3 )

+ 5)x 3 ,

means

(8

means

(8 · 5)x 3 +3 ,

which equals

13x 3 .

which equals 40x 6 .

= amn.

OBJECTIVE 3 Use the rule ( am)n

We can simplify an expression such as (5 2 ) 4 with the product rule for exponents, as follows.

(52)4 =

52 . 52 . 52 . 52

= 52+2+2+2 = 58

Definition of exponent Product rule Add.

Observe that 2 · 4 = 8. This example suggests power rule (a) for exponents. Power Rule (a) for Exponents

For any positive integers m and n, (a111 ) 11 = a11111 • (Raise a power to a power by multiplying exponents.) Example:

(3 4 ) 2

=

34 · 2

=

3s

l!:m. NOW TRY "7 EXERCISE 4

li:f4MQlll Using Power Rule (a)

Use power rule (a) for exponents to simplify. (a) ( 41)s (b) (y4) 7

Use power rule (a) for exponents to simplify. (a) (25) 3

(c) (x2)5

(b) ( 57) 2

= 25· 3

= 57· 2

=

= 215

= 514

=xlO

OBJECTIVE 4 Use the rule ( ab

r

x2·5

Power rule (a) Multiply. NOWTRYg

= ambm.

Consider the following.

(4x) 3 = ( 4x)( 4x)( 4x)

Definition of exponent

= (4 · 4 · 4)(x · x · x)

Commutative and associative properties Definition of exponent

This example suggests power rule (b) for exponents. Power Rule (b) for Exponents

r

For any positive integer m, (ah = a111b"'. (Raise a product to a power by raising each factor to the power.) NOW TRY ANSWERS 4. (a) 4Js

(b)

y28

Example:

(2p)5

= 25p5

SECTION 5.1

i:.NOWTRY

~EXERCISES Use power rule (b) for exponents to simplify. (a) (-5ab) 3 (c)

(b) (4t3p5) 2

lf)ij&!Qllj

The Product Rule and Power Rules for Exponents

345

Using Power Rule (b)

Use power rule (b) for exponents to simplify. (a) (3xy) 2

(b) 3(xy) 2

= 32x2y2

(-3) 6

= 3( x2y2)

Power ru le (b)

= 9x2y2

Power rule (b) Multiply.

(c) (2m2p3) 4

= 2 4(m2) 4(p3) 4

Power rule (b)

= 24m8p 12

Power rule (a)

Compare parts (a) and (b). Pay attention to the use of parentheses.

= 16m8p1 2 (e) ( - 4) 8

(d) ( - 56) 3

=( -1 ·56) 3

-a = - 1·a

= ( - 1)3. (56) 3

Power rule (b) ~~

= - 1 ·5 18

Power rule (a)

= - 5 '8

Multiply .

=( - 1 ·4) 8

-a= - 1 ·a

= ( - 1)8( 48)

Power ru le (b)

Raise - 1 to the designated power.

1

· a =a NOW TRY~

0

CAUTION

Power rule (b) does not apply to a sum.

(4x )2 =42x 2 , but (4 + x) 2 ¥- 42 + x 2 .

OBJECTIVE 5

Use the rule

(E-r = :: .

Consider the following. Definit ion of exponent

2·2·2·2 3 .3 . 3.3

Mu ltip ly fractions .

24 Definition of exponent

34

This example suggests power rule (c) for exponents.

Power Rule (c) for Exponents

For any positive integer m,

a"' = -b"' (-a)m b

(where b =I= 0).

(Raise a quotient to a power by raising both numerator and denominator to that power.) NOW TRY ANSWERS 5. (a) - 125a3b3 (b) l6t6p 10 (c) 36

Example:

( ~3 ) 2 -- 5322

346

CHAPTER 5

Exponents and Polynomials

1!£MMQl#ill Using Power Rule (c)

/:.NOW TRY

~EXERCISES

Use power rule (c) for exponents to simplify.

Use power rule (c) for exponents to simplify .

(a) (a)

(~)3

(b)

(~)5 (q

~

(1)5

0)

EmD

(})4

(b)

(c) ( :

25 35

Power rule (c)

14 54

32 243

Simplify.

--

In general,

m4

n4 (where n # 0)

1 625

In Example 6(b), we used the fact that 14 1n = 1 ,

)4 NOWTRYg

= 1 because 1 · 1 · 1 · 1 = 1.

for any integer n.

Rules for Exponents

For positive integers m and n, the following hold true. Examples

Product rule Power rules (a) (b) (c)

a111 • a" = a111 +" (a 111 )11 = a11111 (ab) 111 = a111b111 ( ~b)"' -- a111 bm

62 • 65 = 62+5 = 67 (34)2

= 34·2 = 38

(2p )5 = 25p5

(where b

+ 0)

OBJECTIVE 6 Use combinations of the rules for exponents.

IJ:MMQlll Using Combinations of the Rules Simplify. (a)

(b) (5x) 3(5x) 4

(1)2 ·

23

= (5x ) 7

22 23

== - 2 · 3

1

NOW TRY ANSWERS 27

6. (a ) 64

(b)

iqs

•

1

Another correct way to simplify this expression follows. Multiply fractions.

( 5x )3 ( 5x ) 4

22+3 32

Product rule

25 32

Add.

32 9

Power rule (b)

Power rule (c)

22 . 23 32

Product rule

Alternative

solution Power rule (b) Commutative property

= 53+4x3+4

Product rule Add the exponents.

Apply the exponents.

SECTION 5.1 l:.NOWTRY

"7 EXERCISE 7

(~)3 ·3

2

34 7

(c) ( 2x2y3 ) 4( 3.xy2) 3

Simplify. (a)

The Product Rule and Power Rules for Exponents

(b) (8k) 5 (8k) 4

(c) (x4y )5( -2x2y5) 3

= 24(x2)4(y3 ) 4 • 33x3( y2 )3 = 24xsy12 • 33x3y6

Power rule (b}

= 24 . 33 . x Bx3y l2y6

Commutative and associative properties

= 16. 27 .

Apply the exponents; product rule

xlly l S

= 432x11y1s

Power rule (a)

Multiply.

Think of the negative sign as a factor of - 1.

= ( -1

0

. x3y) 2( -1 . x5y4 ) 3

- a = - 1 ·a

= ( - 1)2(x3 )2y2 . ( - 1)3(x5) 3( y4 )3

Power rule (b}

= (-1 )2x6y2 • (-1 )3x 1sy12

Power rule (a)

= (-l) sx21y14

Product rule

= - x21y 14

Simplify; ( -1 ) 5 = - 1

NOW TRY~

Be aware of the d istinction between (2y) 3 and 2y3 .

CAUTION

(2y) 3 = 2y . 2y . 2y = 8y 3 2y 3 = 2 . y . y . y

The base is 2y. The base is y .

OBJECTIVE 7 Apply the rules for exponents in a geometry application.

l(f;M!Qlj:I

Using Area Formulas

Find an expression that represents the area, in square units, of (a)

FIGURE 1

(b) FIGURE 2.

5x3

Assume x > 0, m > 0.

h

6x4 FIGURE 1

FIGURE 2

(a) For FIGURE 1, use the formula for the area of a rectangle.

NOW TRY ANSWERS 7. (a)

~~~ (b) 89k9

(c) - 8x26y20

Si

= LW

Area formula

Si

= (6x4 ) ( 5x3 )

Substitute.

Si = 6 · 5 · x 4 + 3

Commutative property; product rule

Si = 30x7

Multiply. Add t he exponents.

and

348

CHAPTER 5

Exponents and Po lynomials

I:. NOW TRY "7 EXERCISE 8

(b )

is a triangle with base 6m4 and height 3m3 .

FIGURE 2

Find an expression that represents the area, in square units, of the fig ure. Assume

x > 0.

~ IOx8

NOW TRY ANSWER 8. 15x

1 2

.511

= - bh

.511

= -( 6m4 )(3m3 )

Area formula

1

6m 4

Substitute.

2

FIGURE 2 (repeated)

1

.511

= - (6 · 3 ·

.511

= 9m7

2

m4 +3 )

Properties of real numbers; product rule Multiply. Add the exponents.

NOW TRY

g

15

5.1 Exercises O Video solutions for select problems available in MyLab Math

0

FOR

EXTRA

HELP

Mylab Math

Concept Check Determine whether each statement is true or false. If false, correct the right-hand side of the statement.

1. 3 3

=

5. - 22 = 4

6. 2 3

•

(~)2 = :2

24

3. (a2)3 =as

4.

24 = 4 7

7. (3x) 2 = 6x

8. ( - x) 2 = x 2

2. (-2) 4

9

=

Write each expression in exponential form. See Example 1.

9. w . w . w . w . w . w

11.

10. t . t . t . t . t . t . t

(~)(~)(~)(~)(~)(~)

12.

(~)(~ )(~ )(~)(~)

13. (- 4 )( -4 )( - 4 )( - 4)

14. (-3)(-3)(-3)(-3)(-3)(-3)

15. (- 7y)( - 7y)( - 7y)( - 7y)

16. (- 8p)( - 8p)( - 8p)( - 8p)( - 8p)

17. Explain how the expressions ( -3) 4 and - 34 are different. 18. Explain how the expressions (5x) 3 and 5x3 are different.

Identify the base and the exponent of each expression. In Exercises 19-26, also evaluate. See Example 2.

19. 35

20. 27

21. (-3) 5

22. ( - 2) 7

23. (- 6) 2

24. (- 9) 2

25. - 62

26. - 92

27. (-6x) 4

28. ( - 8x) 4

29. - 6x4

30. - 8x4

Concept Check Simplify each expression.

31. 82 • 85 =

32. 5m2 • 2m6 = (5 . - - ) . (m- . m- )

s-+-

= _ _ m-+-

Use the product rule for exponents to simplify each expression, in exponential form. See Example 3.

34. 36 • 37

33. 5 2 • 56 36. 5 3

•

58

•

52

37.

t 3 • t 8 • t 13

if possible. Write each answer 35. 42 • 47 • 4 3 38. n5 • n6 • n 9

SECTION 5.1

The Product Rule and Power Rules for Exponents

39. (-8r4 )(7r3 )

40. (l0a7 )(-4a3)

41. (-6p5)( - 7p5)

42. (- 5w8 )(-9w8 )

43. (5x2)( - 2x3)( 3x4)

44. ( 12y3)(4y)(-3y5)

45. (-2x4)(-5x6 )(-3x) 48. 4 12 + 45

46. ( - 10y2 )( - 2y 5)(-8y)

47. 38

49. 5 8 • 39

50. 63 • 89

349

+ 39

Use the power rules for exponents to simplify each expression. Leave answers in exponential form in Exercises 51-66. Assume that variables in denominators are not zero. See Examples 4-6. 51. (43) 2

52. (83) 6

53. (t4) 5

54. ( y6)5

55. (7r) 3

57. (5xy) 5

58. (9pq ) 6

59. (-5) 4

56. ( l lx) 4 60. ( -6 ) 6

61. ( - 5) 5

62. (-6) 3

63. 8(qr) 3

64. 4(vw ) 5

65.

(~)

(~ )3

67.

(~)3

68.

(~)5

69.

71.

(~)3

72.

(~)4

73. ( -

75. ( -2x2y) 3

76. ( -5m4p2)3

Concept Check example,

8

3 66. (12) 7 70.

~)5

Gr

74. (

4p ) 3 -q

78. (4x3y5) 4

77. (3a3b2) 2

If a negative number is raised to an even power, the result is positive. For (-3) 4 = (- 3)( -3 )( -3)(-3) = + 81-thatis, 81.

lf a negative number is raised to an odd power, the result is negative. For example, (-3) 5 = ( -3 )( -3 )( - 3)( - 3)( - 3) = - 243. Without actually performing the computation, state whether the power is positive or negative. 79. (a) ( -4) 3 80. (a) ( -4)

4

(b) ( -4) 4

(-4) 2

(b)

(-4) 3( -4) 8

(c) ( -4) 12

(d) ( -4) 13

(c) ( - 4 ) 12( - 4 ) 16

(d) (-4)(-4) 96

Simplify each expression. Leave answers in exponential form in Exercises 81-90. See Example 7. 3

~)

5

~)6

(~) • 9 3

81.

(%) (%)2

82. (

84.

(~)4. 33

85. (2x)9(2x)3

86. ( 6y) 5 ( 6y) 8

87. (6x2y3)5

88. (5r5 t 6 ) 7

89. (x2)3(x3)5

90. (y4) 5(y3) 5

91. (2w2x3y) 2(x4y) 5

92. (3x4y2z)3(yz4)5

•

• (

83.

93. (-r4s) 2(-r2s3)5 95. (

5a2b5)3 6(c c

-

Concept Check

~

0)

96. (

6x3y9 )4 T

(z ~ 0)

Determine whether each statement is true or false.

97. (a) ( -3) 3 = 33

98. (a) (-2) 2( -2) 4 = 26

(b) (-3) 4 = 34

(b) ( -2 )5( -2)5 = 210

(-3) 5 =

(c) ( -2) 8( -2 )9 = 211

(c)

35

(d) (-3) 6 =3 6

(d) (-2) 10(-2)20=230

2

350

CHAPTER 5

Exponents and Polynomials

99. Concept Check A student simplified the expression ( 102 ) 3 as 10006 . This is incorrect. WHAT WENT WRONG? Simplify correctly.

100. Concept Check A student simplified the expression (3x2y3) 4 as 3 · 4x8y 12 , or 12.x8y 12 . This is incorrect. WHAT WENT WRONG? Simplify correctly. Find an expression that represents the area, in square units, of each figure. See Example 8. (If necessary, refer to the formulas at the back of this text. The symbol I in the figures indicates 90° angles.)

101.

~[_ _ _ _~I

102. 3x2

4x 3

103.

4m4

f

-~~S-P_2

104.

____

2p5

G

Find an expression that represents the volume, in cubic units, of each figure. (If necessary, refer to the formulas at the back of this text.)

105.

5x2 5x

106.

2

I I /

9xy 3

5~

~ 4x y4

5x2

2

JJ.- -

/

/

Compound interest is interest paid on both the principal and the interest earned earlier. The formula for compound interest is A = P( 1 +

r)",

where A is the amount accumulated from a principal of P dollars left untouched for n years with an annual interest rate r (expressed as a decimal). Use the preceding formula and a calculator to find A to the nearest cent.

107. P = $250, r = 0.04, n = 5

108. P = $400, r = 0.04, n = 3

109. P = $1500, r = 0.015, n = 6

110. P = $2000, r = 0.015, n = 4

Integer Exponents and the Quotient Rule OBJECTIVES

1 Use 0 as an exponent. 2 Use negative numbers as exponents.

3 Use the quotient rule for exponents.

4 Use combinations of t he rules for exponents.

Consider the following list.

= 16 23 = 8 24

22 = 4

As exponents decrease by 1, the results are divided by 2 each time.

Each time we decrease the exponent by 1, the value is divided by 2 (the base). Using this pattern, we can continue the list to lesser and lesser integer exponents.

SECTION 5.2

Integer Exponents and the Quotient Rule

351

21 = 2 This suggests the justification of the definition to follow.

2° = 1 We continue the pattern here.

2- 3 = ~ 8

From this list, it appears that we should define 2° as 1 and bases raised to negative exponents as reciprocals of those bases.

OBJECTIVE 1

Use 0 as an exponent.

The definitions of 0 and negative exponents must be consistent with the rules for exponents from the preceding section. For example, if we define 6° to be 1, then 6° · 62 = 1 · 62 = 62

and

6° · 62 = 6°+2 = 6 2 ,

and we see that the product rule is satisfied . The power rules are also valid for a 0 exponent. Thus, we define a 0 exponent as follows.

Zero Exponent

For any nonzero real number a,

Example:

lif!f!1MQlll

,;;_NOW TRY

~EXERCISE 1 Evaluate.

= 1.

=1

Using Zero Exponents

Evaluate. (a) 60°

(a) 6°

(b) - 12° (c) (- 12x) 0

17°

a0

(b) ( - 60)0 = 1

=1

(c) - 60° = - (1)

(x ,= 0)

(e) 6y0

(d) 140 - 120

(g) 8°

=- 1

= 6 ( 1 ) = 6 (y =F 0)

+ 11° = 1 + 1 = 2

(d)y0 = 1

(y# O)

(f) ( 6y) 0 = 1

(y =F 0)

(h) -8° - 11° = - 1 - 1 = -2

NOW TRY~

0

CAUTION Look again at Examples 1 (b) and 1 (c). In (-60) 0 , the base is -60, and because any nonzero base raised to the 0 exponent is 1, (-60) 0 = 1. In -60°, which can be written -(60) 0 , the base is 60, so - 60° = - 1.

OBJECTIVE 2

From the lists at the beginning of this section, 2- 1 = ~' 2-2 = ~'and 2-3 = ~ . We can make a conjecture that 2- 11 should equal ~ . Is the product rule valid in such cases? For example,

NOW TRY ANSWERS I. (a) I

(b) - I

(c) I

Use negative numbers as exponents.

(d) 0

352

CHAPTER 5

Exponents and Polynomials

The expression 6- 2 behaves as if it were the reciprocal of 62, because their product is 1. The reciprocal of 62 is also ~, leading us to define 6- 2 as ~, and generalize accordingly.

Negative Exponents

1

a-" - - a"'

For any nonzero real number a and any integer n, Example:

By definition, a-" and a" are reciprocals. a" · a- " = a" · -

1

a"

Because 111

= 1

= 1, the definition of a-" can also be written as follows . a- 11

= _!_ = .!'.'.. = a"

a"

1 ( 3

For example,

l(MMQl#j

(.!_)" a

)-2 32_ =

Using Negative Exponents

Write with positive exponents and simplify. Assume that all variables represent nonzero real numbers. (a) 3- 2

(c) (

1

-

-

1

- 32

)-3

2

=

1

-

1

9

23

=

1

(b) 5- 3 = - = 53 125

-

8

~ and 2 are reciprocals. (Reciprocals have a product of 1.)

Notice that we can change the base to its reciprocal if we also change the sign of the exponent. (d)

(~r4 =

(e)

(~r5 ~ and 4 :?. are

(1)5

3

Power rule (c)

3s 4s

Power rule (c)

Apply the exponents.

243 1024

Apply the exponents.

?. and ?.2 are

(%)4

5

54 24 625 16

reciprocals.

=

reciprocals.

Integer Exponents and the Quotient Rule

SECTION 5.2

(g) 3p- 2

/:.NOW TRY

~EXERCISE2

1

Write with positive exponents and simplify. Assume that all variables represent nonzero real numbers. (a) 2- 3 (c)

(~r4

(b)

(~r2

(d)

3-2+ 4-2

(f) x - 6y3

(e) p- 4

353

4

2 2

4

Apply the exponents. Find a common denominator.

4

Multiply.

Subtract.

4 1 (h) -=4

x

x3 1n = 1,

1

for any integer n.

y4

x3

y4

Power rule (c)

~ and x are reciprocals.

= x4

In general,

Multiply.

1

NOW TRY~

---a" a-" .

0

CAUTION A negative exponent does not indicate a negative number. Negative exponents lead to reciprocals. Expression

Example 1 1

3- 2 = - 2 = 3 9

Not negative

1 1 - 3- 2 = - - = - 32

t

t

t9

Negative

Consider the following. 2- 3

3-4 23 Definition of negative exponent

34 -

23

- -

34

34 = -

23 34 23

NOW TRY ANSWERS 2. (a)

I

S

(b) 49

25 (d) 144

(e) -

(c)

I

p

4

16

81 3

(f)

~6 x

· -

Therefore,

2-3 3-4

34 23·

ba means a ~. b . To d ivide, multiply by the reciprocal of the divisor. Multiply.

354

CHAPTER 5

Exponents and Polynomials

Negative-to-Positive Rules for Exponents

For any nonzero numbers a and band any integers m and n, the following hold true.

a-m b" -= ba"' 11

24

Examples:

3s

and

(~)-m = (~)m

and

(~r3 = (~Y

~NOW TRY

li:f4MQl#I Changing from Negative to Positive Exponents

Write with positive exponents and simplify. Assume that all variables represent nonzero real numbers.

Write with positive exponents and simplify. Assume that all variables represent nonzero real numbers. 4-2 m- s 53 125 p' p (a) (b) 5- 3 42 pl ms ms 16

\:7 EXERCISE 3

(a)

5-3

n- 4

6- 2

(b) m-2

x2y-3 (c)

sz-4

(d)

(;s)-3

a- 2b (c) - -3 3d -

bd 3 3a 2

Notice that b in the numerator and the coefficient 3 in the denominator are not affected.

(d) (;yr4

=

(;)4

Negative-to-positive rule

24y4 Power rules (b) and (c)

x4

16y 4 x4

@CAUTION

Apply the exponent.

NOW TRY

Be careful. We cannot use the rule

a-m b" b -n =

;.n to change negative exponents to

positive exponents if the exponents occur in a sum or difference of terms.

1

Example:

5-2 + 3-1 _ _ written with positive exponent s is 7 2 3

1

-+S2 3 1

7 -23

OBJECTIVE 3 Use the quotient rule for exponents. Consider the following. 65 63

-= NOW TRY ANSWERS 36 m2 3. (a) (b) - 4 125 n 27s3

(d) -

x 2z4

(c) -

5y 3

6. 6 . 6 . 6. 6 6. 6 . 6

=62

Divide out the common factors.

The difference of the exponents, 5 - 3 = 2, is the exponent in the quotient. 2 6 6 · 6 = _!_ = 6_2 Divide out the Al so, 64 6 . 6 . 6 . 6 62 · common factors.

r3

Here, 2 - 4

g

= -2. These examples suggest the quotient rule for exponents.

SECTION 5.2

Integer Exponents and the Quotient Rule

355

Quotient Rule for Exponents

For any nonzero real number a and any integers m and n,

am a" -

-

am-n

.

(Keep the same base and subtract the exponents.)

58 54

- = 58-4 = 54

Example:

CAUTION A common error is to write ~= 18 - 4 = 14 . This is incorrect. By the quotient rule, the quotient must have the same base, 5, just like the product in the product rule.

0

58 = 58-4 = 54 54

-

We can confirm this by writing out the factors.

5s

54

5 · 5 · 5 · 5 ·5· 5 · 5 · 5 5.5.5.5

--------- =

54

/:.NOW TRY

lif:f;lMQlll

Simplify. Assume that all variables represent nonzero real numbers. 63 t4 (a) 64 (b) i- s

Simplify. Assume that all variables represent nonzero real numbers.

"7 EXERCISE 4

(p + q) -3 (c) (p

+ q) - 1

(p =F -q )

Using the Quotient Rule

58 = 5 8- 6 = 52 = 25 56 ~,

(a) -

42 1 1 = 42 - 5 = 4- 3 = - = 45 43 64

(b) -

'\. Keep the same base. ]

5-3 = 5- 3 ( c) 5-7

( - 7)

=

54 = 625

Be careful with signs.

52.xy- 3 (d) - - 3- 1x- 2y2

5

(d) q-3 q

= q5-(-3) = q8

(m + n) - 2 (f) (m + n) - 4 32 34

xs x3

= 3 2-4 • xs-3 = 3 - 2x 2 x2

Quotient rule Subtract. Definition of negative exponent

32

= (m + n) -2-(-4) = (m + n)-2+ 4 = (m + n) 2 (m ~ - n) The restriction m ~ -n is necessary to prevent a denominator of 0 in the original expression. Division by 0 is undefined.

x2 Apply the exponent.

9

7x-3y2 (h) 2-1x2y-s

(g) 3x- 5

1

NOW TRY ANSWERS 4. (a)

I

6

(b) 3

75x (d) -

y5

t9

(c) (p

+ q)

4

Avoid the error of applying -5 to3.

=3 · x5 3 xs

- 5 applies only to x.

7 . 2'y2y5 x2x3

Negative-topositive ru le

Multiply.

14y 7

Multiply; product rule

xs

NOW TRY

g

356

CHAPTER 5

Exponents and Polynomials

The definitions and rules for exponents are summarized here.

Summary of Definitions and Rules for Exponents

For all integers m and n, and all real numbers a and b for which the following are defined, these definitions and rules hold true. Examples 74 . 75 = 74+5 = 79

Product rule a6

Zero exponent

(-3) 0 = 1

= 1 1

5- 3 =

a-"= a"

Negative exponent

am

-= a"

Quotient rule

_!_

53

2

_2 = 22- 5 = 2 - 3 = _1 25 23

am-11

(42)3 = 42· 3 = 46

Power rules (a)

(a"')"

=

a"'"

(b)

(ah)"'

= a"'b"'

(c)

(~)"' = b"'

(3k)4

= 34k4

a"'

a-m b-"

Negative-topositive rules

b"

= a"'

(~)-"' = (~)"'

OBJECTIVE 4 Use combinations of the rules for exponents.

1$MMQl4j

Using Combinations of Rules

Simplify. Assume that all variables represent nonzero real numbers. ( 42)3 (a) - 45

(b) x 2 • x- 6

46 45 = 46-5

Power rule (a)

•

= x2+(-6)+(-1)

Product ru le

= x -5

Add.

Quotient rule

=4 1

Subtract.

=4

a 1 = a, foralla.

(c) (2x) 3(2x) 2

x- 1

x5

(2x) 3 (2x) 2 = 23x3 . 22x2

Definition of negative exponent

Alternative solution

= (2x) 5

Product ru le

= 25x5

Power rule {b)

= 23+2x3+2

Product rule

= 32x5

25

= 25x5

Add exponents.

= 32x5

25 = 32

= 32

Power rule {b)

SECTION 5.2

Integer Exponents and the Quotient Rule

l:.NOWTRY

"7 EXERCISE 5

(d) (2;3r4

Simplify . Assume that all var iables represent nonzero real numbers. 3

=

1s

(b) ( 4t) 5 ( 4t) - 3

(a)

(33)4

(c)

c~4r3

(5k) - 6 (5k) 8 (e) (5k) 7 (5k) - 4 (f)

(e)

( 3x- 2 ) - 3 4- ly3

(2~3 )4

Negative-topositive rule

--

3-3x6 43y-9

Power rules (a)-(c)

54 4x12 2

Power rules (a)-(c)

x6y9 43. 33

Negative-topositive rule

625 16x 12

x6y9 Apply the exponents.

43

1728

( 4m)-3 (3mt4 4-3m- 3 3-4m- 4

(7y)-3+4 (7y) 12+(-10)

Power rule (b)

(7y)2

(b) 16t2 c 17

(d) 32a 11 b 9

(c)

1000

Y, 343 2

I (e) 5k

5.2 Exercises O

Video solutions for select problems available in MyLab Math

Quotient rule

34m l -43

Subtract.

81m 64

Apply the exponents.

Product rule

(7y) I

Negative-topositive rule

--

43

--

5. (a) 27

33 = 64 . 27 = 1728

(g) (7y) 12(7y) - IO

34m4- 3

-

•

(7y) -3(7y)4

34m4 43m3

NOW TRY ANSWERS

357

Add.

= (7y) 1- 2

Quotient rule

= (7y) - I

Subtract.

1 7y

a-1 = .!. a' for a ?6 0. NOW TRY~

Imm

Because the steps can be done in several different orders, there are many correct ways to simplify expressions like those in Example 5. For instance, see the alternative solution for Example 5(c).

0

FOR EXTRA HELP

Concept Check

1. (- 2) - 3

Concept Check 511

9.

58

Mylab Math Determine whether each expression is positive, negative, or 0.

2.( - 3) - 2

3. - 24

4. - 36

7. 1 - 5°

8. I - 7°

Simplify each expression. 6-5

10. 6-2

=5- - -

=6--(-

= 5-

= 6-

1

6-

l

358

CHAPTER 5

Exponent s and Polynomials

Decide whether each expression is equal to 0, 1, or -1. See Example 1.

11. 9°

12. 3°

13. (-2) 0

14. (- 12) 0

15. - 8°

16. - 6°

17. - ( - 6) 0

18. - (- 13) 0

19. (-4) 0 -4°

20. (-1 1) 0 - 11°

21. (3x)O

23. s0 - 12°

24. 6° - 13°

010 25. 120

(x

~

22. ( -5t)O

0)

(t

~

0)

os 26. 20

Concept Check Match each expression in Column I with the equivalent expression in Column II. Choices in Column II may be used once, more than once, or not at all. (Jn Exercise 27, x ~ 0.)

I 27. (a) XO

II

I 2-4 28. (a)

A. 0

II A. 8

(b) - xO

B. 1

(b) ( - 2) - 4

B. 16

(c) 7x 0

c.

(c) z-4

I c. --

(d) (7x)O

D. 7

(e) - 7x0

E. - 7

(f) (- 7x)O

I F. 7

- 1

16

l (d) 2-4

D. - 8

(e)

E. -16

- z-4 I

(f) ( - 2) - 4

F.

16

Evaluate each expression. See Examples 1 and 2.

29. 6°

+ s0

4

32. 5- 4

33.

35. (6r2 7

36. (2Y3 -3

38. (- 4 ) - 3

39. 3x0

41. 5- 1 + r

31. 4- 3

30. 4° + 2°

34.

c 2r

37. ( - 3) - 4

(x

~

40. - 5t0

0)

42. 6- 1 + Z- 1

1

44. 6- 2 - 3- 1

45.

CY' 2

3 3r c

+ (2Y - ' 3

(t

~

0)

2 - Z- 1

43.

r

46.

CY' (

4

+ -Y' 3

3

Simplify each expression. Assume that all variables represent nonzero real numbers and no denominators are zero. See Examples 2-4.

58 47. 5s

48. 106 103

94 49. 9s

3- 2 51. 5- 3

4-3 52. 3- 2

53.

6- 3

55.

62 l

59. 6- 3

4- 2

56.

5

5=I x'2

73 50. 74 6 54. 6- 2

y4

43

57. -=3

58. --=6

l

2 61. -=4

62.

60. 5- 2

x

r

y

3

-=s s

SECTION 5.2

r5

a6

67. --=4

68. -=4

r

Integer Exponents and the Quotient Rule

38y5

70. I02" 3 y

a

(a+b) - 3

-5 -4

72. f__2_ 9r- 3

(x + y )-

73. (a+ b) -4

(x + 2y )- 3

(x + y) - 8 74.

75.

9

359

(x + 2y) -

(p - 3q) -2

76.

5

(p - 3q) -

4

Simplify each expression. Assume that all variables represent nonzero real numbers. See

Example s .

(74)3

77.

79

(3x) - 2

81. (4x) - 3

(2y)- 3 82. (5y)- 4

85. (6x) 4(6x) - 3 (m- 8n- 4) 2

88.

91.

m2n5

e~-2r3

(8x) - 8(8x ) 9 94. (8x) 13(Sx) - 11

97. Concept eh ec k

( x- ly)- 2 -2 z

84.

(p- 4qr3 --=3 r

(m1n) - 2 -4 3 m n

86. ( 10y) 9 ( lOy)- 8

87.

(X- 1y2z) - 2 89· ( - 3 3 ) - 1 x yz

( a2b3c4) -4 90. (a- 2b- 3c-4) - 5

92.

x-y

83.

(2r ) - 4 (2r) 5 93. ( 2r) 9 ( 2r )- 7

( wz- 5) - 2 -3wz

( - 9p) 16 ( - 9p )- 16 96. ( - 9p) - 4l (-9p) 42

( -4y) 8( - 4y) - 8 95. ( - 4y) - 26( - 4y)27 163

· lifi1ed 22 .incorrect Iy as shown . A stu dent s1mp

.!£ = (~)3-2 = 22

31

=

8

2

WHAT WENT WRONG? Give the correct answer.

98. Concept Check

A student simplified - 54 incorrectly as shown. -54 = ( - 5) 4 = 625

WHAT WENT WRONG? Give the correct answer.

Extending Skills Simplify each expression. Assume that all variables represent nonzero real numbers.

99.

(4a 2b3)- 2(2ab- 1) 3 (a3b ) -4

( m6n ) - 2( m2n- 2) 3

100.

(2y- lz2) 2(3y- 2z- 3) 3

101.

(y3z2) - l

102.

(9- 1z- 2x) - 1( 4z 2x4) - 2

103.

(5z- 2x-3)2

104.

m- ln- 2 ( 3p- 2q3 )2( 5p- 1q- 4) - 1 ( p2q- 2) - 3

( 4- 1a- 1b- 2) - 2( 5a- 3b4) - 2 (3a- 3b- 5) 2

360

CHAPTER 5

Exponent s and Po lynomials

SUMMARY EXERCISES

Applying the Rules for Exponents 1. Concept Check Match each expression (a)-(j) in Column I with the equivalent expression A-J in Column II. Choices in Column II may be used once, more than once, or not at all. I

+ 2°

II

(b) 2 1 • 2°

A. 0

B. 1

(c) 2° - 2- 1

(d)2 1 - 2°

c.

D. 2

(e) 2° · 2- 2

(f) 2 1 • 2 1

E. 2

F. 4

(h) 2° . 2°

G. - 2

H. - 4

(a) 2°

(g)

r

2 -

r

1

-1 1

I.

1

J.

4

4

Simplify each expression. Assume that all variables represent nonzero real numbers.

2. 5- 1 + 6- 1

3. - 4° + ( - 4) 0

4. 5- 2

5. - ( - 19°)

6. - ( - 13) 0

7.

8. 8- 1 + 6- 1

9. 52° - (- 8) 0

0 13

21.

16.

( ~)-2 k t-

22.

1 ( ~)3X-

23. (

(3 - lx- 3y )- 1(2x2y- 3) 2 ~-----~

( 5x- 2y2 )- 2

0

13

10. - ( - 80)0

c" ( c2)4

15. (c3) 3(c2)- 6

+ 6- 2

2

4

4

- 9[2 ) -2 2

9x

(TIX-3) - z(x4) - 6

34. - - - - - r 1x- 3

r 4x-3 ) - 2

35. ( 3- 3[

6

2 2 mn- p)- 2(mn- p)3 40. ( m2np4 m2np4

SECTION 5.3

Scientific Notation

361

Scientific Notation OBJECTIVES

OBJECTIVE 1 Express numbers in scientific notation.

1 Express numbers in scientific notation.

Numbers occurring in science are often extremely large (such as the distance from Earth to the sun, 93,000,000 mi) or extremely small (the wavelength of blue light, approximately 0.000000475 m). Because of the difficulty of working with many zeros, scientists often express such numbers with exponents using scientific notation.

2 Convert numbers in scientific notation to standard notation. 3 Use scientific notation in calculations.

Scientific Notation

A number is written in scientific notation when it is expressed in the form

x 10",

a

where 1 ::; Ia I < 10 and n is an integer.

VOCABULARY D scientific notation D standard notation

In scientific notation, there is always one nonzero digit before the decimal point. Scientific notation ,---"-...

x 3.19 x 3.19 x 3.19 x 3.19 x 3.19 x 3.19

mlJDI

10 1 = 3.19 102 = 3.19 103 = 3.19 1=

3.19

10- 2 =

3.19

10-

10- 3

= 3. 19

x 10 = 31.9 '-' x 100 = 319. x 1000 = 3190. x 0.1 = 0.319 '-' x 0.01 = 0.0319 x 0.001 = 0.003 19 V'J

'-A.A./

'-.A./

'-"-A/

Decimal point moves 1 place to the right. Decimal point moves 2 places to the right. Decimal point moves 3 places to the right. Decimal point moves 1 place to the left. Decimal point moves 2 places to the left. Decimal point moves 3 places to the left.

In scientific notation, the multiplication cross x is commonly used.

A number in scientific notation is always written with the decimal point after the first nonzero digit and then multiplied by the appropriate power of 10. Example:

56,200 is written 5.62 X 104 because 56,200

= 5.62 x

10,000 = 5.62

x 104 .

Additional examples:

and

42,000,000

is written

0.000586

is written

2,000,000,000

is written

x 107 • 5.86 x 10- 4 • 2 x 109.

4.2

It is not necessary to write 2.0.

To write a number in scientific notation, follow these steps. (For a negative number, follow these steps using the absolute value of the number. Then make the result negative.)

362

CHAPTER 5

Exponents and Polynomials

Converting a Positive Number to Scientific Notation

Step 1

Position the decimal point. Place a caret 11 to the right of the first nonzero digit, where the decimal point will be placed.

Step 2

Determine the numeral for the exponent. Count the number of digits from the decimal point to the caret. This number gives the absolute value of the exponent on 10.

Step 3

Determine the sign for the exponent. Decide whether multiplying by 1011 should make the result of Step 1 greater or less. • The exponent should be positive to make the result greater. • The exponent should be negative to make the result less.

/:.NOW TRY

1£fi1MQlll Using Scientific Notation

Write each number in scientific notation.

Write each number in scientific notation.

"7 EXERCISE 1 (a) 12,600,000

(a) 93,000,000

Step 1

(b) 0.00027 (c) -0.0000341

Place a caret to the right of the 9 (the first nonzero digit) to mark the new location of the decimal point.

9 A 3,000,000 Step2

Count from the decimal point, which is understood to be after the last 0, to the caret.

9.3,000,000. :"(2.7559

101) X 10- 3

X

= 2.7559 x

Multiply. Use the product rule.

10- 2

Write 27.559 in scientific notation. Product rule

= 0.027559

Write in standard notation.

Thus, 700,000 nanometers would measure 2.7559 X 10- 2 in.,

or

0.027559 in.

NOW TRY

g

ll£f!1MQl4j Using Scientific Notation to Solve an Application As of November 30, 2017, the gross federal debt was about $2.0590 X 10 13 (which is more than $20 trillion). The population of the United States was approximately 326 million that year. About how much would each person have had to contribute in order to pay off the federal debt? (Data from www.usgovernmentdebt.us; www.census.gov) Divide to obtain the per person contribution. 2.0590 x 10 13 326,000,000 2.0590 x 10 13 3.26 x 108

= 2.0590

X

3.26

= 0.63 160 X NOW TRY ANSWERS 4. 3. 1496 x 10- 1 in., or 0.31496 in. 5. 4. 1 X 10- 3 mi2 , or 0.0041 mi2

5.3 Exercises O Video solutions for select problems available in MyLab Math

105 105

= 63,160

Write 326 million in scientific notation.

Quotient rule Divide. Round to 5 decimal places. Write in standard notation.

Each person would have to pay about $63,160. FOR EXTRA HELP

0

NOW TRY

g

Mylab Math

Concept Check Match each number written in scientific notation in Column I with the correct choice from Column II. Not all choices in Column II will be used. I

II

1. (a) 4.6 x 10-

4

4

(b) 4.6 x 10

(c) 4.6

X

(d) 4.6 X

105 10- 5

A. 46,000

I

2. (a) 1 x

II

109

x106

B. 460,000

(b) 1

c.

(c) 1 X 108

0.00046

D. 0.000046 E. 4600

(d) 1 x 10

10

A. 1 billion B. 100 million

c.

1 million

D. 10 billion

E. 100 billion

SECTION 5.3

Scientific Notation

365

Concept Check Determine whether or not each number is written in scientific notation as defined in Objective 1. If it is not, write it as such. 3. 4.56 x 104

4. 7.34 x 106

5. 5,600,000

7. 0.8 x 102

8. 0.9 x 10

9. 0.004

3

11. Concept Check

6. 34,000 10. 0.0007

Write each number in scientific notation.

(a) 63,000

The first nonzero digit is - - · The decimal point should be moved _ _ places. 63,000

= --

x 10-

(b) 0.0571

The first nonzero digit is - - · The decimal point should be moved _ _ places. 0.0571

= --

x 10-

12. Concept Check Write each number in standard notation. (a) 4.2 x 103

Move the decimal point _ _ places to the _ _ . 4.2 x 103 = - (b) 6.42

x 10- 3

Move the decimal point _ _ places to the _ _ . 6.42 x 10- 3

= --

Write each number in scientific notation. See Example 1. 13. 5,876,000,000

14. 9,994,000,000

15. 82,350

16. 78,330

17. 0.000007

18. 0.0000004

19. 0.00203

20. 0.0000578

21. - 13,000,000

22. - 25,000,000,000 23. - 0.006

24. - 0.01234

Write each number in standard notation. See Example 2. 25. 7.5 x 105

26. 8.8 x 106

27. 5.677 x 10 12

29. 1 x 10 12

30. 1 x 10

31. 6.21

33. 7.8 x

34. 8.9

37. -4

10- 4

x 10- 3

7

X

10- 5

38. - 6 x 10- 4

x 100

28. 8.766 x 109 32. 8.56 x 10°

x 10- 10

35. 5.134X 10- 9

36. 7.123

39. -8.1 x 105

40. -9.6 x 106

Each statement contains a number in boldface italic type. If the number is in scientific notation, write it in standard notation. If the number is not in scientific notation, write it as such. See Examples 1 and 2. 41. A muon is an atomic particle closely related to an electron. The half-life of a muon is about 2 millionths (2 X 10- 6 ) of a second. (Data from www2.fisica.unlp.edu.ar) 42. There are 13 red balls and 39 black balls in a box. Mix them up and draw 13 out one at a time without returning any ball ... the probability that the 13 drawings each will produce a red ball is . .. 1.6 X 10- 12• (Data from Weaver, W., Lady Luck.) 43. An electron and a positron attract each other in two ways: the electromagnetic attraction of their opposite electric charges, and the gravitational attraction of their two masses. The electromagnetic attraction is 4,200, 000,000, OOO, OOO, OOO, 000,000, OOO, OOO, OOO, OOO, OOO, OOO

times as strong as the gravitational. (Data from Asimov, I., Isaac Asimov's Book of Facts.)

366

CHAPTER 5

Expone nts and Po lynomials 44. The name "googol" applies to the number ~~~~~~~~~~~~~~~~~~~

OOO, OOO, OOO, OOO, OOO, 000,000, OOO, OOO, OOO, 000,000, OOO, OOO, 000.

The Web search engine Google honors this number. Sergey Brin, president and cofounder of Google, Inc., was a mathematics major. He chose the name Google to describe the vast reach of this search engine. (Data from The Gazette.)

Perfo rm the indicated operations. Write each answer (a) in scientific notation and (b) in standard notation. See Example 3. 45. (2 x 10 8 )(3 x 10 3 )

46. ( 4 x 107 ) (3 x 103 )

x 104 )(3 x 102 )

48. (8 x 105 )( 2 x 103)

47. (5

49. (3 x 10- 4 )( - 2 x 108)

50. (4 x 10- 3 )( - 2 x 107 )

51. (6 x 103 ) (4 x 10- 2)

52. (7 x 105 )(3 x 10- 4 )

53. (9 x 104 )(7 x 10- 7 )

54. (6 x 104 )( 8 x 10-s )

55.

58.

9 X 10- 5

3 x 10-

l

15 x 104 3 - 3 x 10

12 x 10- 4 3 4 x 10-

57.

8 x 103 2 - 2 x 10

2.6 x 10- 3 · 2 x 102

60 ·

9.5 x 10- 1 5 x 10 3 -4.5 x 104 2 1.5 x 10-

56. 59

61. - -2

3 x 109 62. - -5

63.

- 7 2 x 103 64. - · -6.0 x 10- 1

- 8 x 10- 4 65. - - - - 4 x 103

- 5 x 10- 6 66. - 2 x 102

4 X 105

8

x 10

6

x 10

Calculators can express numbers in scientific notation. The displays often use notation such as 5 .4 E 3

to represent

5 .4 X 103 .

Similarly, 5.4E- 3 represents 5.4 X 10- 3. Predict the display the calculator would give fo r the expression shown in each screen. 67.

69.

rilo'i'i'i!!""i"i'"M"i't-!i"''i'i'i' 0 . 00000047

.,

ri'·!ii'i'""'i"i'l"""i"iH!!i'i'i'

·~

(8E5)/(4E-2 )

68

·

V'iWll""i''i"n.tOi"i'i''Oi'i'i'

l0 . 000021

70.

1 .1·······

·~

J

W''i'i'i"""i''i""'l1'i"iHOi'i'i' ·1 J

l(9E-4) / (3E3)

Extending Skills Use scientific notation to calculate the result in each expression. Write answers in scientific notation.

73 ·

75 · 77 ·

650,000,000( 0.0000032) 0.00002 0.00000072 ( 0.00023 ) 0.0000000 18 0.0000016( 240,000,000) 0.00002(0.0032)

74 · 76

78

3,400,000,000( 0.000075 ) 0.00025 0.00000008 1( 0.000036 )

•

0.00000048 0 .0000 15( 42,000,000)

• 0.000009(0.000005)

SECTION 5.3

Scientific Notation

367

Use scientific notation to calculate the answer to each problem. See Examples 3-5. 79. The Double Helix Nebula, a conglomeration of dust and gas stretching across the center of the Milky Way galaxy, is 25,000 light-years from Earth. If one light-year is about 6,000,000,000,000 mi, about how many miles is the Double Helix Nebula from Earth? (Data from www.spitzer.caltech.edu) 80. Pollux, one of the brightest stars in the night sky, is 33.7 light-years from Earth. If one Light-year is about 6,000,000,000,000 mi (that is, 6 trillion mi), about how many miles is Pollux from Earth? (Data from The World Almanac and Book of Facts.)

81. In 2017, the population of the United States was about 326.4 million. To the nearest dollar, calculate how much each person in the United States would have had to contribute in order to make one person a trillionaire (that is, to give that person $1,000,000,000,000). (Data from U.S. Census Bureau.) 82. In 2016, the U.S . government collected about $5712 per person in individual income taxes. If the population of the United States at that time was 323,000,000, how much did the government collect in taxes for 2016? (Data from www.usgovemmentrevenue.com) 83. Before Congress raised the debt limit in 2017, it was $1.98 X 10 13• To the nearest dollar, how much was this for every man, woman, and child in the country? Use 326 million as the population of the United States. (Data from www.washingtonpost.com; www.census.gov) 84. In 2016, the state of Minnesota had about 7.33 X 104 farms with an average of3.53 X 102 acres per farm . What was the total number of acres devoted to farmland in Minnesota that year? (Data from U.S. Department of Agriculture.) 85. Light travels at a speed of 1.86 X 105 mi per sec. When Venus is 6.68 X 107 mi from the sun, how Jong (in seconds) does it take light to travel from the sun to Venus? (Data from The World Almanac and Book of Facts.) 86. The distance to Earth from Pluto is 4.58 X 109 km. Pioneer JO transmitted radio signals from Pluto to Earth at the speed of light, 3.00 x 105 km per sec. About how long (in seconds) did it take for the signals to reach Earth? 87. During the 2016- 2017 season, Broadway shows grossed a total of $ 1.45 X 109 . Total attendance for the season was 1.33 X 107 . What was the average ticket price (to the nearest cent) for a Broadway show? (Data from The Broadway League.) 88. In 2016, $1.14 X 10 10 was spent to attend motion pictures in the United States and Canada. The total number of tickets sold was 1.32 billion. What was the average ticket price (to the nearest cent) for a movie? (Data from Motion Picture Association of America.) 89. In 2017, the world's fastest computer could perform 93,014,600,000,000,000 calculations per second. How many calculations could it perform per minute? Per hour? (Data from www.top500.org) 90. In 2017, the fastest computer in the United States could handle 17 .59 quadrillion calculations per second. (Hint: 1 quadrillion = 1 X 10 15.) How many calculations could it perform per minute? Per hour? (Data from www.top500.org)

368

CHAPTER 5

Exponents and Polynomials

RELATING CONCEPTS For Individual or Group Work (Exercises 91-95)

In 1935, Charles F. Richter devised a scale to compare the intensities of earthquakes. The intensity of an earthquake is measured relative to the intensity of a standard zero-level earthquake of intensity 10 . The relationship is equivalent to I

= /0 X 10'1,

where R is the Richter scale measure.

For example, if an earthquake has magnitude 5.0 on the Richter scale, then its intensity is calculated as I

= /0 X 105 0 = /0 X 100,000,

which is I 00,000 times as intense as a zero-level earthquake. /11te11sity 10

x

Riclzter Scale

10° 10 x 10 1 10 x 102 10 x 103 10 x 104 10 x 105 10 x 10 6 10 x 107 10 x 10 8 2

0

4

3

5

7

6

8

To compare two earthquakes, such as one that measures 8.0 to one that measures 5.0, calculate the ratio of their intensities. intensity 8.0

10 X 108.o

intensity 5.0

/ 0 X I 05·0

108 = -

105

=

108 - 5

=

103

=

1000

An earthquake that measures 8.0 is 1000 times as intense as one that measures 5.0. Use the information in the table to work Exercises 91-95 in order. Year

Earthquake Location

Richter Scale Measurement

1960

Valdivia, Chile

9.5

2010

Maule region, Chile

8.8

2007

Southern Sumatra, Indonesia

8.5

2015

Gorkha district, Nepal

7.8

2015

Farkhar, Afghanistan

7.5

2018

Mooreland, Oklahoma

3.8

Data from earthquake.usgs.gov

91. Compare the intensity of the 1960 Valdivia earthquake with that of the 2007 Southern Sumatra earthquake. 92. Compare the intensity of the 2010 Maule earthquake with that of 2015 Gorkha earthquake. 93. Compare the intensity of the 1960 Valdivia earthquake with that of the 2015 Farkhar earthquake. 94. Compare the intensity of the 2010 Maule earthquake with that of the 2018 Mooreland earthquake. 95. Compare the intensity of the 2015 Gorkha earthquake with that of the 2018 Mooreland earthquake.

Adding, Subtracting, and Graphing Polynomials

SECTION 5.4

369

Adding, Subtracting, and Graphing Polynomials OBJECTIVES

OBJECTIVE 1 Identify terms and coefficients.

1 Identify terms and coefficients.

Recall that in an expression such as 4x3 + 6x 2 + Sx + 8,

2 Combine like terms. 3 Describe polynomials using appropriate vocabulary. 4 Evaluate polynomials.

the quantities 4x 3 , 6x 2 , Sx, and 8 are terms. In the leading (or first) term 4x3 , the number 4 is the numerical coefficient, or simply the coefficient, of x 3 . In the same way, 6 is the coefficient of x 2 in the term 6x 2 , and 5 is the coefficient of x in the term Sx. The constant term 8 can be thought of as

5 Add and subtract polynomials. 6 Graph equations defined by polynomials of degree 2.

8 · 1 = 8x 0

x 0 = 1,

because

so 8 is the coefficient in the term 8. Other examples are given in the table. T Terms and Their Coefficients Term

Numerical Coefficient

6

D D D D D D D D D D D D D

term leading term numerical coefficient (coefficient) like terms unlike terms polynomial descending powers degree of a term degree of a polynomial monomial binomial trinomial parabola

D vertex

~EXERCISE 1 Identify the coefficient of each term in the expression. Give the number of terms. t - l0t2

34r 3

34 - 26

- k = - 1k

- 1

r = 1r 3

3x

3

3 = 3x 1x

x

8

1

1

3 = 3 = 3x

l:t:fAlf'i!Qlll

3

Identifying Coefficients

Identify the coefficient of each term in the expression. Give the number of terms. (a) x - 6x 4

+3

can be written as

x,

-6x

(b) 5 - v3

4

,

and

lx

+ (-

t

There are three terms:

D axis of symmetry (axis)

i:.NOWTRY

- 7

26x5yz4

-

VOCABULARY

6

- 7y

t

Sv0 + (- l v3 ) .

t

LJ

There are two terms.

t The coefficients are 1, -6, and 3.

3.

can be written as

+ 3x0

6x 4 )

t

The coefficients are 5 and -1.

NOW TRY

g

OBJECTIVE 2 Combine like terms. Recall that like terms have exactly the same variables, with the same exponents on the variables. Only the coefficients may differ. 19m5

and

14m5

6y9, - 37y 9, and y9 NOW TRY ANSWER 1. I ; - 10; two terms

3pq 2xy2

and and

-2pq - xy2

7x Examples of like terms

7y

and

z4 and z 2pq

and

-4xy2

Examples of unlike terms

2p

and

5x 2y

370

CHAPTER 5

Exponents and Polynomials

Using the distributive property, we combine like terms by adding or subtracting their coefficients.

li£MMQl#j Combining Like Terms

/:.NOW TRY

~EXERCISE2 Simplify by combining like terms. (a) -

7x6

+

Simplify by combining like terms.

2 (b) x - - x 5 (c) 3x 2

-

x2

+ 2x

=

+ 6x 3 (-4 + 6)x 3

=

2x 3

(a) - 4x 3

12x6

(c) Y +

(b) 9x 6

ac + be =(a + b )c

=

2 3

+ 9rs 13 + 9)rs

(d) 8rs - 13rs

-y (1+ l)y =-ys +

2 3

=

3

= (8 -

y = 1y

= 4rs Note how the distributive property justifies the procedure for combining like terms.

1 = ~ ;Distributive property

Add the fractions.

+ Sm + 4m2 = ( 12 + 4 )m2 + Sm

(e) 12m2

(f) Su + ll v

These are unlike terms. They cannot be combined.

,---------.

=

+ x6 ~ (9 - 14 + 1 )x 6 14x 6

= - 4x 6

-y

= ly

-

Stop here. These are unlike terms.

16m2 +Sm

NOW TRY~

0

CAUTION In Example 2(e), we cannot combine 16m 2 and 5m because the exponents on the variables are different. Unlike terms have different variables or different exponents

on the same variables.

OBJECTIVE 3 Describe polynomials using appropriate vocabulary. Polynomial in x

A polynomial in x is a term or the sum of a finite number of terms of the form

ax", for any real number a and any whole number n.

For example, the expression 16x8

7x 6

-

+ Sx 4

-

3x2

Polynomial in x (fhe 4 can be written as 4x 0 .)

+4

is a polynomial in x. This polynomial is written in descending powers of the variable because the exponents on x decrease from left to right. By contrast, 2x3 NOW TRY ANSWERS 2. (a) Sx 6

3

(b) 5x

(c) 2x2

+ 2x

-

x2

4 x'

+-

or

2x3

-

x2

+ 4x- 1,

Not a polynomial

is not a polynomial in x. A variable appears in a denominator or as a factor to a negative power in a numerator.

SECTION 5.4

Adding, Subtracting, and Graphing Polynomials

371

l1Iilll

We can define a polynomial using any variable, not just x , as in Example 2(e). A polynomial may have terms with more than one variable, as in Example 2(d).

The degree of a term is the sum of the exponents on the variables. The degree of a polynomial is the greatest degree of any nonzero term of the polynomial.

T Degrees of Terms and Polynomials Term

Degree

3x 4

4

5x, or 5x 1

1

- 7,or -

7x0

Polynomial 3x 4

Degree

5x2 + 6

4

5x + 7 x5

0

2+1=3

2x2y, or 2x2y 1

-

5y2

+

3x 6 -

1 7

6

+ xy - 2x2y

3

Some polynomials with a specific number of terms have special names. • A polynomial with exactly one term is a monomial. (Mono- means "one," as in monorail.)

9m,

- 6y 5 , x 2 , and

6

Monomials

• A polynomial with exactly two terms is a binomial. (Bi- means "two," as in bicycle.)

-9x4

+ 9x3 ,

8m2

+ 6m,

and

3t - 10

Binomials

• A polynomial with exactly three terms is a trinomial. (Tri- means "three," as in triangle.)

9m3 /:.NOW TRY

~EXERCISE3 Simplify, give the degree, and tell whether the simplified polynomial is a monomial, a binomial, a trinomial, or none of these.

+ 2x - 4 + 4x3 x 8 - x 7 + 2x8 -3x2 + 8x - 6x + 4

(a) 3x2

(b) (c)

(d)

x3

-

4m2

(b) 5x3 ; degree 3; monomial

(c) 3x8

-

x 1 ; degree 8; binomial

(d) -3x 2 + 2x + 4; degree 2; trinomial

'

and - 3z5

-

z2

+z

Trinomials

For each polynomial, first simplify, if possible. Then give the degree and tell whether the simplified polynomial is a monomial, a binomial, a trinomial, or none of these. (a) 2x3

+5

The polynomial cannot be simplified. It is a binomial of degree 3.

(b) 6x - 8x + l 3x

= llx

Combine like terms to simplify.

The degree is 1 (because x = x 1). The simplified polynomial is a monomial.

+ 2xy

=xy NOW TRY ANSWERS

8 3

l(f!JMQIH Classifying Polynomials

(c) 4.xy - 5.xy

3. (a) The polynomial cannot be simplified; degree 2; trinomial

19 3

+ 6, - y2 + - y + 5

Combine like terms to simplify.

The degree is 2 (because xy = x 1y 1, and 1 + 1 = 2). The simplified polynomial is a monomial. (d) 2x2

-

3x

=

2x2

+ 8x + Sx -

12 12

Combine like terms.

The degree is 2. The simplified polynomial is a trinomial.

NOW TRY~

372

Exponents and Polynomials

CHAPTER 5

OBJECTIVE 4 Evaluate polynomials. When we evaluate a polynomial, we find its value. A polynomial usually represents different numbers for different values of the variable.

l(MMQlll

/:.NOW TRY

~EXERCISE4 Find the value for t

4t 3 - t2

-

= -

t

3.

Evaluating a Polynomial

Find the value of 3x4

+ 5x3 - 4x - 4 for (a) x = -2 and (b) x = 3.

+ 5x 3 - 4x - 4 = 3( -2) 4 + 5( -2) 3 - 4( -2) - 4 = 3( 16) + 5 (-8) - 4(-2) - 4 = 48 - 40 + 8 - 4

3x4

(a) Use parentheses to avoid errors.

=

Apply the exponents. Multiply.

12

Add and subtract.

+ 5x 3 - 4x - 4 = 3(3) 4 + 5(3) 3 - 4 (3) - 4 = 3(81) + 5 (27) - 4(3) - 4 = 243 + 135 - 12 - 4 = 362

3x4

(b) [ Replacexwith3

Substitute - 2 for x.

7'

Let x = 3. Apply the exponents. Multiply. Add and subtract.

0

NOW TRY

g

CAUTION Use parentheses around the numbers that are substituted for the variable, as in Example 4. Be particularly careful when substituting a negative number for a variable that is raised to a power, or a sign error may result.

OBJECTIVE 5 Add and subtract polynomials. Adding Polynomials

To add two polynomials, combine (add) like terms.

l(MMQlltj

Adding Polynomials Vertically

Find each sum. (a) Add 6x3

-

+ 3 and -2x3 + 7x2 - 5. 6x3 - 4x 2 + 3 Write like terms in columns. + (-2x3 + 7x2 - 5)

4x2

Now add, column by column. Add the coefficients only. Do not add the exponents.

NOW TRY ANSWER 4. - 11 4

.__~~ 6x 3

-2x 3 4x 3

-4x2 7x2 3x 2

3

-5 -2

Add the three sums together to obtain the answer. 4x3 + 3x2 + (-2) = 4x 3 + 3x2 - 2 ~ Final sum

Adding, Subtracting, and Graphing Polynomials

SECTION 5.4 l:.NOWTRY

"7 EXERCISE 5 Find each sum. (a) Add 7y 3 - 4y 2 + 2 and -6y

3

+ 5y

(b) Add -5x4 x 3 - 5x.

2 -

-

(b) Add 2x2 - 4x + 3 and x 3 + 5x. Write like terms in columns and add column by column.

2x2

4x + 3

-

+ (x3

+ 5x 3 2 x + 2x + x + 3

3. 2x

373

+ 3 and

Leave spaces for missing terms.

NOW TRY~

The polynomials in Example 5 also can be added horizontally.

l:.NOWTRY

"7 EXERCISE 6 Add 10x4 - 3x2 - x and x4 - 3x2 + 5x horizontally.

i:+;f!!MQlllll Adding Polynomials Horizontally Find each sum. (a) Add 6x3

(6x3

-

4x2

t

+ 3 and -2x3 + 7x2 - 5.

4x2

-

+ 3) + ( -2x3 + 7x2

t

1

-

5) = 4x 3 + 3x2 1'

+ 5x . (2x - 4x + 3) + (x3 + 5x) = x 3 + 2x2 - 4x + 5x + 3

(b) Add 2x2

-

-

2

Same answer as found in Example 5(a)

4x + 3 and x 3 2

Commutative property

=x3 +2x2 + x +3

See Example 5(b).

The difference x - y is defined as x adding x and the opposite of y.)

Examples:

NOW TRY~

+ ( - y ). (We find the difference x -

y by

+ (-2), which equals 5. -8 - (-2) is equivalent to -8 + 2, which equals -6. 7 - 2 is equivalent to

7

A similar method is used to subtract polynomials.

Subtracting Polynomials

To subtract two polynomials, change the sign of each term in the subtrahend (second polynomial) and add the result to the minuend (first polynomial)-that is, add the opposite of each term of the second polynomial to the first polynomial.

lif:f!1MQlll Subtracting Polynomials Horizontally Perform each subtraction. (a) (5x - 2) - (3x - 8) = ( 5x

- 2)

+ [ - (3x - 8) ]

Definition of subtraction

~

NOW TRY ANSWERS 5. (a) y 3 + y 2 - I (b) - 5x4

6. I Ix4 -

+ x3 - 7x + 3

6x2

+ 4x

= (5x - 2) + [ - 1( 3x - 8)] = (5x - 2) + ( - 3x + 8)

Distributive property

= 2x + 6

Combine like terms.

- a = - 1a

37 4

Exponents and Polynomials

CHAPTER 5

/:.NOW TRY

~EXERCISE 7 Perform each subtraction. (a) (3x - 8) - (5x - 9) (b) ( 4t4

t2

-

- ( 5t4

CHECK

To check a subtraction problem, use the following fact.

If a - b = c,

3t2

a

= b

+ c.

+ 6. Check as follows. 8) + (2x + 6)

We found that (5x - 2) - (3x - 8) = 2x

+ 7) -

then

( 3x -

+ 1) (b)

= 5x - 2 .I Be careful to write Subtract 6x3 - 4x2 + 2 from l lx3 + 2x2 - 8. the problem in the correct order. (llx3 + 2x2 - 8) - (6x3 - 4x2 + 2) = ( l lx 3 + 2x2 - 8) + ( - 6x3 + 4x2 - 2) = 5x3 + 6x2 - 10 Combine like terms.

NOW TRY

g

Subtraction can also be done in columns. /:.NOW TRY

~EXERCISES Subtract by columns.

1£MMQll:I Subtracting Polynomials Vertically (14y 3

Subtract by columns: 14y 3 -

( 12.x2 - 9x + 4) - ( - 10x2 - 3x + 7)

-

(2y3 -

6y

2

7y

2

6y 2

-

+ 2y -

+ 2y - 5 - 4y + 6)

5) - (2y 3

-

7y 2

-

4y

+ 6) .

Arrange like terms in columns.

Change all signs in the second row (the subtrahend), and then add.

+ 2y - 5 + ( - 2y 3 + 7y + 4y - 6) l 2y 3 + y 2 + 6y - 11 14y 3

-

6y 2

2

/:.NOW TRY

~EXERCISE9 Perform the indicated operations.

( 6p4

-

8p 3

- (-7p 4

+ (p4

-

+ 2p - 1) + 6p2 - 12) 3p + 8)

NOW TRY

g

NOW TRY

g

NOW TRY

g

Perform the indicated operations.

(4 - x + 3x2 )

-

(2 - 3x + 5x2 ) + (8 + 2x - 4x2 )

Rewrite, using the definition of subtraction.

( 4 - x + 3x2 )

-

(2 - 3x + 5x2 ) + (8 + 2x - 4x2 )

= (4

- x + 3x2 ) + ( -2 + 3x - 5x2 ) + (8 + 2x - 4x2 )

= (2

+ 2x - 2x2 ) + (8 + 2x -

10

+ 4x -

6x 2

4x2 )

Combine like terms. Combine like terms.

lif!t'MHIDI Adding and Subtracting Multivariable Polynomials

/:.NOW TRY

~EXERCISE 10 Subtract.

Add or subtract as indicated.

+ 2ab - b) + (3a - ab + b) = 4a + 2ab - b + 3a - ab + b

(a) ( 4a

+ y2)

- (6x 2 - 7xy

Add .

lif!f4MQl#i Adding and Subtracting More Than Two Polynomials

=

(4x2 - 2xy

Change all signs.

+ 2y2 )

=

NOW TRY ANSWERS 7. (a) - 2x + I (b) -t4 + 212 + 6 8. 22x2 - 6x - 3 9. 14p4 - 8p3 - 6p2 - p + 19 10. -2x2 + 5.xy - y2

(b) (2x2y

7a + ab

Combine like terms.

+ 3.xy + y 2 )

-

(3x 2y - xy - 2y 2 )

= 2x2y + 3.xy + y 2 = -x2y + 4xy + 3y2

3x2y

+ xy + 2y 2

Be careful with signs. The coefficient of zyis 1.

SECTION 5.4

375

Adding, Subtracting, and Graphing Polynomials

OBJECTIVE 6 Graph equations defined by polynomials of degree 2. Earlier we graphed linear equations (which are actually polynomial equations of degree 1). By plotting points, we can graph polynomial equations of degree 2. /:.NOW TRY

~EXERCISE 11 Graph y = - x 2 - I.

l(MMQllll

Graphing Equations Defined by Polynomials of Degree 2

Graph each equation.

(a) y = x 2 It is easier to select values for x and find corresponding y-values. Selecting x = 2 and substituting in y = x 2 gives y = 22 = 4.

The point (2, 4 ) is on the graph of y = x 2. (Recall that in an ordered pair such as (2, 4), the x-value comes first and the y-value second.) We show some ordered pairs that satisfy y = x 2 in the table with FIGURE 3. If we plot the ordered pairs from the table on a coordinate system and draw a smooth curve through them, we obtain the graph shown in FIGURE3. The graph of y = x 2 is the graph of a function, because each input x is related to just one output y. The curve in FIGURE 3 is a parabola.

x y

Y Axis ofsymmetry

3 9 2 4 1 1 0 0 -1 1 -2 4 -3 9 T

+

T

t

t

t

l-

+

FIGURE3

• The point (0, 0), the lowest point on this graph, is the vertex of the parabola. • The vertical line through the vertex (the y-axis here) is the axis of symmetry, or simply the axis, of the parabola. This axis is a line of symmetry for the graph. If the graph is folded on this line, the two halves will coincide. (b) y = - x 2

+3

Plot points to obtain the graph. For example, let x y

= - x2 + 3

y

(-2) 2

=

-

= -4 + 3 y = -1 y

= - x2 + 3 y = - 02 + 3 y

+3

y =0+3 y= 3

The points ( -2, -1 ), (0, 3), and several others are shown in the table that accompanies the graph in FIGURE 4. The vertex (0, 3) is the highest point on this graph. The graph opens downward because x 2 has a negative coefficient.

=

-2 and x

x y

=

0. y

-2 -1 -1 2

0 1

3 2

2 -1

FIGURE4

NOW

TRY~

NOW TRY ANSWER x 11 . •- . y-2 .

5 2 y = -x -

m!III

A// equations defined by polynomials of degree 2 have parabolas as their graphs. When graphing, find points until the vertex and points on either side of it are located. (In this section, all parabolas have their vertices on the x -axis or the y-axis.)

376

CHAPTER 5

Exponents and Polynomials

5.4 Exercises O Video solutions for select problems available in MyLab Math

0

FOR

EXTRA

HELP

Mylab Math

Concept Check

Complete each statement.

1. In the term 4x 6 , the coefficient of x 6 is _ _ and the exponent is _ _ . 2. The expression 4x3

-

5x2 has exactly (one I two I three) term(s).

3. The degree of the term - 3x9 is _ _ . 4. The polynomial 4x2 + y 2 (is I is not) an example of a trinomial.

5. When x 2 + 10 is evaluated for x is _ _ .

+ 3x3 -

6. 5x-

3, the result is _ _ . For x

=

=

-3, the result

7x is a trinomial of degree 6.

7. Combining like terms in -3xy - 2xy

+ 5xy gives--·

8. _ _ is a monomial with coefficient 8, in the variable x, having degree 5. Identify the coefficient of each term in the expression, and give the number of terms. See Example 1.

9. 6x4

11.

10. -9y 5

13. -19r2

14. 2y 3

r

-

-

y

t4

15. x

12. s7

+ 8x2 + 5x 3

16. v - 2v 3

-

v1

In each polynomial, simplify by combining like terms whenever possible. Write results that have more than one term in descending powers of the variable. See Example 2 and Objective 3.

18. -4y 3 + 3y 3

17. -3m5 + 5m5 20. 9y 2

+ (-19y2 )

0.5m2

21. 0.2m 5

-

9x3 + 10x3

5x5

24. 6x3

+ 4a8 -

3a2

27. -1.5x2 + 5.3x2 - 3.8x2

-

29. - 4y 2 + 3y 2 - 2y 2 + y 2 1 31. - - tu1

3

2 1 tu 5

+-

1 1 tu 15

+-

8 - tu1

+ 0.9y 2

25. - 4p1 + 8p7 + 5p9

30. 3r5 -

+ (-3r 5 )

22. -0.9y

-

23. - 3x5 + 3x5 26. - 3a8

19. 2r5

8r5

-

28. 8.6y4

-

10.3y4 + 1.7y4

+ r 5 + 2r5

3 1 7 1 32. - - p2q - - p2q + - p2q - - p2q 4 3 12 6

5

For each polynomial, first simplify, if possible, and write the result in descending powers of the variable. Then give the degree and tell whether the simplified polynomial is a monomial, a binomial, a trinomial, or none of these. See Example 3. 33. 6x4 4

35. 5m

5

37. - x 4 3

-

3m2 + 6m 4

-

2 - x4

4

39. 0.8x

34. 7t 3

9x

-

-

7m 3

36. 6p

4 38. -r6 5

3

-

41. -1 1ab

0.3x 4

-

+ 2ab -

43. Concept Check

3t

-

5

+ 4p3 - 8p5 +

l0p2

1 5

+ - r6

0.5x 4 + 7

40. l .2t3

4ab

42. 5xy + 13xy - 12xy

-

0.9t3 - 0.3t3

+9

A student incorrectly evaluated the polynomial below for x = -2.

WHAT WENT WRONG? Give the correct answer. - x 2 + 3x + 4 =

[ -(-2 )]2 + 3(-2)

=4-6+4 =2

+4

SECTION 5.4

44. Concept Check

A student incorrectly evaluated the polynomial below for x

WHAT WENT WRONG? Give the correct answer. 2x3

+ 8x - 5

=

[2( - 1)]3 + 8( -1 ) - 5

=

(-2) 3

=

-8- 8- 5

=

- 21

Find the value of each polynomial fo r (a) x

+ 5x -

48. - 2x2

-

8- 5

2 and (b) x

=

=

46. x 2 + 5x - 10

45. 2x2 + 5x + 1

49. 2x5 - 4x4

1

+ 5x3 -

x2

- 1. See Example 4. 47. -3x2 + 14x - 2 50. x4 - 6x3 + x 2 - x

Add. See Ex amples 5 and 6.

51.

54.

2x2 - 4x + (3x 2 + 2x)

52.

4a 3 - 4a2 - 4 + (6a3 + 5a2 - 8)

55.

57. 9m3

- 5y 3 + 3y

+

(8y 3 -

3m2 + Sm

53.

+

4y)

4 2

56.

+ (-2Ix 2 - 3-Ix + -32)

(2m2 -

7Y

+6

2m - 4 ) I

-

7

sY + 9

+ (~y2 - ~y + ~)

5m2 + 4m - 8 and - 3m3 + 6m 2 - 6

-

58. 12r5 + l l r4

7r3 - 2r2 and - 8r5 + 3r 3 + 2r2

-

Subtract. See Example 8.

59.

5y 3 - 3y2 - (2y3 + 8y2)

60.

- 6t3 + 4t2 - (8t3 - 6t2 )

61.

l 2x4 - x 2 + x - (8x4 + 3x2 - 3x)

62.

13y5 - y 3 - 8y2 - ( 7y5 + 5y3 + y2 )

63.

I 2m3 - 8m2 + 6m + 7 - ( - 3m3 + 5m 2 - 2m - 4)

64.

5a4 - (-6a4

-

+

3a3 + 2a2 - a + 6 a 3 - a 2 +a - I )

Perform each indicated operation. S ee Examples 6, 7, and 9.

65. (8m2 - 7m) - (3m2 + 7m - 6)

66. (x2 + x) - (3x2 + 2x - 1)

67. ( l 6x 3 - x 2 + 3x) + ( - l 2x3 + 3x2 + 2x) 68. ( - 2b6 + 3b4

-

b2) + (b6 + 2b4 + 2b2)

69. Subtract 9x2 - 3x + 7 from - 2x2 - 6x + 4. 70. Subtract - 5w3 + 5w2 - 7 from 6w3 + 8w + 5. 71. (9a4

-

3a2 + 2) + (4a4

-

377

Adding, Subtracting, and Graphing Polynomials

4a2 + 2) + ( - 12a4 + 6a 2 - 3)

72. (4m2 - 3m + 2) + (5m2 + 13m - 4) + ( - 16m2 - 4m + 3) 73. [ (8m2 + 4m - 7) - (2m2 - 5m + 2) ] - (m2 +m + 1) 74. [ (9b3 - 4b2 + 3b + 2) - (-2b 3 - 3b2 + b) ] - (8b3 + 6b + 4) 75. [ (3x2 - 2x + 7) - (4x 2 + 2x - 3)] - [ (9x2 + 4x - 6) + (-4x 2 + 4x + 4)] 7 6. [ ( 6t 2 - 3t + 1) - ( l 2t2 + 2t - 6) J - [ ( 4t2 - 3t - 8) + ( - 6t2 + lOt - 12) J

=

- 1.

378

CHAPTER 5

Exponents and Polynomials

Concept Check Answer each of the following. 77. Without actua lly performing the operations, mentally determine the coefficient of the x 2 -term in the simplified form of

( - 4x2

+ 2x - 3) - ( - 2x2 + x - 1) + ( -

Sx2

+ 3x - 4 ).

78. Without actually performing the operations, mentally determine the coefficient of the x-term in the simplified form of ( - Sx2

3x

-

+ 2)

- (4x2

-

3x + S) - (- 2x2

-

x

+ 7) .

Add or subtract as indicated. See Example 10.

79. (6b

+ 3c) + (-2b - Sc)

81. (4x

+ 2xy - 3) - ( - 2x + 3xy + 4)

83.

(2c4 d

+

84. (3k2h3

3c 2d 2 -

4d2 ) -

+ Skh + 6k3h2 )

80. (-St+ 13s) +(St - 3s)

(c4 d -

+

(2k 2h 3

82. (Sab

-

9kh

+ 2a - 3b) - (6ab - 2a + 3b)

5d2 )

Sc2d 2 -

+ k3h2)

Find a polynomial that represents the perimeter of each rectangle, square, or triangle.

89.

.4~312+ 21 + 7

61+0

90. ~-

L _ _ _ _ _ J2p +

5

9p 3 + 2p 2 + I

512 + 2

Find (a) a polynomial that represents the perimeter of each triangle and (b) the degree measures of the angles of the triangle. (Hint: The sum of the measures of the angles of any triangle is 1S0°.)

92.

91.

- 81 2 s + 6ts 2 + ts

Extending Skills

Perform each indicated operation.

93. Find the difference of the sum of 5x2 and - x 2 + 4x - 6. 94. Subtract the sum of9t3

-

+ 2x -

3 and x 2

-

Sx

+ 2 and the sum of 7x 2 - 3x + 6

3t + Sand t 2 - St+ 4 from the sum ofl2t + Sand t2

-

Graph each equation by completing the table of values. See Example II .

95. y = x 2 x -2

-

y

4

96. y = x 2 x

-

y

6

97. y = 2x2 - 1 x

-2

-2

-1

- 1

-1

0

0

0

2

2

2

y

lOt

+ 3.

SECTION 5.4

98. y

101. y

=

2.x2 + 2

x -2

y

Adding, Subtracting, and Graphing Polynomials

99. y

=

-x2

x -2

+4

100. y

y

=

-x2

x

y

-1

- 1

0

0

0

2

2

2

(x + 3) 2

; I

-5

I

+2

-2

-1

=

379

102. y = (x - 4) 2

-4

I

-3

RELATING CONCEPTS

I

-2

I -1

For Individual or Group Work (Exercises 103- 106)

The following binomial models the distance in feet that a car going approximately 68 mph will skid in t seconds.

lOOt - 13t2 When we evaluate this binomial for a value oft, we obtain a value for distance. This illustrates the concept of a function-for each input of a time, we obtain one and only one output for distance. Exercises 103-106 further illustrate the function concept with polynomials. Work them in order. 103. Evaluate the given binomial lOOt - 13t2 fort = 5. Use the result to fill in the blanks: In _ _ seconds, the car will skid _ _ feet.

104. If one "dog" year is estimated to be about seven "human" years, the monomial 7x gives the dog's age in human years for x dog years. Evaluate this monomial for x = 9. Use the result to fill in the blanks: If a dog is _ _ in dog years,

then it is _ _ in human years. 105. If it costs $15 plus $2 per day to rent a chain saw, the binomial

2x

+ 15

gives the cost in dollars to rent the chain saw for x days. Evaluate this binomial for x = 6. Use the result to fill in the blanks: If the saw is rented for _ _ days,

then the cost is _ _ dollars. 106. If an object is projected upward under certain conditions, its height in feet is given by the trinomial - 16t2

+ 60t + 80,

where t is in seconds. Evaluate this trinomial fort = 2.5. Use the result to fill in the blanks: If _ _ seconds have elapsed, then the height of the object is _ _ feet.

380

CHAPTER 5

Exponents and Polynomials

Multiplying Polynomials OBJECTIVES

OBJECTIVE 1 Multiply monomials.

1 Multiply monomials.

Recall that we multiply monomials using the product rule for exponents.

2 Multiply a monomial and a polynomial.

li:fdMQlll Multiplying Monomials

3 Multiply two polynomials. 4 Multiply binomials using the FOIL method.

VOCABULARY

Find each product. (a) 8m2 ( -9m )

0

D FOIL method D outer product D inner product

(b) 4x 3y 2 (2x 2 y)

= 8(- 9) · m2m 1

= 4(2)

= - 72m2 + 1

=

= -72m3

= 8x5y3

CAUTION

· x 3x 2 • y2y1

8x3+2y2 + 1

Commutative and associative properties Multiply; product rule Add.

NOW TRY

g

Do not confuse addition of terms with multiplication of terms.

7q5 + 2q5

l!:m. NOW TRY

= (7 + 2)q5

Distributive property

= 7 •

= 9q5

Add.

= 14q 1o

2q5 +5

Commutative property; product rule Multiply. Add.

~EXERCISE 1

Find each product. (a) - 6x( 5x3 )

OBJECTIVE 2 Multiply a monomial and a polynomial.

(b) 6mn3 (12mn)

We use the distributive property and multiplication of monomials.

l!:m. NOW TRY

lif!MMQl#j Multiplying Monomials and Polynomials

Find the product.

Find each product.

~EXERCISE 2

-3x5 (2x3

-

5x2

+ 10)

~

(a) 4x2 (3x

+ 5)

a(b + c ) = ab + ac

= 4x (3x) + 4x = l2x 3 + 20x 2 2

2

(5)

Distributive property Multiply monomials.

+ 3m 2 + 2m - 1) = -8m3 (4m3 ) + ( -8m 3 )(3m2 ) + ( -8m3 )(2m) + ( -8m3 )( -1 ) = -32m6 - 24m5 - 16m4 + 8m3

(b) - 8m3 ( 4m3

Distributive property Multiply monomials.

NOW TRY

g

OBJECTIVE 3 Multiply two polynomials.

To find the product of the polynomials x 2 + 3x + 5 and x - 4, we can think of x - 4 as a single quantity and use the distributive property as follows.

+ 3x + 5) (x - 4 ) = x 2 (x - 4) + 3x(x - 4) + 5 (x - 4) = x 2 (x) + x 2 (- 4) + 3x(x) + 3x (-4) +

(x 2

NOW TRY ANSWERS 1. (a) - 30x4 (b) 72m 2n4 2. - 6x8 + 15x7 - 30x5

Distributive property

5(x) + 5(-4) Distributive property again

= x3 = x3

-

4x2

-

x2 -

+ 3x2 - 12x + 5x - 20 7x - 20

Multiply monomials. Combine like terms.

SECTION 5.5

Multiplying Polynomials

381

Multiplying Polynomials

To multiply two polynomials, multiply each term of the first polynomial by each term of the second polynomial. Then combine like terms.

li!f!1MQIU Multiplying Two Polynomials

/:m.NOW TRY

~EXERCISE3 Multiply.

(x2 - 4)(2.x2

Multiply (m 2 + 5)(4m3 -

2m2 + 4m).

-

( m2 + 5 ) ( 4m3 _ 2m2

Sx + 3)

+ 4m)

= m2 (4m3 ) + m 2( -

-

Distributive properly Multiply monomials.

2m4

-

Combi ne like terms. ~ NOW TRY~

li!f!1MQlll Multiplying Polynomials Vertically

i:m.NOW TRY ~EXERCISE4 Multiply.

5t2

4m5

+ m2 (4m) + 5(4m 3 ) + 5(- 2m2 ) + 5(4m)

+ 4m3 + 20m3 - 10m2 + 20m 2m4 + 24m 3 - 10m2 + 20m

= 4m5 =

2m2)

Multiply each term of the first polynomial by each t erm of the second.

Multiply (x3

+ 2x2 + 4x +

7t + 4 2t - 6

x3

1)(3x + 5 ) vertically.

+ 2x 2 + 4x + 1 3x + 5

Write the polynomials vertically.

Begin by multiplying each of the terms in the top row by 5.

+1 3x + 5 3 2 5x + 10x + 20x + 5 x3

+

+

2x 2

4x

5(x 3 + 2x2 + 4x + 1)

Now multiply each term in the top row by 3x. Then add like terms.

+

x3 Place like terms in columns so they can be combined.

9x

-

2

12.x

3x

4

+

+ + + + llx3 + 22x 2 + 23x + 5 5x 3 6x3

Th is process is similar to multiplication of whole numbers.

3x(x 3 + 2x 2 + 4x + 1) Add in columns.

NOW TRY~

li!f!1MQl1j Multiplying Polynomials Vertically (Fractional Coefficients)

/:m.NOW TRY

~EXERCISES Find the product of 3

3x

4

4x + 1 3x + 5 10x 2 + 20x + 5 12x 2 + 3x 2x2

Find the product of 4m 3 I 2

+ 3 and 3x

2

-

3·

2m2 + 4m and ~ m2

-

4m3

-

10m3

-

2m2 + 4m ~mz + ~ 2

NOW TRY ANSWERS 3. 2x4 - 5x 3 - 5x 2 + 20x - 12 4. IOt3 - 44t2 + 50t - 24 5. 3x 5 - 4x4 - 6x3 + 9x2 - 2

4

2m5 -

m

2m5

m4

-

+ ~.

5m2

2

+

+ 2m + 12m3 - 5m2 +

lOm

Terms of top row are multiplied by~. Terms of to p row are multiplied by ~m 2 •

3

lOm

Add in columns.

NOW TRY

~

We can use a rectangle to model polynomial multiplication. For example, to find ( 2x + 1) ( 3x + 2 ), we label a rectangle, as shown below on the left. Then we put the product of each pair of monomials in the appropriate box, as shown on the right. 3x

~

2

1 1 --_ , _ _ _ _ _ _ _ ,

3x

2

382

CHAPTER 5

Exponents and Polynomials

The product of the binomials is the sum of the four monomial products. 3x

+ 1)(3x + 2) = 6x2 + 4x + 3x + 2 = 6x2 + 7x + 2

2

(2x

~ 1-1-::-2_,___:x----< This approach can be extended to polynom ials with any number of terms.

Combine like terms.

OBJECTIVE 4 Multiply binomials using the FOIL method. When multiplying binomials, the FOIL method reduces the rectangle method to a systematic approach without the rectangle. Consider this example.

(x +3)(x +5)

= x(x + 5) + 3(x + 5) = x(x) + x(5) + 3(x ) + 3(5)

Distribut ive property Distribut ive property again

= x + 5x + 3x + 15 = x2 + 8x + 15 2

Multiply. Combine like terms.

The letters of the word FOIL refer to the positions of the terms.

~

(x+3)(x +5 )

Multiply the First terms: x (x) .

(x +3)(x+ 5)

Multiply the Outer terms: x(5).

t

t

t

~

t

F 0

This is the outer product.

~

(x+3)(x +5 )

Multiply the Inner terms: 3{x).

Li

I

This is the inner product.

~

(x+3)(x +5 )

Multiply the Last terms: 3(5).

t

t

L

We add the outer product, 5x, and the inner product, 3x, to obtain 8x so that the three terms of the answer can be written without extra steps.

(x + 3)(x + 5 )

= x 2 + 8x +

15

FOIL Method for Multiplying Binomials

Step 1

Multiply the two First terms of the binomials to obtain the first term of the product.

Step 2

Find the Outer product and the Inner product and combine them (when possible) to obtain the middle term of the product.

Step 3

Multiply the two Last terms of the binomials to obtain the last term of the product.

Step 4

Add the terms found in Steps 1-3.

F

Example:

=

x2

L

~

=

(x +3 )(x +5)

1\ ;!/

0 =

15 (x + 3)(x + 5) = x 2 + 8x + 15

5x

8x

Combine like terms.

SECTION 5.5 l:.NOWTRY

lf)ij&!Qllil

Use the FOIL method to find the product.

Use the FOIL method to find the product (x

"7 EXERCISE 6 (t - 6)(t + 5)

Multip lying Polynomials

Using the FOIL Method

I

+ 8)(x - 6) . M ultiply the First terms: x(x) = x 2 . Find the Outer product: x( - 6) = - 6x. } Find the Inner product: 8(x) = 8x.

Step 3

L

M ultiply the Last terms:

Step 4

The product (x

Step 1

F

Step 2

0

x2

Last

-48

(x + 8)(x - 6)

(x+8)(x - 6)

\1~/

~Yx/

(x + 8)(x - 6) = x 2 + 2x - 48

- 6x

Outer

Combine like terms. NOW TRY

2x

I:. NOW TRY "7 EXERCISE 7

itf!JMQlff

Multiply.

M ultiply (9x - 2)(3y

+ 1).

First

(9x - 2)(3y + 1)

27xy

Outer

(9x - 2)(3y+ l )

9x

Inner

(9x - 2 )(3y + 1)

- 6y

Last

(9x - 2 )(3y + 1)

-2

F

The product (9x - 2)(3y

I:. NOW TRY "7 EXERCISE 8

if$JMQll:i

Find each product.

Find each product. (a) (2k

~

Using the FOIL Method

(7y - 3)(2x + 5)

(3p - 5q)(4p - q) 5x2 (3x + l)(x - 5)

- 6x + Bx = 2x Add the terms found in Steps 1-3.

~

~

Shortcut:

Combine:

8( -6) = - 48.

+ 8)(x - 6) is x 2 + 2x - 48.

First

(a) (b)

383

These unlike terms cannot be combined.

L

0

NOW TRY~

+ 1) is 27xy + 9x - 6y - 2.

Using the FOIL Method

+ Sy)(k + 3y) F

0

I

L

= 2k(k) + 2k(3y) + Sy(k) + 5y(3y)

= 2k 2 + 6ky +Sky + = 2k 2 + llky + 15y2 (b) (7p

15y2

Combine li ke terms.

+ 2q)(3p - q)

=

21p 2

-

Multiply.

(c) 2x2 (x - 3)(3x FOIL method

pq - 2q2

+ 4)

= 2x2 (3x 2 - Sx - 12) = 6x4 - 10x 3 - 24x 2

FOIL method Distributive property NOW TRY~

NOW TRY ANSWERS 6. 12 - t - 30 7. 14yx + 35y - 6x - 15 8. (a) 12p2 - 23pq + 5q2 (b) 15x4 - 70x3 - 25x2

mJD

Alternatively, the factors in Example 8(c) can be multiplied as follows.

2x2 (x - 3) (3x + 4) = (2x3 - 6x 2 ) (3x =

6x4 -

1Ox3 -

Multiply 2x2 and x - 3 first.

+ 4)

24x 2

Multiply that product and 3x + 4. FOIL method; The same answer resu lts.

384

CHAPTER 5

Exponents a nd Polynomials

5.5 Exercises 0 Video solutions for select problems available in MyLab Math

0

FOR

EXTRA

HELP

Mylab Math

Concept Check

Match each product in Column I with the correct polynomial in Column II.

I

II

3

7

1. (a) 5x ( 6x (b) -

(c)

)

5x7 (6x3 )

2 16x9

(c) (x -S )(x -4 )

II 2 A. x + 9x + 20 B. x2 - 9x + 20 C. x2 - x - 20

D. -30x 10

(d) (x + S)(x- 4 )

D. x2 + x - 20

B.

c.

(5x7 ) 3

(d) (- 6x 3 ) 3

Concept Check

I

A. 125x 21

2. (a) (x - 5)(x +4 )

30x 10 -

(b) (x + 5 )(x + 4 )

Fill in each blank with the correct response.

3. In multiplying a monomial by a polynomial, such as in

4x(3x2

+ 7x3 )

=

4x(3x2 )

+ 4x(7x 3 ),

the first property that is used is the _ _ _ _ property. 4. The FOIL method can only be used to multiply two polynomials when both polynomials are _ __

5. The product 2x2 (- 3x5 ) has exactly 6. The product (a + b )(c performed.

+ d)

term(s) after the multiplication is performed.

has exactly

term(s) after the multiplication is

Find each product. See Example 1. 8. 10p 2( 5p 3 )

7. 5y4 (3y 7 )

9. - 15a4 ( - 2as)

10. -3m6 ( -5m4 )

11. 5p( 3q2 )

12. 4a 3 (3b2)

13. -6m3 (3n2)

14. 9r3 ( - 2s2)

15. y 5 • 9y . y 4

16. x 2 · 3x3

17. (4x 3 )(2x2) ( - x5)

18. (7t5)(3t4) ( - t8)

•

2x

Find each product. See Example 2. 19. 2m (3m 22. 4x(3

25. 2y 3 (3

+ 2)

20. 4x(5x

+ 2x + 5x3 ) + 8r - 9)

29. 3a2(2a2 - 4ab + 5b 2)

Concept Check

24. -7y(3

+ 4p2 )

+ 5y 2 - 2y 3 )

26. 2m4 (6 + Sm + 3m2) 28. - 9a5 (- 3a6

-

2a 4 + 8a 2 )

30. 4z3 (8z2 + Szy - 3y2)

Multiply.

31. 7m3n2(3m2 + 2mn - n3 ) =

21. 3p(-2p3

23. -8z(2z + 3z2 + 3z 3 )

+ 2y + 5y 4 )

27. - 4r 3( - 7r2

+ 3)

7m3n 2( - - ) + 7m 3n2(- -)

+ 7m3n2 ( - - )

32. 2p 2q(3p2q 2 - 5p + 2q2) = _

_

(3p2q2)

+ _ _ (- 5p )

+ _ _ (2q2)

Find each product. See Examples 3-5. 33. (6x + 1)(2x2 + 4x + 1)

34. (9a + 2 ) (9a2 + a + 1 )

35. (9y - 2 )(8y2 - 6y + 1)

36. (2r - 1)(3r 2 + 4r - 4 )

SECTION 5.5

37. (4m + 3)(Sm3 39. (2x - 1)(3x5

-

4m2 + m - S)

-

2x3 + x 2

41. (Sx 2 + 2x + l )(x2 43. (6x4

-

4x 2 +

-

38. (2y + 8)(3y4 40. ( 2a + 3) (a4

2x + 3)

-

2y 2 + 1) a3 + a 2 - a + 1)

-

42. (2m 2 + m - 3)(m2

3x + S)

-

8x)(~x + 3)

385

Multiplying Polynomials

44. (8y6 + 4y4

-

4m + S)

12y2 )(~y 2 + 2)

-

Find each product using the rectangle method shown in the text. Determine the individual terms that should appear on the b lanks or in the rectangles, and then give the final product.

45. (x+3)(x+4) x 4

46. (x

+ S)(x + 2)

~ EE Product: _ _ _ _ __

Product: _ _ _ __

47. (2x+ l )(x2 +3x+2) 3x

x2

48. (x + 4)(3x2 + 2x + 1) 2x

2

~I I I I

===I I I I Product: _ _ _ __

Product: _ _ _ _ __

Concept Check

For each product, find and simplify the following.

(a) Product offirst terms

_

(_

(b) Outer product

)

_

(d) Product of last terms

_

(_

(_

(c) Inner product

)

_

(_

)

(e) Complete product in simplified form

)

49. (2p - S) (3p

so.

+ 7)

(2p - S)(2p

+ S)

Find each product. See Examples 6-8.

51. (m + 7)(m + S)

52. (n + 9)(n + 3)

53. ( n - 1) ( n + 4)

54. (t - 3)(t + 8)

55. (2x + 3)(6x - 4)

56. (3y + S)(8y - 6)

57. (9 + t) (9 - t)

58. ( 10 + r) ( lO - r)

59. (3x - 2)(3x - 2)

60. (4m + 3)(4m + 3)

61. (Sa + 1)(2a + 7)

62. (b + 8)(6b - 2)

63. (6 - Sm )(2 + 3m)

64. (8 - 3a)(2 +a)

65. (S - 3x)(4 + x )

66. (6 - Sx )( 2 + x )

67. (3t - 4s) (t + 3s)

68. (2m - 3n)(m + Sn )

69. (4x + 3)(2y - 1)

70. (Sx + 7)(3y - 8)

71. (3x + 2y)(Sx - 3y)

72. (Sa+ 3b)(Sa - 4b)

73. 3y3(2y + 3)(y - S)

74. 2x2 (2x - S )(x + 3)

75. -8r3 (Sr 2 + 2)(Sr2

-

2)

76. -St4 (2t4 + 1)(2t4

-

1)

77. - 3r(r - l )( r 2 + r + 1)

386

CHAPTER 5

Exponents and Polynomials

78. Concept Check

Multiply each of the following.

(a) (x -1 )(x2 + x+ 1)

(b) (y - 2)(y 2 + 2y +4 )

Describe the final product as "the _ _ _ _ of two _ _ __

D

Find polynomials that represent, in appropriate units, (a) the area and (b) the perimeter of each square or rectangle. (If necessary, refer to the formulas at the back of this text.)

80. 6x+ 2

Find each product. Recall that a 2 = a · a and a 3 = a2 · a.

Extending Skills 81. (x

+ 7) 2

82. (m

84. (b - IO)(b

87. (5k

+ 10)

+ 3q) 2

90. (p - 3) 3 93. -3a(3a

85. (2p - 5) 2

86. (3m - 1) 2

88. (Sm+ 3n) 2

89. (m - 5) 3

91. (2a

+ l )(a - 4)

95. 7(4m - 3)(2m

+ 1)

97. (3r - 2s ) 4 99. 3p3(2p2 +5p)(p3 +2p + 1) 101. -2x5 (3x2 103. ( 3p

2+

+ 2x - 5)(4x + 2)

~q )( 2p2 - ~q)

83. (a - 4) (a + 4)

+ 6) 2

+ 1) 3

92. (3m 94. - 4r(3r

+

1) 3

+ 2 )(2r - 5)

96. 5(3k - 7)(5k + 2) 98. (2z -5y) 4 100. 5k2(k 3

-

3)(k 2 - k

+ 4)

102. -4x3 (3x4 + 2x2 - x )(-2x+ 1)

2 104. ( 2x +

~y)( 3x2 - ~y)

The figures in Exercises 105-108 are composed of triangles, squares, rectangles, and circles. Find a polynomial that represents the area, in square units, of each shaded region. In Exercises 107 and 108, leave 7T in the answers. (If necessary, refer to the formulas at the back of this text.) 106.

2x+ 5

: ~ ]HI 107.

108.

5x + I

~~·3

SECTION 5.6

Special Products

387

RELATING CONCEPTS For Individual or Group Work (Exercises 109-114)

Work Exercises 109-114 in order. (All units are in f eet.) 109. Find a polynomial that represents the area, in square feet, of the rectangle.

3x + 6

10 ~6_ _ _~

110. Suppose we know that the area of the rectangle is 600 ft2 . Use this information and the polynomial from Exercise 109 to write an equation in x, and solve it. 111. Refer to Exercise 110. What are the dimensions of the rectangle? 112. Use the result of Exercise 111 to find the perimeter of the rectangle.

113. Suppose the rectangle represents a strip of lawn and it costs $0.75 per square foot to lay sod on the lawn. How much will it cost to sod the entire lawn? 114. Suppose it costs $20.50 per linear foot for fencing. How much will it cost to fence the entire lawn?

Special Products OBJECTIVES

OBJECTIVE 1 Square binomials.

1 Square binomials.

lif!fi'MQlll Squaring a Binomial Find (m + 3) 2 .

2 Find the product of the sum and difference of two terms. 3 Find greater powers of binomials.

This is the

answer.

(m + 3) 2 means

(m + 3)(m + 3). (m + 3)(m + 3) = m 2 + 3m + 3m + 9 FOIL method Combine like terms. = m2 + 6m + 9

This result has the squares of the first and the last terms of the binomial.

m2 VOCABULARY

D difference of two squares D conjugates

= m2

and

32

=9

The middle term of the trinomial, 6m, is twice the product of the two terms of the binomial, m and 3. This is true because when we used the FOIL method above, the outer and inner products were m( 3) and 3 ( m), and

m(3) + 3(m) equals 2(m)(3). /:.NOW TRY EXERCISE 1

Thus,

(m

+ 3)

m2

2

"7

Find (x

+ 5) 2 .

+

6m

t

t

Squarem.

2(m)(3)

+

9.

t Square 3.

Square of a Binomial

The square of a binomial is a trinomial consisting of the square of the first term

+

twice the product of the two terms

+

For x and y, the following hold true. NOW TRY ANSWER 1. x 2

+ 10x + 25

(x + y)2 = x2 + 2xy + y2 (x _ y)2 = x2 _ 2xy + y2

the square of the last term

NOW TRY~

388

CHAPTER 5

Exponents and Polynomials

/:.NOW TRY

~EXERCISE2 Square each binomial.

1!£MMQl#j Squaring Binomials Square each binomial.

1) 2

(a) (3x -

(x - y ) 2 = x2 - 2 · x · y + y 2

t t

(b) ( 4p - Sq) 2 (c)

(6t -

~) 2

(a) ( t - 8) 2

+ 2) 2 m(2m + 3) 2

(d) - (3y (e)

t tt t

t

= t2 - 2( t)( 8) + 82 = t2 - 16t + 64

(b) ( 5z - 1) 2

Be careful to square 5z correctly.

= (5z ) 2 - 2(5z )( l ) + 12 = 5 2z 2 - lOz + 1 = 25z 2 - lOz + 1

(c) ( 3b

+ 5r) 2

=

(3b) 2

Be careful to square 3b and 5rcorrectly.

(x _ y )2 = x2 _ 2xy + y2 (5z) 2 = 52z 2 = 25z 2 by power rule (b)

(d) (2a - 9x) 2

+ 2( 3b )(5r) + (5r) 2

= 9b2 + 30br + 25r2

(e) ( 4m

+ ~)2

= (2a) 2

-

= 4a2 -

36ax

2(2a)(9x)

+ (9x) 2

+ 8 lx 2

(f) -(2x - 3)2

= -[(2x) 2 -

= (4m) 2 + 2(4m)(~) + (~)2

= - (4x2

1

=-

= 16m2 + 4m + -

4x 2

-

2(2x)(3)

12x

+ 32 ]

+ 9)

+ 12x - 9

4

(g) x ( 4x - 3)

2

= x ( I 6x = 16x3 -

2

~ Remember themiddle term,2(4x)(3). ) -

24x

24x2

+ 9)

+ 9x

Square the binomial. Distributive property

NOWTRY

g

In the square of a sum, all of the terms are positive. (See Example 2(c).) In the square of a difference, the middle term is negative. (See Example 2(a).)

Q CAUTION

A common error in squaring a binomial is to forget the middle term of the product. In general, remember the following.

+ y) 2 = x 2 + 2xy + y 2, (x - y)2 = x2 - 2xy + y2,

(x

not x 2 + y 2. not x2 - y2.

OBJECTIVE 2 Find the product of the sum and difference of two terms. In binomial products of the form (x + y) (x - y), one binomial is a sum of two terms. The other is a difference of the same two terms. Consider the following. NOW TRY ANSWERS 2. (a) 9x2 - 6x + I (b) 16p 2 - 40pq + 25q2 (c) 3612

-

41

(x + 2)(x - 2)

= x2 = x2 -

+ .!.9

(d) - 9y 2 - 12y - 4 (e) 41113 + 121112 + 9111

Thus, the product of x

2x

4

+ y and x -

+ 2x - 4

FOIL method Combine like terms.

y is a difference of two squares.

SECTION 5.6

Special Products

389

Product of a Sum and Difference of Two Terms

The product of a sum and difference of two terms is a binomial consisting of the square of the first term

the square of the second term.

For x and y, the following holds true. (x

~NOW TRY

~EXERCISE 3 Find the product.

(t + IO)(t - 10)

ltf!JM!QIU (a) (x

(b)

This is the product of a sum 4 ) ..,__ and difference of two terms. x 2 - 42 (x + y )(x - y ) = x 2 - y 2 -

16

Square 4 .

(~-w )(~+ w)

=

(x

4 9

Square 3 .

+ 2)(x - 2) = x(x2 - 4) = x 3 - 4x

(c)

(d)

+ y )(x

- y ) = x2 - y 2

2

(c) x (x

Fi nd the product of the sum and difference of two terms. NOW TRY~

Distributive property

l::WJMQlll Finding the Product of a Sum and Difference of Two Terms Find each product. (x + y ) (x -

+ 6)

(sr-~)(sr+~) y(3y + 1)(3y - I ) - 5(p + q2)(p - q2)

Commutative property

(~)2 - w 2

= --w2

(b)

y2

-

Finding the Product of a Sum and Difference of Two Terms

=(~ + w)(~ - w)

(a) (4x - 6)(4x

=x2

+ 4 ) (x = x2

~EXERCISE4 Find each product.

- y)

Find each product.

=

~NOW TRY

+ y)(x

t

y)

t t

t

(a) (Sm + 3 ) (Sm _ 3 )..,_ This is the product of a sum Be careful to square 5m correctly.

= ( Sm )2 - 32 = 2Sm2 - 9

and difference of two terms. (x + y )(x - y ) = x 2- y 2 Apply the exponents.

(b) (4x + y)(4x - y)

= (4x) 2 NOW TRY ANSWERS 3. 12 - 100 4. (a) 16x2

-

36

(b) 25r2

-

~ 25

(c) 9y 3

(d) -

-

y

5p2

+

5q4

=

(d) p (2p

16x2

-

y2 y2

(4x ) 2 = 42x 2 = 16x 2

+ 1)(2p - 1) p

16 (e) -3 (x

+ y ) (x 2

y2 )

= -3 (x2 - y4)

=p(4p2 - l)

= 4p3 -

1 =z2 __

Distributive property

= -3x2 + 3y4

NOW TRY~

390

CHAPTER 5

Exponents and Po lynomials

l1IiI]]

The expressions x are conjugates.

Example:

+ 2 and x

x

+y

and x - y, the sum and difference of the same two terms,

- 2 are conjugates.

OBJECTIVE 3 Find greater powers of binomials. ~NOWTRY

~EXERCISE 5 Find the product. (2m - 1) 3

l(f!1MQl4j

Finding Greater Powers of Binomials

Find each product. (a) (x+5) 3 =

(x

+ 5) (x + 5) 2

= (x + 5)(x2 +

a 3 = a·a2

lOx + 25 )

Square the binomial.

=

x 3 + 10x2 + 25x + 5x2 + 50x + 125

Multiply polynomials.

=

x 3 + 15x2 + 75x + 125

Combine like terms.

(b) (2y - 3) 4

- 3) 2(2y - 3) 2

= (2y

= (4y 2 = 16y

4

-

12y + 9) ( 4y2

-

48y3 + 36y 2

+ 36y 2

- 108y

= 16y

4

(c) - 2r(r

a 4 = a2 · a2

96y 3

-

-

12y + 9)

-

48y3 + 144y2

-

108y 2

+ 216y

Square each binomial.

-

Multiply polynomials.

+ 81 216y + 81

Combine like terms.

+ 2) 3

=-2r(r+2)(r+2) 2

a3= a·a2

= - 2r( r + 2) ( r + 4r + 4) = -2r(r3 + 4r 2 + 4r + 2r2 + = -2r(r3 + 6r2 + 12r + 8)

Square the binomial.

2

NOW TRY ANSWER 5. 8m3 - I 2m2 + 6m - I

5.6 Exercises O Video solutions for select problems available in MyLab Math

= FOR EXTRA HELP

-2r4 -

0

12r3 -

24r2 -

8r + 8)

Multip ly polynomials. Combine like terms.

16r

Distributive property

NOW TRY

g

Mylab Math

1. Concept Check

A student incorrectly squared (a

(a+ WHAT WENT WRONG? 2. Concept Check

b) 2 =

a2

+ b) as fo llows.

WRONG

Give the correct answer.

A student incorrectly squared (x - y) as follows. (x - y) 2 = x 2

WHAT WENT WRONG? 3. Concept Check

+

b2

+ y2

WRONG

Give the correct answer.

Consider the square of the binomial 4x + 3:

( 4x

+ 3) 2.

(a) What is the first term of the binomial? Square it. (b) Find twice the product of the two terms of the binomial:

(c) What is the last term of the binomial? Square it. (d) Use the results of parts (a)-(c) to fi nd ( 4x

+ 3 )2.

2(__ )( __ )

= __.

SECTION 5.6

4. Concept Check

Consider the product of (7x

(7x

391

Special Products

+ 3y) and (7x - 3y):

+ 3y) (7x -

3y).

(a) What is the first term of each binomial factor? Square it. (b) What is the product of the outer terms? The inner terms? Add them. (c) What are the last terms of the binomial factors? Multiply them. (d) Use the results of parts (a)-(c) to find (7x

+ 3y) (7x -

3y ).

Find each product. See Examples I and 2.

8. (z - S) 2

5. (m + 2) 2

6. (x + 8) 2

9. (x+2y) 2

10. (p - 3m) 2

11. (Sp + 2q) 2

12. (Sa+ 3b) 2

14. (9x - 4) 2

15. ( 4a + Sb) 2

16. (9y

13. (4x - 3)2 17. ( 6m -

~n )2

18. (sx +

7. (r -3) 2

~Y)2

19.

(1

2x + 1 )

3

+ 4z ) 2

)2

1 1 20. ( 4x + S

2

21. 2(x + 6) 2

22. 4(x + 3) 2

23. t(3t - 1) 2

24. x(2x + S) 2

25. 3t( 4t + 1) 2

26. 2x(7x - 2) 2

27. - (4r - 2) 2

28. - (3y - 8) 2

Find each product. See Examples 3 and 4.

29. (k

+ S)(k - S)

30. (a+ 8)(a - 8) 33. (Sx

+ 2)(Sx -

35. (Sy+ 3x)(Sy - 3x)

36. (3x

+ 4y)(3x -

38. ( 13r + 2z)(13r - 2z)

39. (2x2

32. (7 - 2x)(7

41.

+ 2x)

-

S)(2x2

31. (4-3t)(4+3t) 2) 4y)

+ S)

34. (2m

+ S)(2m -

37. (lOx

+ 3y)(10x -

(~- x)(~+x)

+ 1)

46. p(3p

47. - S(a - b 3 )(a

+ b3 )

48. - 6(r - s4 )(r + s 4 )

~(2k 1

51. - - ( lOx 100

+ 7)(3p - 7)

1 50. 3(3m - S)(3m

1)(2k + 1)

+ 10)( 10x - 10)

3y)

40. (9y 2 - 2)(9y 2 + 2)

45. q(Sq - l)(Sq

49.

S)

+ S)

1

52. - 200 (20y + 20)(20y - 20)

Find each product. See Example 5.

53. (x + 1) 3

54. (y + 2) 3

55. (t - 3) 3

56. (m - S) 3

57. (r + S) 3

58. (p + 3) 3

59. (2a + 1) 3

60. (3m+ 1) 3

61. ( 4x - 1) 4

62. (2x - 1) 4

63. (3r - 2t) 4

64. (2z + Sy ) 4

65. 2x(x + 1) 3

66. 3y(y+2) 3

67. -4t(t + 3)3

68. -Sr(r + 1) 3

69. (x +y) 2 (x-y) 2

70. (s + 2) 2 (s - 2) 2

392

CHAPTER 5

Exponent s and Polynomials

The special product (x

+ y)(x - y)

=

x2

-

y 2 can be used to perform some multiplications.

51 x 49

Examples:

=

(50

102 x 98

+ 1)(50 - 1)

= 502 -

12

=

(100 + 2)(100 - 2)

=

1002

-

22

=

2500 - 1

=

10,000 - 4

=

2499

=

9996

Use this method to calculate each product mentally.

71. 101 x 99

72. 103 x 97

73. 201 x 199

74. 301 x 299

1 1 75. 20- x 192 2

76. 30- x 293 3

1

2

Find a polynomial that represents the area, in square units, of each figure. (If necessary, refer to the formulas at the back of this text.) 77.

~ m-2n 78. m+2n

0

79. 6p+q

&,_,

l

3a + 2

6p + q

80.

81.

G

82.

3x+ I

Ji4 _o

\

5x+ 3

@ I

Refer to the cube shown here.

83. Find a polynomial that represents the volume of the cube (in cubic units).

-

84. If the value of x is 6, what is the volume of the cube (in cubic units)? RELATING CONCEPTS For Individual or Group Work (Exercises 85-94) Refer to the figure, and justify the special product

a

b

~

(a+

b)2

= a2

+ 2ab + b2.

b

85. Express the area of the large square as the square of a binomial. 86. Give the monomial that represents the area of the red square. a

87. Give the monomial that represents the sum of the areas of the blue rectangles. 88. Give the monomial that represents the area of the yellow square.

89. What is the sum of the monomials obtained in Exercises 86-88? 90. Why must the binomial square (Exercise 85) equal the polynomial (Exercise 89)? Apply the special product (a

+ b) 2

=

a2

+ 2ab + b2 to a purely numerical problem.

91. Evaluate 35 2, using either traditional paper-and-pencil methods or a calculator. 92. The number 35 can be written as 30 + 5. Therefore, 35 2 = (30 + 5) 2 . Use the special product for squaring a binomial with a = 30 and b = 5 to write an expression for (30 + 5)2. Do not simplify at this time. 93. Use the order of operations to simplify the expression found in Exercise 92. 94. How do the answers in Exercises 91 and 93 compare?

x+2

Dividing Polynomials

SECTION 5.7

393

Dividing Polynomials OBJECTIVES

OBJECTIVE 1 Divide a polynomial by a monomial.

1 Divide a polynomial by a monomial.

We add two fractions with a common denominator as follows.

a c

2 Divide a polynomial by a polynomial. 3 Apply polynomial division in a geometry problem.

b c

a+b c

-+- = - In reverse, this statement gives a rule for dividing a polynomial by a monomial. Dividing a Polynomial by a Monomial

To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.

a+ b a b -- = - + c

2+5 3

c

2 5 +3 3

Examples: - - = -

(where c

c

x

+ 3z

+ 0)

x 2y

3z 2y

- - = - + - (y =F 0)

and

2y

The parts of a division problem are named as follows.

+ 6x - - - - = 2x + 1

Dividend ~ 12x2 Divisor ~

~NOW TRY

"7 EXERCISE 1 Divide 16a6

-

12a4 by 4a2•

lif!f!1MQlll Divide 5m5

-

6x

Dividing a Polynomial by a Monomial

10m3 by 5m2 . 5m5

10m3

-

A fraction bar means division.

2

5m

5m5 5m2

= m3 CHECK

~Quotient

Multiply

5m2

•

(m3

t

10m 3

Use the preceding rule, with + replaced by - .

5m2 2m -

2m)

Quotient rule

= Sm5 -

t

Divisor

10m 3 ~

./

Original polynomial (Dividend)

Quotient

10

3

Because division by 0 is undefined, the quotient sms 5-m 2 m is undefined if 5m2 = 0, or m = 0. From now on, we assume that no denominators are 0. NOW TRY

g

lif$WQlfj

Dividing a Polynomial by a Monomial

Divide. l6a5

NOW TRY ANSWER 1. 4a4 - 3a2

-

12a4 4a 3

l 6a 5 4a3

+ 8a2 This becomes ~. not 2a.

l 2a4

8a2

4a

4a3

= -- - -+3 2

= 4a2 - 3a + -

a

Divide each term by 4a3 . 2 aa s 2 - 3 = - a2- 3 = 2a- 1 = -

4a

4

a

394

CHAPTER 5

Exponents and Polynomials

~NOW TRY Divide.

+ 24x4 -

36x5

polynomial because a~ has a variable in the denominator. While the sum, difference, and product of two polynomials are always polynomials, the quotient of two polynomials may not be a polynomial.

The quotient 4a2

"7 EXERCISE 2 12x3

6x4

CHECK

3a

-

4a 3 ( 4a2

+ a~ is not a

~)

3a +

-

Divisor x Quotient should equal Dividend.

= 4a 3 (4a 2 ) + 4a 3( -3a ) + 4a 3 (~) = 16a5 ~NOW TRY

"7 EXERCISE 3 Divide 7y4

-

40y5

+ 8a2

12a4

Distributive property Dividend

./

NOW TRY

g

l+:f4MQl#I Dividing a Polynomial by a Monomial (Negative Coefficient) + 100y 2

by - 5y 2•

Divide - 7x 3

+

12x4

-

4x by - 4x.

Write the dividend polynomial in descending powers.

12x4

-

7x 3 - 4x

4x

-

Write in descending powers before dividing.

12x 4 - 4x

7x 3 - 4x

4x - 4x

--------

CHECK

7x2

=

-3x 3

-

=

-3x 3

+-

_

- (-1)

4

7x2 4

+1

7 2 -4x( -3x 3 + : +

1)

=-

4x(-3x 3 )

= 12x 4 -

-

Divide each term by - 4x.

Quotient rule Be careful with signs, and be sure to include 1 in the answer.

Divisor x Quotient should equal Dividend.

72 4x( : )

-

4x(l)

Distributive property

7x 3 - 4x ./

Dividend

~NOWTRY

l+:f4MQlll Dividing a Polynomial by a Monomial

Divide

Divide -180x4y 10

"7 EXERCISE 4 35m5n4

49m2n3

-

+ I2mn

-180x4y 10

+ 150x3y 8 - 120x 2y 6 + 90xy4

-

NOWTRYg

lOOy by -30xy 2 •

+ 150x3y 8 - 120x2y 6 + 90xy 4 - lOOy -30xy 2

by 7m2n.

-180x4y 10 - 30xy

150x3y 8 - 30xy

- - - 2- + - - -2

= 6x3y 8 NOW TRY ANSWERS

Sx 2y 6 + 4xy4

120x2y 6 - 30xy

90xy4 - 30xy

- - -2 + - - -2

-

lOOy 30xy2

Divide each term by - 30xy 2 .

10

-

3y 2 + 3xy

Check by multiplying the divisor by the quotient.

NOWTRYg

2

2. 6x + 4 - -

x

3.

8y 3

7y2

- -

4. 5m3n 3

-

5

- 20

7n2 + ~

7m

OBJECTIVE 2 Divide a polynomial by a polynomial. We use a method of "long division" to divide a polynomial by a polynomial (other than a monomial). Both polynomials must first be written in descending powers.

SECTION 5.7

Dividing Whole Numbers

Dividing Polynomials

395

Dividing Polynomials

Step 1 4x 2

-

14x

+ 15 by 2x + 3.

2x + 3)8x3

-

4x2

-

Divide 8x3

Divide 6696 by 27. 27)6696

-

14x + 15

Step2

= 4x

8x3 divided by 2x

= ]2

66 divided by 27

2

. --~

4x 2 (2x + 3) = 8x3 + 12x2

2 . 27 = 54

)6~96

I L

2x +

-

---+------,

4x 2

- (x - 1) x+2

@-?>

m

Multiplying and Dividing Rational Expressions

x- 1 - (x + 2) '

Multiply.

Note the operation. If the operation is division, use the definition of division to rewrite it as multiplication.

Step 2

Multiply numerators and multiply denominators.

Step 3

Factor numerators and denominators completely.

Step 4

Write in lowest terms using the fundamental property.

Note: Steps 2 and 3 may be interchanged based on personal preference.

x+2

Distribute the negative sign in the denominator.

x- 1 -x - 2

- - - (4 -

\/3) ( 4 + \/3)

2(4 + \/3)

2(4 + \/3)

42 - (\/3)2

13

Rationalizing a Binomial Denominator NOW TRY ANSWERS 2. (a) 23 (c) 9

sV7

(b) 43 - 30Vz

+ 6vY + y

3. (a) 540

(b) 10 - x

To rationalize a binomial denominator, where at least one of those terms is a square root radical, multiply the numerator and denominator by the conjugate of the denominator.

602

CHAPTER 8

Roots and Radicals

~NOWTRY

1!£MMQlll Using Conjugates to Rationalize Denominators

Rationalize each denominator. In part (c), assume that k 2'. 0.

Rationalize each denominator. In part (c), assume that x;:::: 0.

"7 EXERCISE 4

(a)

6

5

• ~

(b) • ~

4+v3

(c) •

9

r.

vk - 6

+ V7

(a)

v7 -2

(k

"" 36)

5

• ;-;

3 + v5

5(3 - Vs) (3 + Vs) (3 - Vs)

Multiply the numerator and denominator by the conjugate of the denominator.

5(3 - Vs) 32- (Vs) 2

(x + y )(x - y ) = x2

5(3 - Vs)

-

y2

32 = 9; (Vs )2 = 5

9- 5

5(3 - Vs)

Subtract in the denominator.

4 We leave the numerator in factored form. This makes it to easier to determine whether the expression is in lowest terms.

6+

(b)

V2

Vl -5 (6 + V2) (V2 + 5) (V2- 5) (V2 + 5)

Multiply the numerator and denominator by the conjugate of the denominat or.

6Vl + 30 + 2 + 5Vl

FOIL method; (x + y )(x - y ) = x2

2 - 25 ll Vl + 32 - 23

y2

Combine like terms.

- (11 V2 + 32)

x -x -=-y y

23 Be careful. Distribute the - sign to both terms in the numerator.

-

- 11V2 - 32

Distributive property

23

The last three lines above give three equivalent forms of the answer.

4

(c)

3+ NOW TRY ANSWERS

6(4 4. (a)

Vx 4 (3 -

v'3)

13

(3 + Vx )(3 - Vx )

17 +1V7 (b)

3

9( Vk+ 6) (c)

k - 36

Vx)

4(3 We assume here that x ~ 9.

Vx)

9- x

Multiply by

32

= 9;

3_ 3

Vx Vx

(Vx) 2 = x

= 1.

NOW TRY~

SECTION 8.5

More Simplifying and Operations with Radicals

~NOW TRY

OBJECTIVE 3 Write radical expressions with quotients in lowest terms.

Write the quotient in lowest terms.

lif:t!JMQllj

\:7 EXERCISE 5

12V6 + 28

603

Writing a Radical Quot ient in Lowest Terms

Write the quotient in lowest terms.

3VJ +9

20

Each term in the numerator and denominator has a common factor of 3.

12

3(VJ+3) 3(4)

v3+3

= 1 · ---

4

v3+3

Factor first.

Now divide out the common factor; ~ = 1.

NOW TRY~

Identity property; lowest terms

4

0

CAUTION An expression like the one in Example 5 can be simplified only by factoring a common factor from the denominator and each term of the numerator. First factor, and then divide out the common factor.

NOW TRY ANSWER

3\/6+7

Example:

Factor

4 +8V5 4

5. - - 5

8.5 Exercises O Video solutions for select problems available in Mylab Math

FOR

EXTRA HELP

O

as

4(1 +2V5)

to obtain

4

1 + 2\/5.

Mylab Math

In this exercise set, we assume that variables are such that no negative numbers appear as radicals in square roots and no denominators are 0. Concept Check mediate steps. 1.

V25 + V64

Concept Check 5.

Perform the operations mentally, and write the answers without doing inter-

2.

ViOO - v49

3.

v8. v'2

6.

(v'28 - v14)(v'28 + v14)

4.

V6 · V6

Find each product mentally.

(Vx + vY)(Vx- vY)

7. Concept Check

A student simplified the expression -37 - 2Vi5

by combining -37 and -2 to obtain the expression - 39Vi5, which is incorrect. WHAT WENT WRONG? 8. Concept Check

A student incorrectly squared the binomial

V7 + 5 as follows.

WHAT WENT WRONG? Square the binomial correctly. Find each product. Refer to the five guidelines given in this section to be sure answers are simplified. See Examples 1-3.

9.

v's(\/3 - V7)

10.

V7(\/10 - \/3)

11.

V3(Vi8 - VsO)

604

CHAPTER 8

Roots and Radicals

12.

v2( v20 - VsO)

15.

3Vi4 . v2 -

18.

(4Vs + 2)(2-Vs + 4)

21.

(V6 -

24. (4 Vs

1)

2Vs(3Vs + v2)

13.

v2s

16. 7'\f6.

2

+ 5 )2

V3 - 2Vi8

19.

(s - V7) 2

22.

(V7 -

25.

(sV7 - 2V3) 2

2)

14.

3V7(2V7 + 4Vs)

17.

(2V6 + 3)(3V6 + 7)

2

23.

(2V7 + 3)2

26.

(sv2 - 3\13) 2

27. (Va+l) 2

28.

(\/Y+ 4) 2

Vx) 2

30.

(12 -Y,:) 2

29. (7 -

Vil) 2

20. ( 6 -

31.

(s - v2)(s + v2)

32.

(3 - Vs)(3 + Vs)

33.

(Vs - V7)(v8 + V7)

34.

(Vi2 - Vil)(Vi2 +Vil)

36.

(vt - Vi3)(vt + Vi3)

38.

(\13 + Vs)(VlS - Vs)

35. (

vY - vlo)( vY + vlo)

37. (v2 + V3)(V6 - v2) 39.

(\/lo - vs)(vs + v20)

40.

(V6 - V3)(V3 + Vi2)

41.

(Vs + V30)(V6 + \13 )

42.

(vlo - v20)(v2 - Vs)

43.

(sV7 - 2V3)(3V7 + 4\13)

44.

(1\/lo + sv2)(3vlo - 3v2)

45.

(3Vt + V7)(2Vt - Vi4)

46.

(r Jz - V3)(Vz - vs)

47. (~ + \12,;)(~ - \12,;)

48.

(v4P - V3k)(v4P + V3k)

49. Concept Check Determine the expression by which we should multiply the numerator and denominator to rationalize each denominator. (a)

5

• r,:

+ v3

3

• ~

(b) • /,

v6 - v5

(c) •

r

5

vx - 1

50. Concept Check Determine the greatest common factor of the terms in the numerator and denominator of each expression.

12 + 8Vs 16

(a)

(b)

18Vl - 24 6

51. Concept Check A student tried to rationalize the denominator of

4

V3 + \13.

4+

2

+ \13

by multiplying by

4

WHAT WENT WRONG? By what should he multiply?

(c)

6+

V3

3

52. Concept Check A student incorrectly simplified the expression

6

+ 30\/il 6

by dividing out 6 from the first term of the numerator and the denominator to obtain 1 + 30Vll. WHAT WENT WRONG? Give the correct simplified form.

SECTION 8.5

605

More Simplifying and Operations with Radicals

Rationalize each denominator Write quotients in lowest terms. See Examples 4 and 5. 1

53.

1

54.

2 + Vs

V15 Vi8

V12

58.

62.

v2 +3 65. V3 3- 1

66.

2V6 + 1 69. v2 2+5

3v'2-4 70. V3 3+2

V3

63.

6+

2\/3

60.

71.

VlOs

64.

vY

9Vs 6v'2 - 6

5-v2 68. V3 3+3

V7 + v2

72.

V3 - v2 75.

2\/3 2-ViO

6-Vs 67. v2 2+2

12

74.

Vx

5- V6

3 + 3\13

Vs+ 2 V3 23

38

56.

\/3 + 5

v2+V3

8 4+

59.

v2-1

Vs 61. v2 V3 2+ 3

7 2 - vJl

4+

57. V3 3+ 1

73.

55.

V6+v'5 V3 + v'5

1

76.

Vx+vY

2

Vx - vY

Write each quotient in lowest terms. See Example 5. 77.

81.

5V7 - 10 5 12 - 2Vi0 4

Extending Skills

85.

V4(V2 -

8 6Vs - 9 7 . 3

79.

82. 9 - 6v2 12

83.

16 +

Vl2s

8

24

86. Vs(4Vs -

3)

88. 6ef9(2ef9 -

(Vi - 1) (V4 + 3)

91. (Vs -

80. 4yl6 + 6 10 4

V75

·

25 + 10

Peiform each operation and express the answer in simplest form.

87. 2ef2(3efs + 5 ~) 89.

2\/3 + 10 8

90. ( '-Y9

V4)(V25 + V20 + \Yi6)

92.

V25) €7)

+ 5 ) ( V3 -

4)

(V4 + V2)(\Yi6 - V8 + V4)

Solve each problem. 93. The radius r of the circular top or bottom of a tin can with surface area Sand height h is given by r =

Vh2 + 0.64S ------ h +

2

f- r --j

T 1 h

(a) What radius should be used to make a can with height 12 in. and surface area 400 in. 2 ? (b) What radius should be used to make a can with height 6 in. and surface area 200 in .2 ? Round the answer to the nearest tenth. 94. If an investment of P dollars grows to A dollars in 2 yr, the annual rate of return on the investment is given by r =

VA - VP VP

(a) Rationalize the denominator to simplify the expression. (b) Find the annual rate of return r (as a percent) if $50,000 increases to $54,080.

606

CHAPTER 8

Roots and Radicals

RELATING CONCEPTS For Individual or Group Work (Exercises 95-100)

Work Exercises 95- 100 in order, to see why a common student error is indeed an error.

95. Use the distributive property to write 6(5 + 3x) as a sum. 96. The answer in Exercise 95 should be 30 + l 8x. Why can we not combine these two terms to obtain 48x? 97. Find the product. (2Vi0 + 5V2)(3Vi0- 3V2) 98. The answer in Exercise 97 should be 30 + l8Vs. Many students will, in error, try to combine these terms to obtain 48 Vs. Why is this wrong? 99. Write the expression similar to 30 + I 8x that simplifies to 48x. Then write the expression similar to 30 + l8Vs that simplifies to 48Vs. 100. Write a short explanation of the similarities between combining like terms and combining like radicals.

SUMMARY EXERCISES

Applying Operations with Radicals Perform all indicated operations, and express each answer in simplest form. Assume that all variables represent positive real numbers.

sViO

1. 5Vi0 -

4. \/98 - vn+

7.

V50

l + V2

2. Vs(Vs -V3)

3. ( l +

5. (3Vs - 2V7) 2

6.

8

8.

1- Vl

V3) (2 - V6)

3

V6

9. (\/3 + 6)(\/3 - 6)

V7-Vs

1

10.

Yrt + V33

13.

V66 -

5

1

12

V9 3 9

11. V8x 3y 5z 6

12.

14.~3x

15. 6 \/3 5\/i2

-4

8Vs0

18 \/6 - Vs . \/6+ Vs

16.

V25 2 25

17.

19.

Vlli - Vih

20. (5 + 3\/3)

22.

Vi4 +

23. V7+V7

24. 9\/24 - 2\/54

5 26. Vs

27.

Vs

25. ~ · Jf 28.

8 4 - Vx

(x =P 16)

29.

\Y4 3 4

1%

2

21.

Hi l

\Y24 + 6Vsl

30. (7 + Vx)

2

SECTION 8.6

Solving Equations with Radicals

607

Solving Equations with Radicals OBJECTIVES

OBJECTIVE 1 Solve radical equations having square root radicals.

1

A radical equation is an equation having a variable in the radicand.

Solve radical equations having square root radicals.

2 Identify equations with no solutions.

Vx=6, Vx+l=3, 3Vx=~

Radical equations

To solve radical equations, we use the squaring property of equality.

3 Solve equations by Squaring Property of Equality

squaring a binomial.

4 Solve radical equations

If each side of a given equation is squared, then all solutions of the original equation are among the solutions of the squared equation.

having cube root radicals.

0

CAUTION Using the squaring property can give a new equation with more solutions than the original equation. For example, starting w ith x = 4 and squaring each side gives x 2 =42 ,

VOCABULARY D radical equation D extraneous solution

or

x 2 =16.

This last equation, x 2 = 16, has two solutions, 4 or -4, while the original equation, x = 4 , has only one solution, 4. Because of this possibility, checking is more than just a guard against algebraic errors when solving an equation with radicals. It is an essential part of the solution process. Alf proposed solutions from the squared equation must be checked in the original equation.

The squaring property allows us to eliminate the radicals in an equation. Then we can solve the resulting equation, which will be either linear or quadratic. /:.NOW TRY

ltf!1MQlll

Solve~ =6.

Solve

~EXERCISE 1

Using the Squaring Property of Equality

Vx+l = 3. Vx+l = 3

(Vx+l) 2= 32 This equation is linear.

x + l=9

Proposed solution ____,. x

CHECK A check is essential.

=8

On the left, ( Va)2 = a. Subtract 1.

Vx+l = 3

Original equation

v8+1 ~3

Let x = 8.

\19~3

3=3 ./ NOW TRY ANSWER 1. {41}

To eliminate the radical, use the squaring property and square each side.

Add . True

Because this statement is true, { 8} is the solution set of Vx+l = 3. In this case, the equation obtained by squaring had just one solution, which also satisfied the original equation. NOW TRY ~

608

CHAPTER 8

Roots and Radicals

~NOW TRY

"7 EXERCISE 2 Solve 4Vx = Y lOx

+ 12.

1!£MMQl#j Using the Squaring Property with a Radical on Each Side Solve 3 Vx = v:;+S. We need to eliminate both radicals. 3Vx= v:;+8 (3Vx) 2= ( v:;-+8)2 32( Vx) 2= ( v:;+S)2 Be careful here.

9x = x 8x

Proposed solution ____,.. x

=8

Subtractx.

=

Divide by 8.

~~ 3(1) 1= This is not the solution.

On the left, (ab ) 2 = a2b2.

(Vx) 2 = x; (~) 2 = x + 8

+8

1

3Vx = Vx+8 3vl 1= \/i+8

CHECK

Squaring property

Original equation Letx = 1.

V9

3=3

./

True

Because a true statement results, the solution set is { 1}.

NOW TRY

g

0

CAUTION The final result obtained in the check is not the solution. In Example 2, the solution set is { 1 } , not { 3 } .

OBJECTIVE 2 Identify equations w ith no solutions. Not all radical equations have real number solutions. ~NOW TRY

"7 EXERCISE 3 Solve Vx = - 6.

1!£MMQl#I Using the Squaring Property When One Side Is Negative Solve Vx = -3. Vx= -3 (Vx) 2= (- 3)2 Proposed sol ution ____,.. x

CHECK

=9

Squaring property Apply the exponents.

Vx = - 3

Original equation

\!9 1= - 3

Let x = 9.

3 = -3

False

Because the statement 3 = - 3 is false, the number 9 is not a solution of the given equation. It is an extraneous solution and must be rejected. In fact, Vx = - 3 has no real number solution. The solution set is 0 . Do notwrite { 0 } to represent the empty set.

NOW TRY

g

NOW TRY ANSWERS

2. {2} 3. 0

miIJ] Because

Vx

represents the principal or nonnegative square root of x, we might have seen immediately in Example 3 that there is no real number solution.

SECTION 8.6

Solving Equations with Radicals

609

Solving a Radical Equation

l:.NOWTRY

"7 EXERCISE 4 Solve t

= V~ t 2-+3t-+ 9.

Step 1

Isolate a radical. Arrange the terms so that a radical is isolated on one side of the equation.

Step 2

Square each side.

Step 3

Combine like terms.

Step 4

Repeat Steps 1-3, if there is still a term with a radical.

Step 5

Solve the equation. Find all proposed solutions.

Step 6

Check all proposed solutions in the original equation. Write the solution set.

l!f$JMQlll

Using the Squaring Property with a Quadratic Expression

Solve x

= v'x2 + Sx + 10.

Step 1

The radical is already isolated on the right side of the equation.

Step 2

Square each side. x2 = x2

Squaring property 2

On the right, ( Va) = a. Subtract x 2 .

This step is not needed. No remaining terms contain radicals. - 10 = Sx

Step 5

Proposed solution --- -

Step 6

= x 2 + Sx + 10

0 = Sx + 10

Step 3 Step 4

(\Ix 2 + Sx + 10) 2

CHECK

The principal square root of a quantity cannot be negative.

Subtract 10.

2=x x =

Divide by 5.

Vx 2 + Sx + 10

Orig inal equation

-2 :1= v'(-2) 2 + 5(-2) + 10 -2 :1=

-2

v4 -

10 + 10

Let x = - 2. Multiply.

=2

False

Because substituting -2 for x leads to a false statement, the equation has no real NOW TRY ~ number solution. The solution set is 0.

OBJECTIVE 3 Solve equations by squaring a binomial. Recall the rules for squaring binomials.

(x + y ) 2 = x 2 + 2xy + y 2 and (x - y ) 2 = x 2

-

2xy + y 2

We apply the second rule in Example 5 on the next page when finding (x - 3) 2 .

(x - 3) 2

Remember the middle tenm when squaring

= x2 - 2x(3) +3 2 NOW TRY ANSWER 4. 0

= x 2 -6x +9 Verify the result by multiplying (x - 3)(x - 3) using the FOIL method.

610

CHAPTER 8

Roots and Radicals

/:.NOW TRY

~EXERCISE 5

Sol ve~ =x - 5.

1!£f4MQl4j Using the Squaring Property When One Side Has Two Terms Solve~ =x- 3. ~=x-3 (~)2 = (x- 3)2 Be careful

2.x-3= x 2 -6x + 9

squaring

Square each side. (va) 2 = a; (x _ y )2 = x2 _ 2xy + y 2

This equation is quadratic because of the x 2-term. Standard form __..,,. x 2

-

8x

+ 12 = 0

(x-6)(x-2)=0 x -6=0 or Proposed solutions --7 x = 6 CHECK

Subtract 2x, add 3 , and interchange s ides. Factor.

x-2=0

Zero-factor property

x= 2

Solve each equation.

or

~= x - 3

~ = x -3

v2(6) - 3 =1 6 - 3

Let x

=

6.

v2( 2) - 3 =1 2 - 3

Let x = 2.

V4""=3 =1 -1 Vi =1 -1

Vi2-=3 =13

V9 =13 3=3 ./

1= - 1

True

Only 6 is a valid solution. (2 is extraneous.) The solution set is { 6} .

False NOW TRY

g

1!£f4MQl#j Rewriting an Equation before Using the Squaring Property Solve

\/9x -

1 = 2.x.

If we begin by squaring each side, we obtain the following.

(\19x _ l )2 = (2.x) 2 9x-2 \/9x + 1 = 4x2

This equation still contains a radical.

We must apply Step 1 here and isolate the radical before squaring each side.

\/9x This is a key step.

1

= 2x

Original equation

v9x=2x+l

Add 1 to isolate the radical. (Step 1)

( v9x) 2 = (2.x + 1) 2 No terms contain radicals.

4x2

-

9x = 4x2 + 4x + 1 Sx

+ 1=0

( 4x - 1) (x - 1) 4x - 1 = 0 NOW TRY ANSWER 5. { 12}

Proposed solutions __..,,. x

1

=4

Square each side. (Step 2) ( va)2 = a; (x + y)2 = x2 + 2xy + y2

Standard form (Step 3)

=0

Factor. (Step 5; Step 4 is not needed .)

or

x - 1=0

or

x= I

Zero-factor property Solve each equation.

SECTION 8.6

CHECK

~NOW TRY

~- 1 = 2.x

Solving Equations with Radicals

~-1=2x

(Step 6)

611

(Step 6)

~EXERCISES

Solve

Ynx - 3 =

2x.

1

Letx = 4 .

Rf)-11=2(±) 3 ? 1 - -1= 2 2 1 2

V9(i) -

11= 2( 1)

v1=~

1 ./ 2

2

True

=

2

./

Both proposed solutions check, so the solution set is { ~, 1}.

0

CAUTION

Let x = 1.

True

NOW TRY

~

When squaring each side of the equation V9x=2x+ 1,

See Exam ple 6.

the entire binomial 2x + 1 must be squared to obtain 4x2 + 4x + 1 . It is incorrect to square the 2x and the 1 separately and w rite 4x2 + 1 .

Solving some radical equations, such as

V21+x = 3 + Vx, requires squaring both sides of the equation a second time to eliminate all radical terms.

~NOW TRY

~EXERCISE 7 Solve

Vx + 2 = Vx+S.

lif!f!1MQIU Using the Squaring Property Twice Solve \/21+x = 3 + Vx. A radical is isolated on the left. (Step 1) V2l+x = 3 + Vx (\/21+x) 2= (3 + Vx) 2

Square each side. (Step 2)

21 +x = 9+ 6Vx +x ~

6Vx 2= Vx

12 =

22= (Vx)2 Proposed solution ----;;. 4

CHECK

=x

Subtract 9. Subtract x. (Step 3) Divide by 6. Square each side again. (Step 4) Apply the exponents. (Step 5)

V21+x = 3 + Vx v21+4 1= 3 + V4 \!251=3+2

Original equation (Step 6) Let x = 4. Simplify.

NOW TRY ANSWERS

5

6. u.3} 7. {I}

The solution set is { 4}.

=5

./

True NOW TRY~

612

CHAPTER 8

Roots and Radicals

OBJECTIVE 4 Solve radical equations having cube root radicals. We do this by extending the concept of raising both sides of an equation to a power. Instead of squaring each side, we cube each side, using the fact that

(Va)3 =

l!:m. NOW TRY

lif:NMQlj:i

Solve each equation.

Solve each equation.

~EXERCISES (a)

Cb)

V8x-=3 = % V2x2 = V110x -

(a) 12

a,

for any real number a.

Solving Equations with Cube Root Radicals

Th=€x+i (Th) 3= (€x+i)3 Sx

3x

=

Cube each side.

+1

Apply the exponents.

2x = 1

Subtract 3x.

1

x=-

Divide by 2.

2

CHECK

Th =€x+i

Orig inal equation 1

Let x = 2·

1%~~ \[%= \[% ~ The solution set is

Multiply; 1 = ~

True

{fl. W= V126x+ 27

(b)

( W) 3 = ( V26x + x2 x2

-

27) 3

= 26x + 27

26x - 27 = 0

+ 1= x

CHECK

0

or

Factor.

x - 27 = 0

= - 1 or

x

W = v26x +27 V(=1)2 ~ V26( -

Apply the exponents. Standard form

(x + l)(x - 27) = 0 x

Cube each side.

1) + 27

= 27

Zero-factor property Solve each equation.

w = v26x + 27 V(27)2 ~ V26( 27 ) + n

Let x = - 1.

Yfi29 ~ Yfi29 NOW TRY ANSWERS

8. (a)

m

(b)

{2, 3}

1 =1

~

True

9

=9

Both proposed solutions check, so the solution set is { -1 , 27}.

~

Let x = 27.

True NOW TRY

g

SECTION 8.6

8.6 Exercises 0 Video solutions for select problems available in MyLab Math

0

FOR EXTRA HELP

613

Solving Equations with Radicals

Mylab Math

Concept Check

Fill in the blanks to complete the following.

1. To solve an equation involving a radical, such as \/2x"-=l = 5, use the _ _ __ , all property of equality. This property says that if each side of an equation is solutions of the equation are among the solutions of the squared equation. 2. Solving some radical equations involves squaring a binomial using the following rules. (x + y)2= _ __

(x -y)2= _ __

Solve each equation. See Examples 1-4.

3.

Vx =

6. Vt + 3 = 10

Vx = 10 Vi = - 5

5.

Vw -

8.

V'P = - s

10. ~ = 5

11.

v,:-=-4 =

14. ~ = 5

4.

7

1.

9.

Vx+2 =

12.

Vx-=-12 =

3

13. V4-=t= 7

15.

V2t+3 =

0

16. ~

18.

V6t+4 =

-

3

4

7

=

9

17.

\!3x""=8 =

19. V lOx - 8 = 3Vx

20.

\!l7t="4 = 4 Vt

21. 5Vx = v10x + 15

22. 4Vx= v2ox- 16

23. ~ =

24. ~ = ~

25. k

27. 7x = V49x2 + 2x - 10

28. 6x = V36x2 +5x - 5

29. ~ = ~

31. ~ =

32. V3m + 3 =

30.

3

Vx+2 = vh"=5

=

0

=

Vk 2 -

33. Concept Check Consider the following incorrect "solution."

5k - 15

v4x+l

V x2 -

2x - 6

V5m"=l

Vh+l =x -7

-(x - 1) = 16

Square each side.

- x + 1 = 16 =

=

Vh+l

34. Concept Check The first step in solving the equation

- ~= -4

x

26. x

-2

is to square each side of the equation. When a student did this, he incorrectly obtained the following.

Distributive property

- 15

Solve for x.

2x + 1 =x2 +49

Solution set: { -15} WHAT WENT WRONG? Give the correct solution set.

WHAT WENT WRONG? Square each side of the original equation correctly.

Solve each equation. See Examples 5 and 6.

35. V 5x + 11 = x + 3 38.

vTx""+3 = x -

5

36. ~ = x+ 1

37.

Vh+l = x -

39. V3x + 10 = 2x - 5

40.

V4t+13 = 2t - 1 V6t+7 + 3 = t + 5

41. ~ - l = x

42.

44. V lOz + 24 + 1 = x + 5

45. 2~=x - l

47.

-ylh + 4 = x

50. Vx + 3 = x - 9

-Vx+l -

1= x

43.

7

46. 3Vx+l3 = x + 9

V3x + 6 = x

49. Vx + 9 = x + 3

51. 3~=x - 2

52. 2~=x - 4

48.

614

CHAPTER 8

Roots and Radicals

Solve each equation. See Example 7.

53. Vx + 6 55.

=

Vx+n

54. Vx - 4

v;+l- ~=

57. ~

1

+ Vx+2 =

5

Vx-=-32

56.

v'h+3 + v;+l =

1

58.

vS+l + Vx+4 =

3

60.~+ ~= 4

59.~-~ =2

62. ~ +

+ 11 + Vx+6 = 2

6t. v2x

=

Vx+l6 =

7

Solve each equation. See Example 8.

63. ~ =

V5x+2

64.

65. ~ =

V's+7x

66. ~ =

67. V3x2

69. 'V'x 2

-

9x + 8 =

efx

68.

+ 24x = 3

'0"4x = ~ V5x2 -

70. 'V'x2

vs=-7x 6x

+ 6x =

+2

=

efx

2

Solve each problem.

71. The square root of the sum of a number and 4 is 5. Find the number. 72. A certain number is the same as the square root of the product of 8 and the number. Find the number. 73. Three times the square root of 2 equals the square root of the sum of some number and 10. Find the number.

74. The negative square root of a number equals that number decreased by 2. Find the number. Solve each problem. Give answers to the nearest whole number. 75. To estimate the speed at which a car was traveling at the time of an accident, a police officer drives the car under conditions similar to those during which the accident took place and then skids to a stop. If the car is driven at 30 mph, then the speed s at the time of the accident is given by

s=30~, where a is the length of the skid marks left at the time of the accident and p is the length of the skid marks in the police test. Find s for the following values of a and p. (a) a= 862 ft; p = 156 ft

(b) a

= 382 ft; p = 96 ft

(c) a = 84 ft ; p = 26 ft

76. A formula for calculating the distance d in miles one can see from an airplane to the horizon on a clear day is

d

=

l.22Vx,

I'

where x is the altitude of the plane in feet. How far can one see to the horizon in a plane flying at the following altitudes? (a) 15,000 ft

(b) 18,000 ft

(c) 24,000 ft

I I

x l I I I I I

''

',

d ', ' ',

' '

SECTION 8.6

Solving Equations with Radicals

615

On a clear day, the maximum distance in kilometers that can be seen from a tall building is given by the formula sight distance = 111 .7 Vheight of building in kilometers. Use this formula and the conversion equations 1 ft = 0.3048 m and 1 km = 0.62 1371 mi as necessary to solve each problem. Round answers to the nearest mile. (Data from A Sourcebook of Applications of School Mathematics, NCTM.) 77. The London Eye is a unique structure that features 32 observation capsules and has a diameter of 135 m. Does the formula j ustify the claim that on a clear day passengers on the London Eye can see Windsor Castle, 25 mi away? (Data from www.londoneye.com) 78. The Empire State Building in New York City is 1250 ft high. The observation deck, located on the 102nd floor, is at a height of 1050 ft. How far could we see on a clear day from the observation deck? (Data from www.esbnyc.com)

RELATING CONCEPTS For Individual or Group Work

(Exercises 79-84)

The most common formula for the area . 0.

In each equation so far, the exponent 2 appeared with a single variable as its base. We can extend the square root property to solve equations in which the base is a binomial.

lf!f4MQlll

Solving Quadratic Equations of the Form (x

+ b)2 =

k

Solve each equation, and check the solutions. (a)

Use (x- 3) as the base.

(x-3) 2 =16

x- 3

= \/i6

x - 3=4 x =7 CHECK

or x -

= - \/i6 3 = -4

or

x = - 1

or

x- 3

Square root property

Y-i6 =4 Add 3.

Substitute each value in the original equation.

(x - 3) 2 = 16 (7 - 3) 2 =1 16 42

=1 16

16 = 16 ./

(x - 3) 2 = 16 ( -1 - 3) 2 =1 16

Let x = 7.

( -4) 2

Subtract.

=1 16

16 = 16 ./

True

Let x = - 1. Subtract. True

The solution set is { - 1, 7}.

(x+l) 2 =6

(b)

x+l=\/6 x = - 1+ \/6 NOW TRY ANSWERS 2. (a) { ±5} (b) { ± (c) {

±2Vs}

3. no real solution

v'i3}

or or

x + 1 = -\/6 x = - 1 - \/6

Square root property Add - 1.

' The equation in Example 3 has no solution over the real number system. ln the complex number system, which includes numbers whose squares are negative, this equation does have solutions. Such numbers are di scussed in intermediate and college algebra courses.

SECTION 9.1

i:.NOWTRY

CHECK

~EXERCISE4

Solve (x - 2) 2 = 32, and check the solutions.

(x +1) 2 =6

(x +1) 2 =6

( -1+\/6 +1) 2 ~6

6

=6

The solution set is

~EXERCISE5 Solve (2t - 4) 2 = 50. Check

Let x = - 1 +

(-1 - \/6 + 1)2~ 6 2 ( -\/6) ~ 6

Vs.

Simplify.

./ True

vn

Yn

or

3r - 2

3r - 2 = 3v3

or

3r - 2 = - 3v3

=

3r = 2 + 3v3

or

2+3 v3 3

or

k

=-

Square root property

v27 = v9 . \/3 = 3\13

3r= 2-3v3 r=

Add 2.

2-3v3 3

Divide by 3.

(3r - 2) 2 = 27 (3.

2

+

~ v3 -

2

y~

27

Let r =

( 2 + 3 v3 - 2) ~ 27 2

True

The check of the other value is similar. The solution set is

CAUTION

3

Subtract.

27 = 27 ./

0

2+3\/3 .

Multiply.

~ (3v3) 2 ~27

{2 ±

3 3

V3 }.

NOW TRY~

The solutions in Example 5,

2 + 3\/3 and 3

i:.NOWTRY

6. no real solution

NOW TRY~

(3r - 2) 2 = 27

CHECK

5. { 4 ± ~v'2}

./ True

Solve (3r - 2) 2 = 27. Check the solutions.

r=

NOW TRY ANSWERS

Vs.

Simplify.

1:+!4MQl4j Solving a Quadratic Equation of the Form (ax + b )2 =

3r - 2

4. {2 ± 4v'2}

=6

6

Let x = - 1 -

{-1 - \/6, -1 + \/6}, or {-1 ± \/6}.

the solutions.

~EXERCISES Solve (2x + 1) 2 = - 5.

629

Substitute each value in the original equation.

(\/6)2~ 6

i:.NOWTRY

Solving Quadratic Equations by the Square Root Property

2 - 3\/3 3

,

are fractions that cannot be simplified because 3 is not a common factor in the numerator.

1!$VdHll Recognizing When There Is No Real Solution Solve (x + 3) 2 = -9 . Because the square root of - 9 is not a real number, there is no real solution.

NOW TRY~

630

CHAPTER 9

Quadratic Equations

OBJECTIVE 4 Use formulas involving second-degree variables. f:.NOWTRY

~EXERCISE 7 Use the formula in Example 7 to approximate, to the nearest inch, the length of a bass weighing 2.10 lb and having girth 9 in.

lif4MQlll

Finding the Length of a Bass

The weight w in pounds, length L in inches, and girth g in inches of a bass are related by the formula L 2g w = - -. 1200 Approximate, to the nearest inch, the length of a bass weighing 2.20 lb and having girth 10 in. (Data from Sacramento Bee.)

L2g

w = -1200

Given formula

L 2 • 10 2.20 = - 1200

Substitute given values; w = 2.20, g = 10

2640 = 10L2

Multiply by 1200.

L 2 = 264

Divide by 10. Interchange sides.

±\/264 Square root property A calculator gives \/264 = 16.25, so the length of the bass, to the nearest inch, is 16 in. (We reject -\/264 = - 16.25 here, because L represents length.) NOW TRY g L=

NOW TRY ANSWER 7. 17 in.

9.1 Exercises O Video solutions for select problems available in MyLab Math

0

FOR EXTRA HELP

Mylab Math

1. Concept Check A. x + 2y

=0

2. Concept Check

STUDY SKILLS REMINDER Make study cards to help you learn and remember the material in this chapter. Review Study Skill 5, Using Study Cards.

A.

x2

= 25

Concept Check in Column fl.

Which of the following are quadratic equations? B. x 2

8x + I 6

-

=0

C. 2x2

-

5x

=3

=0

Which quadratic equation is in standard form?

B. 3x2

-

x

=4

C. (x - 5) 2

= 16

D. x 2

-

x - 2

=0

Match each equation in Column I with the correct description of its solution

I

ll

3. x 2

=

12

A. No real solution

4. x 2

=

-9

B. Two integer solutions

25 5. x 2 = -

C. Two irrational solutions

6. x 2

D. Two rational solutions that are not integers

36

=

D. x 3 + x 2 + 4

100

7. Concept Check When a student was asked to solve x 2 = 81, she wrote the solution set incorrectly as { 9}. Her teacher did not give her full credit. The student argued that because 92 = 81, her answer had to be correct. WHAT WENT WRONG? Give the correct solution set. 8. Concept Check

When solving a quadratic equation, a student obtained the solutions 3 +2Vs x = - - - or 2

3-2Vs x=--2

and he wrote the solution set incorrectly as { 3 the correct solution set.

± Vs}.

WHAT WENT WRONG? Give

631

Solvi ng Quadratic Equations by the Square Root Property

SECTION 9.1

Solve using the zero-factor property. See Example 1.

9. x 2

-

x - 56

12. x 2

-

6x

15. x 2

-

169 = 0

18.

5x 2 -

=

10. x 2

0

+5= 0

14x = 3

2x - 99

-

=

11. x 2

0

8x + 15

-

13. x 2 = 121

14. x 2 = 144

16. x 2

17. 3x2

19.

400 = 0

-

6x 2

+ 19x + 10 = 0

20.

8x2

-

=

0

13x = 30

+ 18x + 9 = 0

Solve using the square root property. Simplify all radicals. See Examples 2 and 3. 21. x 2 = 81

22. z2 = 169

23. k 2 = 14

24. m2 = 22

25. t2 = 48

26. x 2 = 54

27. x 2 = -100

28. m 2 = -64

25 29. x 2 = 4

30. m 2

36 121

31. x 2 = 0.25

32. w2 = 0.49

33. x 2

-

64 = 0

34. x 2

-

100 = 0

35. r2

36. x 2

37. x 2

+ 16 = 0

38. x 2

+4= 0

41. 2t2

+ 7 = 61

42. 3x2

45. 7x 2 = 4

= -

+ 8 = 80

46. 2x2 = 9

3= 0

-

13 = 0

-

39. 4x2

-

72 = 0

40. 2x2

-

80 = 0

43. 3x2

-

8 = 64

44. 2x2

-

5 = 35

47. 5x2

+4 = 8

48. 7p2

-

5 = 11

Solve using the square root property. Simplify all radicals. See Examples 4-6. 50. (x - 7) 2

49. (x -3) 2 =25 53. (z

52. (p - 5) 2

27

=

40

59. (5 - 2x) 2 = 30

60. (3 - 2x) 2 = 70

2

2

58. (7z - 5) 2 = 25

+ 1) 2 = 18

62. (5z

~ m + 4 )2 = 27

65. ( x -

67. ( x -

(

=

56. (St+ 3) = 36

61. (3k

70.

51. (x - 8) 2

16

+ 5) = -1 3 54. (m + 2) = -17 55. (3x + 2) = 49 2

57. (4x -3) 2 =9

64. (

=

~)2 = ~

+ 6) 2 = 75

i)2

68. ( x - _l_ 5

= ;4

)2 = ~25

63.

2

(~x + 5 )2 = 12

66. ( x 69. ( x

~)2 = 8\

3 1) 2 = 16 +4

2

x

1) 11 +7 = 49

Extending Skills hundredth. 73. (k

71. (4x - 1) 2

48 = O

72. ( 2x - 5) 2

-

180 = 0

Use a calculator to solve each equation. Round answers to the nearest

+ 2.14) 2 = 5.46

75. (2. l lp

-

74. (r - 3.91) 2 = 9.28

+ 3.42) 2 = 9.58

76. (1.7 1m - 6.20) 2 = 5.41

Solve each problem. See Example 7. 77. An expert marksman can hold a silver dollar at forehead level drop it, draw his gun, and shoot the coin as it passes waist level. The distance traveled by a falling object is given by d

=

16t2 ,

where dis the distance (in feet) the object falls in t seconds. If the coin falls 4 ft, use the formula to find the time that elapses between the dropping of the coin and the shot.

632

CHAPTER 9

Quadratic Equations

78. The illumination produced by a Light source depends on the distance from the source. For a particular light source, this relationship can be expressed as 4050

! =7 , where I is the amount of illumination in footcandles and d is the distance from the light source (in feet). How far from the source is the illumination equal to 50 footcandles? 79. The area Sil of a circle with radius r is given by the formula

Sil

80. The surface area S of a sphere with radius r is given by the formula

= 7rr2 .

S = 47rr2 .

If a circle has area 817r in.2 , what is its

If a sphere has surface area 367r ft 2, what

radius?

is its radius?

G

,, I I I I

I

I I I

r I....,...__ I I I I I I \JI

S =41Tr 2

stJ. = 1Tr2

The amount A that P dollars invested at an annual rate of interest r will grow to in 2 yr is

A = P(l

+ r)2.

81. At what interest rate will $100 grow to $104.04 in 2 yr? 82. At what interest rate will $500 grow to $530.45 in 2 yr?

Solving Quadratic Equations by Completing the Square OBJECTIVES

1 Solve quadratic equations by completing the square when the coefficient of the second-degree term is 1.

2 Solve quadratic equations by completing the square when the coefficient of the second-degree term is not 1. 3 Simplify the terms of an equation before solving.

4 Solve applied problems that require quadratic equations.

OBJECTIVE 1 Solve quadratic equations by completing the square when t he coefficient of the second-degree term is 1. The methods we have studied so far are not enough to solve an equation such as

+ 6x + 7 = 0. If we could write the equation in the form (x + 3) 2 = a constant, we could solve it x2

using the square root property. To do that, we need to have a perfect square trinomial on one side of the equation. Recall that a perfect square trinomial has the form x2

+ 2kx + k 2

x2

or

-

2kx

+ k 2,

where k represents a real number.

lif!MMQlll Creating Perfect Square Trinomials Complete each trinomial so that it is a perfect square. Then factor the trinomial. (a) x 2

+ 8x + ___

The perfect square trinomial will have the form x 2 term, 8x, must equal 2k.x. 8x

= 2kx 2 Simplify each expression. 24.

Peiform the indicated operations. 25. (5x5

27.

3x3

-

+

9x4 + 8x2) - (9x 2 + 8x4 10x2 - 7x + 4 x+4

-

26. (2x - 5)(x3

3x5 )

+ 3x2 -

28. x - 3) 2x2 - 4x

2x - 4)

+1

Factor completely. 29. 16x 3

-

30. 2a2 - Sa - 3

48x2y

32. 25m2

31. 16x4 - 1

33. Solve x 2 + 3x - 54

=

-

20m + 4

0.

34. Answer each question. (a) The number of possible hands in contract bridge is about 6,350,000,000. What is this

number in scientific notation? (b) The body of a 150-lb person contains about 2.3 X 10- 4 lb of copper. What is this number in standard notation?

Peiform the indicated operations. Write answers in lowest terms.

2

5

35. - - - - - a-3 2a - 6

2 3 37. - - + --a2 - 4 a2 - 4a + 4

2 36. - - - k k- 1

I a

I b

a

b

-+38.

l

Solve each equation. 1 1 7 39. - - + - = x+3 x 10

40.

2 t-1

_2_ _ _

_

_

t-1

2

41.

Vx+2 =

x - 4

Simplify each expression. 42.

v'iOo

43.

6v6 Vs

44.

)? \j 16

45.

3Vs - 2\/20 + Vi25

Solve each quadratic equation, using the method indicated. 46. 7 - x 2 = 0 (square root property)

48. -

x2

+5 =

2x (completing the square)

50. Graph the equation y = x 2

-

47. (3x

+ 2) 2 =

12 (square root property)

49. 2x(x - 2 ) - 3 = 0 (quadratic formula)

4x . Give the vertex, they-intercept, and any x-intercepts.

ANSWERS TO SELECTED EXERCISES In this section we provide the answers that we think most students will

71. 0. 16;0.167

obtai n when they work the exercises using the methods explained in the

83. 0.008

text. If your answer does not look exactly like the one given here, it is

95. 600%

73. 0.54 75. 0.07

77. 1.17

79. 0.024

81. 0.0625

not necessarily wrong. In many cases, there are equivalent forms of the

85. 79% 87. 2% 89. 0.4% 91. 128% 93. 40% 51 3 I 7 2 3 97. 100 99. 20 101. 50 103. 5, or l 5 105. 40 107. 80% 109. 14% 111. 18.18% 113. 225% 115. 216.6 %

answer that are correct. For example, if the answer section shows~ and

117. 160

your answer is 0.75, you have obtained the right answer, but written it in

125. $119.25; $675.75

119. 4.8

121. 109.2

123. $ 17.80; $ 106.80

127. 19.76 million, or 19,760,000

129. 5%

a different (yet equivalent) form. Unless the directions specify otherwise,

0.75 is just as valid an answer as~·

THE REAL NUMBER SYSTEM

In gene ral, if your answer does not agree with the one given in the text, see whether it can be transformed into the other form. If it can, then

Section 1.1

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

1. false; 32 = 3 · 3 = 9 3. false; A number raised to the first power is that number, so 3 1 = 3. 5. false; 4 + 3(8 - 2) means 4 + 3(6), which simplifies to 4 + 18, or 22. The common e rror leading to 42 is adding 4

PREALGEBRA REVIEW

to 3 and then multiplying by 6. One must follow the rules for order of

Section R.1

15. 144 17. 64

11. A

13. prime

23. prime

19. neither

21. composite; 3 · 19

25. composite; 2 · 2 · 3 1 27. composite; 2 · 2 · 5 · 5 · 5 .

I

5

3

I

6

29. composite; 2 · 7 · 13 · 19 31. 2 33. 6 35. 5 37. 5 39. 5 5 5 6 13 83 51 24 I 41. l 7 43. 6ii 45. 711 47. 5 49. 8 51. 5 53. 35 55. 20 6 6 I 65 5 38 3 21 57. 25 59. 5, or l 5 61. 9 63. fr or 5ii 65. 5• or 75 67. Z' or I 14 JO I I 102 69. 27 71. 3, or 33 73. 12 75. 16 77. 10 79. 18 35 11 84 37 81. 24, or l 24 83. 47, or 147

11 2 8 29 5 85. i5 87. 3 89. 9 91. 24, or l 24

107 43 3 IOI I 5 2 I 93. 144 95. S' or 5g- 97. W' or 520 99. 9 101. 3 103. 4 J7 67 7 11 32 5 I (b) 105. 36 107. 20• or 320 109. ii 111. 9• or 39 113. (a) 2 (c)

~ (d) ~ 115. 6 cups 117. I ~ in. 119.

123. 5 ~ in.

&in.

3 .

129. 3 8 m.

127. 168 yd

2

4,400,000 133. 25 135.

53. 5

67. 10 - (7 - 3) true

75. 45

81. 3

2'

2'

69. 16 :s 16; true

3; true

13. 49 I 16 25. 36 27. 81

~

121. 6 1 8~ ft

63. 9

65. 3. (6 + 4) . 2

71. 61:s60; false 73. 0

77. 66 > 72; false

79. 2

83. Fi ve is less than seventeen; true

2'

2'

0;

3; false

85. Five is not

87. Seven is greater than or equal to fourteen; false

equal to eight; true

89. Fifteen is less than or equal to fifteen; true 91. One-third is equal 93. Two and five-tenths is greater than two and

to three-tenths; false

95. 15 = 5 + 10 97. 9 > 5 - 4

I

2

101. 2 :5 4 103. 20 > 5

99. 16 ¥- 19

4

59. 90 61. 14

46 ; false

fifty-hundredths; false

. .

. .

131. 4 m1lhon, or 4,000,000; 4.4 m1lhon, or

c

109. (a) 14.7 - 40 · 0.13

(b) 9.5

3

4

105. 4 < 5 107. 1.3

:5

2.5

(c) 8.075; walking (5 mph)

(d ) 14.7 - 55 · 0.1 1; 8.65; 7.3525, swimming

111. Alaska, Texas,

113. Alaska, Texas, California, Virginia, Idaho

Californ ia, Idaho

Section 1.2 1. B 3. A (b) 26

Section R.2 (b) 9

(c) 1

(d) 7

25. 15.2 11 27. 116.48 39. 1232.6

49. 0.02329

5. B 7. The exponent refers only to the 4. The correct 9. (a) 11

15. (a) 64

(b) 13

(b) 144

11. (a) 16 (b) 24 13. (a) 16 5 7 7 13 17. (a) 3 (b) 3 19. (a) 8 (b) ii

21. (a) 52

3. (a) 46.25 (b) 46.2 4 64 138 43 3805 (c) 46 (d) 50 5. TO 7. 100 9. 1000 11. 1000 13. 1000 15. 143.094 17. 25.6 1 19. 15.33 21. 2 1.77 23. 81.7 16

59. ~; 0.5

al, @, C), CD

33. The multiplication should be performed before

55. 4 1 57. 95

value is 80.

37. 2.05

11.

23. 1024

125. 8 cakes (There will be some sugar left over.)

5

1. (a) 6

21. 81

the addition. The correct value of the expression is 14. 35. 32 37. 58 49 19 39. 22.2 41. 12 43. 30, or l 30 45. 13 47. 26 49. 4 51. 42

15. composite; 2 · 3 · 5

17. composite; 2 · 2 · 2 · 2 · 2 · 2

CD, C), ell

9.

19. 1000

29. 0.36 31. 0.064

Product refers to multiplication, so the product of I 0 and 2 is 20.

9. C

CD, al

7.

operations.

1. true 3. false; This is an improper fraction. Its value is I. I . 13 . . . 5. false; The fraction 39 1s wntten m lowest terms as 3· 7. false;

51. 1%

61. ~; 75%

(e) 4

29. 739.53

41. 57 11.6 I

53. 20

31. 0.006

43. 94

33. 7. 15

45. 0. 162

I

55. 122%, or 12.5%

63. 4.2 65. 2.25

67. 0.375

35. 2.8

47. 1.2403

57. 0.25 ; 25% 69.

o.5; 0.556

(b) 11 4 23. (a) 25.836 (b) 38.754 25. (a) 24 9 4 13 27. (a) 12 (b) 33 29. (a) 6 (b) 5 31. (a) 3 (b) 6 2 22 28 33. (a) 3 (b) i5 35. (a) 13 (b) 28 37. (a) I (b) 17 (b) 28

39. (a) 3.684 47. 7 - x 59. yes

(b) 8.841

49. x - 8 61. yes

69.2x + 1 =5;2 75. expression

41. 12x 43. x + 9

51. ~

63. yes

53. 6(x - 4)

65. no

45. x - 2 55. yes

57. no

67. x + 8 = 18; 10

3

71.1 6-4x = 13;4 73. 3x=2x+8;8

77. equation

79. equation

81. 75 yr

83. 78 yr

A-1

A-2

Answers to Selected Exercises

Section 1.3 1. 0

15. 120 5. quotient; denominator 7. S, -S

3. positive

(b) A

(c) B

11. 4

(d) B

There are others.

17. true

13. 0

9. (a) A

15. One example is

19. true

Vi3.

3~

++ I I I l + I l +I • -3.8

0

2

-4

41. (a) 3,7

(b) 0,3,7

(c) -9, 0,3,7

-2 I

3

43. (a) 11

numbers. 7

-

9. -2.3, 0, -84, 11 , - 6 (b) 2

47. (a) -8

51. (a) -S.6 (b) S.6 65. -~

63. - II

(e) v 3, 1T

67. 4

(f) All are real numbers. 3

55. - 12 57. -3 59. 3

61. -3

69. l-3.S I 71. - l -6 1 73. IS - 31

75. true 77. true 79. true

101. 14

-~m

103. undefined

67. - 18

81. -2

83. -1

78

95. -25 97. 0

105. 9 + (-9)(2); -9

9

x

81. false 83. true

85. false

127. x - 6

4; 10

=

3

129. x + S (b) yes

49. (a) 4 (b) 4 2

53. 6

65. 6

73. 18 75. 3 77. 7 79. 4 89. 29 91. 47 93. 72

123. --2-; 16 125. 3 = -3; -9

3

(b) 8

63. - 22

107. -4- 2(- 1)(6); 8 109. 1.S(-3.2) - 9; - 13.8 . - 12 . 15+(-3) . 111. 12 [ 9- ( -8 ) ] , 204 113. - 5 +(- 1)'2 115. ~ ,- 1

(f) All are real

(c) 0, 11, -6 .!::

3

45. (a) 2

(e) - v7, v S

(b) 0, 11

61. 13

111. ~( 8-(- 1 )];6 119. o.20(-s · 6);-6 121. G+~)G-~);k

.t::. . 1-

-

(d) - 9 , - 14, -5, 0, 0. 1, 3, S.9, 7

59. 3S

71. -8 87. -3

99. - S

4

-•i 1 24

39. I •I 1•1 lol 1•1 I•

(d)

85. 4

+l+l+ I I I I ++• -6-4-2

I

23. 6 25. -;:; 27. 6

29. -32, - 16, -8, -4, -2, - 1, I , 2, 4, 8, 16, 32

69. 67

31. - 10,97 1 33. -39.73

-6-4-202

5

21. -2.38

31. - 40, - 20, - 10, -8, -S, - 4, -2, - 1, I , 2, 4, S, 8, 10, 20, 40

55. -2 57. - 2

1 5 3 I 23 .!.!:: 23. 2•8 • 14 25. -32,-3,7 27. VS,7T,-V3 3~

19. 0

33. -3 1, - 1, 1, 31 35. 3 37. -7 39. 8 41. -6 43. -4 32 2 15 45. 3,or 103 47. -16 49. 0 51. unde fined 53. - I I

21. false

In Exercises 23-27, answers will vary.

29. 2,216,602

17. -33

140.

-S; - 10

=

131. (a) yes

135. (a) yes 2

s

(b) no

133. (a) no

137. (a) no

2

141. 85

(b) no

142. 85 143. 4

SUMMARY EXERCISES with Real Numbers

144. 6

(b) yes

139. 42 I

145. 2

146. -122

Performing Operations

87. Natural gas service, March to April 89. Shelter, April to May 1. - 16 9. 38

Section 1.4 3. negative

-3

~ -2 ~

2

-5

I I I I I I I I•

-2

-4

17. -3

19. - 9 47. -8

-2

I

49. 0

37. S 39. 2

51. -20

85. 22 87. -2

37

41. -9

53. -3 55. - 4

I

II

101. (- 19+(-4)]+ 14;-9

[~+(-n]+i;-i

111. ( 9 +(-4) ] - 7; -2 117. +3

119. - 4

-s+

127. 17

129. (a) 3.6%

103. ( - 4 +(- 10) ] + 12;-2

2

7

7

109. -2-8;- 10

113. (8-(-S)]- 12; 1 115. -9

121. - 184 m

123. 120°F

125. - 69°F

(b) Americans spent more money than they 135. $323.83

137. 16.38%

143. l 34S ft

145. 136 ft

141. S0,39S ft

19. 4

15. - 1

20. S

21. - S

I

I

5

s

35. -S6 36. 2 37. -13 38.

39. - 1

40. 0

Section 1.6 1. (a) B (i) G

(b) F

(j) H

(c) C

3. yes

(deposit slip)

(d) I

(e) B

(f ) D, F

(g) B

(h) A

5. no 7. no 9. (large deposit) slip; large

11. Subtraction is not associative. I

13. row 1: -S,

3 8

h

I

I

row 2: 10, -10; row 3: 2• -2; row 4: -8, 3; row S: -x, :;:; row 6: y, - :;; ; 15. - 1S; commutative property 19. 6; associative property

commutative property property

107. 4 -(-8); 12

133. $104S.SS

139. - 12.1 8%

28. undefined

opposite; the same

12 + 6; 13

earned, which means they had to dip into savings or increase borrowing.

23. commutative property

27. associative property

25. associati ve property

29. inverse property

31. inverse property

33. identity property 35. commutative property property 45. 2010

39. identity property 47. 400

59. - 0.38

17. 3; 21. 7; associative

37. distributive

41. distributive property 43. l SO

49. 1400 51. 470 53. -9300

55. II

57. 0

61. 1 63. The student made a sign error in the second line.

The expression - 3( 4 - 6) means -3( 4) - 3(-6) , which simplifies to - 12 + 18, or 6. 65. We must multip ly~ by 1 in the fo rm of a fraction, 3 3 3 9 3; 4 · 3 = 12 67. SS 69. 4t + 12 71. 7z - S6 73. -Sr - 24

Section 1.5

3

4

75. -2x - 4 77. -3x + 3 79. 12x + 10 81. -6x + IS

1. positive

3. negative

.

.

I

I

7. - 18 8. 90

14. 8

24. TO 25. -2,or -32 26. 14 27. 13 52 15 29. - 4 30. 37, or 1 37 31. 0 32. 4 33. -7

I

I

89. )2oor312 91. -4,or - 0.2S

95. - 12 97. -S.891 99.

6. 76

13. 0

19

3

131. $ 19,900

I

17

12. 2S

24

59. - 14 61. 10 63. - 4 65. 4 67. 4 69. -8-, or 3 15 7 - 18 71. 8 • or 18 73. 11.6 75. -9.9 77. 10 79. -S 81. 11

57. -8

105.

5

34. -3

13. - 12

23. -5 25. 2 27.

29. - 4 31. 8.9 33. - 6.01 35. 12 45. -7.7

11. -8 3

21. 0

3

93. -6

I

~

5. -8; -6; 2 7. positive 9. negative

83. - 10

6

4. -24 5. - 17

11. 3.33

22. 4,orl 4 23. 9

~4

I I I I I I•

43. 0

3. 0

16. 5, or 15 17. 16• or 116 18. -3

1. negative

15. 2

2. 4 10. 1.02

.

5. positive 0

7. 0 .

9. undefined; O; Examples

include 0 , which is undefined, and I , which equals 0.

11. -28

13. 30

83. -4.Sx - 0.72

85. - l 6y - 20z 87. 24r + 32s - 40y

89. -24x - 9y - 12z 91. - 41 - 3m 93. Sc+ 4d 95. 3q - Sr + Ss

A-3

Answers to Selected Exercises

106. inverse property

Section 1.7 1. B

3. C

property

S. The student made a sign error when applying the

111. 7y + 14

distributive property: 7x - 2(3 - 2x) means 7x - 2(3) - 2(-2x), which simplifies to 7x - 6 + 4x, or I Ix - 6. 11. 5+2x-6y 21. 3

23. I

13. 32+ 12x I

2S. - I

37. un like 39. like

7. 4r + 11

lS. -7+3p 2

27. 2 29. 5 31. -0.5

41. unlike

9. 2 1x - 28y

17. 2-3x

19. - 12

33. I0

43. I 3y 4S. -9x

47. 13b

Sl. -4y S3. 2x + 6 SS. 14 - 7m S7. 6.Sx - 1.8 28 I S9. 9.Sx 61. -3 - 3t 63. 9y 2 6S. 5p2 - 14p3 67. 8x + 15 79. - 16y+63

87. -2x + 4 9S. 4x - 7

73. - 1+3

7S. Sx+ 15

77. 15-9x

119. -2m + 29

120. 4x- 3

121. -2(3x) - 7x; -13x

S. rational numbers, real numbers 28 13 10. -15. or - I i5

14. 77.6

lS. 11

3 I 25 11. -2· or -l 2 12. 36

16. 161 - 36

lOS. 2(3x + 4) - (-4 + 6x); 12 110. 1750 + Sx + 3y (dollars)

17.

I

3. 64 4. 0.001

2. 125

11. true

[ I. I] 1. true

21. x + 6

22. 8 - x

s. 27

6. 17

17. 30

7. 4

18. 60

19. 14 3

23. 6x - 9

24. 12 + 5x

-2

-4

-2

0

2

41. true

43. (a) 9

42. true 5

(b) 6

44. (a) 0

63. 10.31

(b) 0

4S. (a) -6

5

S9. - II

64. - 12

6S. lO

68. [-4 + (-8)) + 13; I 70. [7 - (-5) ) - 5; 7 74. - 10° 7S. 38 80. 10.08

3

86. - 20 87. -4 92. 0

93. - 18

60. - I

66. - 14.22

71. $26.25

88. 11.3

97. - 4 (5)-9;-29 - 20( 12} 100. 15 -(- 15} ;-8 103. identity property

29. 3x + 2 30. 15x - 3

13

78. - 105

6. 0.0926

[I . I ] 9. I 05 s 65; false

8. $56.85; $322. 15

[1.21 11. no

12. yes

[ 1.3] 13. I

+ l• I

~ It

19. I

-2

11 3 lS.-2.6 1 16. -g-,or - 18 17. -2

[ 1.4,1.5] 14.0

20. undefined

I • l•I I I• 18.12

[ 1.6] 21. inverse property [ 1.7] 23. 31 - 5

24. I 3x - 27

8

LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE

Section 2.1

90. 2

79.

k

1. equation ; expression

3. equivalent equations S. (a) expression;

8S. 4

x + 15 (b) expression; z + 7 (c) equation; { - I }

91.

(d) equation;{-17}

7. A,B

13. { -3}

17. { -9}

96. undefined 5 12 98. t;[ l2+(-6) ); 5 99. 8+(-4};3 x

102. 3= -2;-6

104. identity property

S. 0.015

[ 1.4) 2S. 2698 ft

9S. - 423

101. 8x=-24;-3

[R.2] 4. 26.78

22. associative property

73. - $29

84. - 35

89. undefined

3. 48

-2.5

10. 8 > 2; true

67. (-3 1 + 12) + 19;0

77. 36

[ 1.6] 21. C

[1.7) 27. 21x

62. -35, or-! 35

72. - l0°F

82. - 10 83. -24

94. 125

43

61. 7

17. 3

I

28. -2.81 + 6.7

69. - 4 - (-6); 2

76. 23,44 1.76

81. -20

20. -$0.61 trillion 26. C

-4

40. true

46. (a) 7 (b) 7 47. 12 48. -3 49. - 19 SO. -7 29 Sl. - 6 S2. -4 S3. - 17 S4. -36 SS. - 21.8 S6. 14 S7. - 10 S8. - 19

[ 1.5] 16. -70

2S. B

[R. l , R.2) 7. 4± in.

34. irrational numbers, real

(b) 9

lS. II

10. 2

Chapters R and 1 Cumulative Review Exercises

33. natural numbers, whole numbers, 3

36. -9

14. 7, or47 19. 15

24. E

5

37. - 4 38. - I23 I 39. true

numbers 3S. - I 0

9. -6· or -26 2

[1.3- 1.5] 18. 7000 m 23. A

6. - 1.277 5

22. D

-2

4

32. rational numbers, real numbers

4

17

[1.4, 1.5] 8. 4 30

2

I-8 I, or -8

2.5

31. rational numbers, real numbers

integers, rational numbers, real numbers

0

S. -

.

13. 8

12. - 20

[R. l ] 1. 28 2. 6

!

~ 2.25 I + l• I I •I lo+ I •

30.

26. yes

+ I I I•+ 11 • 11 +•

29.

-4 -1

11. 6

20. 13

2S. no

I

28. 4x=8;2

8. 399

14. 13 < 17

2

27. 2x - 6 = 10; 8

18. 24

11.3] 3. + I + I + I I + •

2. false

[ 1.5] 7. 2 + (- 8); I

13. false

12. true

16. ~ ;:,: ~

lS. 5 + 2 ¥- I 0

13. 16

2ly 2

Chapter 1 Test

-6

Chapter 1 Review Exercises

10. 5

8x2 -

19. -47°F 20. 14,776 ft

4. rational numbers, real numbers

1. 625

24

108. 750 + 3y (dollars)

109. IOOO + Sx + 750 + 3y (dollars)

9. 4

I

6. 37

101. (x+3)+5x;6x+3

103. ( 13 + 6x) - (-7x); 13 + 13x

27

114. 4r - Ss

118. -4k + 12

Chapter 1 Mixed Review Exercises

9. 2

97. (4x + 8) + (3x - 2); 7x + 6

107. 1000 + Sx (dollars)

113. 6s + !Sy

117. 16p2 + 2p

122. (5 + 4x) + 8x; 5 + 12x

3

81. 4r+ 15 83. 12k-5 8S. -2y+ 16 14 22 89. - 3 x - 3 91. -23.7y - 12.6 93. -2k - 3

99. (5x+ 1)-(x-7);4x+8

112. -48 + 121

116. 16p2

llS. I I m

108. associative

110. commutative property

3S. like

49. 7k + 15

69. 22-4y 71. - 19p + 16

107. associative property

109. distributive property

lOS. inverse property

23.

{-H

33. { 10.1} 43. {-4}

lS. { 4} 2s.

{J~} 21.

3S.

{-ts}

4S. {-6}

9. { 12}

{ -10}

37. {-3} 47. {-5}

11. {31 }

19. { 6.3}

21. { - 16.9}

29. { - 13 } 39. {2}

31. { 10}

41. {7}

49. {-2}

Sl. {3}

A-4

Answers to Selected Exercises

S3. { - 3}

SS. {O}

S7. {2}

S9. {- 16}

61. {O}

63. {9 }

14. { 12}

6S. {-7}

67. { 2}

69. {1 3}

71. { - 4}

73. {O}

7S. {O }

19. {all real numbers}

{M 79. { 13}

81. {-2}

83. {7}

11.

87. { 13}

89. {29}

91. { 18}

23. {7}

8S. { - 4 }

93. { 12}

9S. Answers will vary. One example is x - 6

-8.

=

1. (a) multiplication property of equality

(b) addition property

(c) multiplication property of equality

(d) addition

property of equality 3. B S. ~ 7. 10 9. -~ 11. - I

23.

13. 6

{¥}

lS. - 4

17. 0. 12

29. { -1!},or {-3.6}

33. {O}

3S. { - 12}

37.

49. {-48}

{-m

69. {o}

21. {6}

27. { -4 }

41. {-30 } 43. {-2.4}

S9.

19. - I

2S. { - 5}

31. { 12}

SI. {72} 61. {3}

11.

17. {-6} 21. { 6}

26. 0

2S. {5}

29. { 25}

30. {all real numbers}

33. {-2}

34.

18. { - 16} 22. {0}

27. 0

28. {- 10.8}

31. { 3}

32. { I }

H*}

4S. {3.5 } S3. { -35}

63. { - 5}

{-H

9. x

11. 8(x + 6)

=

104; 7

=

lS. 3(x - 2)

=

x

47. {- 12.2}

Republicans: 18

SS. {14}

S7. {18}

- 11 - 2x; -3

=

4x

21. Step 1: Republicans; =

50; Democrats: 32;

23. New York: 28 screens; Ohio: 24 screens

2S. Democrats: 46; Republicans: 52 27. Beyonce: $ 169 million; 97 mg; pineapple: 25 mg

11. {-3}

+ 9 = -26; -35

+ 5; 3 3 + 6; 6 17. 4x + 6 = x - 4; 40

Guns N' Roses: $ 131 million

67. {7}

1s. {-6}

13. 5x + 2

+ 14; Democrats; Step 3: (x + 14) + x

Step 2: x

3. A; Distance

S. I ; 16 (or 14); -7 (or -9); x + I

7. complementary; supplementary

19. 3x + (x + 7)

39. {40}

6S. {20}

13. { 18 }

1. C; There cannot be a fractional number of cars. cannot be negative.

m

81. Answers will vary. One example is ~x = - 6.

79. { - 4 }

24. { I }

16. {-6 } 20. { 23.7 }

Section 2.5

Section 2.2 of equality

lS. {-25}

grilled steak: 536.3 kg

29. wins: 67; losses: 15 31. orange:

33. 112 DVDs 3S. onions: 81.3 kg;

37. 1950 Denver nickel: $ 16.00; 1945

Philadelphia nickel: $8.00 39. whole wheat: 25.6 oz; rye: 6.4 oz 41. American: 18 tickets; United: 11 tickets; Southwest: 26 tickets

Section 2.3

43. shortest piece: 15 in. ; middle piece: 20 in.; longest piece: 24 in.

1. Use the addition property of equality to subtract 8 from each side.

4S. gold: 46; silver: 37; bronze: 38

3. Clear parentheses by using the distributive property.

B: 40°; C: 100°

S. (a) identity; B

S9. 10, 11 61. 18 63. 1s, n , 19 6S. 18° 67. 20° 69. 39° 11. so·

(b) conditional; A

1. {4} 9. {-s}

11.

17. {-3}

19. {5}

{-H

31. {O}

29.

39. {5}

41. {8}

(c) contradiction; C

m

13. {4}

21. {O}

23. {- ! } 2S.

m m 27.

3S. 0

37. {O}

43. {O } 4S. {all real numbers} SI. 0

S9. x + 15 ;x - 5

61. 25r 63.

S3. 11 - q

SS.

2.x

47. 0 S7. 65 - h

t

5

boundary of the fig ure.

17. 10 19. 250

63.

U}

11. I

21. 9 23. 480 2S. A

33. { - 32}

41. {all real numbers} S3. {8} 6S. {20}

73. {- 15}

3S. {5}

67. 0

13. 72

lS. 7

27. { 14}

37. { 12}

29. {6}

39. 0

4S. { II } 47. {4}

43. {9} SS. { I }

7S. {120}

=

26

13. Sil = 64

21. r

=

2.6

23. r

v=

1. equation; { - 5}

=

1.5 yr

10

=

v=

52 33.

lS. b

4

=

17. t

5.6

=

2S. Sil = 50.24 27. r

=

41. length: 18 in.; width: 9 in.

9. area

19. B 6

7234.56 3S. 1 = $600 37. p

14

=

29. V =

=

150

$550

43. length: 14 m;

S7. {-2}

69. {2 }

77. {6}

6.

SS. 194.48 fti; 49.42 ft

49. {O}

S9. { - 10 } 61. {3}

3. expression; -3m - 2

equation;{ -~}

8. equation; { 3}

10. expression; 2k - 11

11.

12. { 4 }

9. expression; - ~x + 5 13. { - 5.1}

S3. 154,000 fti

S7. 23,800. 10 ft2 S9. length: 36 in.; volume: 11,664 in.3

61. 48°, 132°

69.s1 °, s1 ° 11.1os , 1os 0

We give one possible answer for Exercises 73-109. There are other correct forms.

71. {all real numbers}

7. expression; - !Oz - 6

{fl

49. perimeter: 5.4 m;

area: 1.8 m2 SI. 10 ft 0

79. { 15,000}

2. expression; 7p - 14 S. equation; {O}

47. two equal sides: 7 m; third side: 10 m

63.70°, 11 0° 6s. ss ,35° 67.30°,60°

SUMMARY EXERCISES Applying Methods for Solving Linear Equations

4. equation; {7}

3. area S. perimeter 7. area

11. P

width: 4 m 4S. shortest: 5 in.; medium: 7 in.; longest: 8 in.

3. 6 S. 12 7. 30 9. 35

SI. { 18}

S7. 17, 19

1. The perimeter of a plane geometric figure is the measure of the outer

31.

Section 2.4

31. {-2}

SS. 10, 12

Section 2.6

39. t

1. 8

47. 36 million mi 49. A and

S3. 10 1, 102

{-D

ls.

33. {all real numbers}

49. {all real numbers}

SI. 68, 69

d

73. t

= -;

83. h

=

7S. b

2Sll

b

=

2S - a - c

lOS. y = 5x - 2

111. W1 + M

77. d

= Wi

3V

C = -:;;

79. H

=

V LW

81. r

=

PtI

P - 2L 2 C-Ax M -C P -2b 93. y= - B - 9S. r = - c97. a= - 2

8S. h

y- b 91. m= - x -

99. b

Sil

h

=

87. b

= 7Tri

=

89. W

= -

9C + 160 103. y = -6x + 4 5 3 I 5x - 3 109. y = 3x - 4

101. F

107. y

= P - a - c

=

+ Ni + I 112.

M = W2

+ Ni + I - W1

0

Answers to Selected Exercises

113. (a) 17

(b) 6

114. M = -3; A negative magic number

(c) 5

21.

23. I I I I I [1 I I >

I I I I [ I I I I > 0 I

indicates that Oakland has been eliminated from winning the division.

5

( 1, oo) -11

1. compare; A, D

18

15

13. 55

3. (a) C

7. 75 to 100;

0.01

(b) D

75 ; 100

5. 1 to 100;

(d) A 100 ; 100

9. 100 to 100;

0.75

4

(c) B

4

1 11. 3

5

17. 15 19. 6 21. 10 lb; $0.749 23. 64 oz; $0.047

15. 2

25. 32 oz; $0.53 1 27. 32 oz; $0.056 29. 263 oz; $0.076 31. true 33. false

35. true

45. {2}

47. {- 1}

57. $8.75 75. $352. 17

39. {7}

37. {35} 49. {5 }

59. $67.50 67. 4 ft

(b) yes

51.

103. 28%

69. 2.7 in.

{ - ~} 53.

43. { 1}

{-2}

~ r:i!~ 4

Shadow

Shadow

105. 32%

107. $3000

{¥}

(b)

31. E 11111111) 0

109. $15.00

{¥}

33. I I

[1

(- 10,oo)

1)

I I I I•

37. El

j

I

I 5

I

E I

)

1j

I I I I I •

45. I I I Ei

[1

I I I I

)

( -5, oo)

( - oo, 2 J

012

I•

5. D

7. C

I

.

I

21. 4 L

27. $2100 at 5%; $900 at 4%

23. 20 mL

41. 3.001 hr

29. $2500 at 6%; $13,500 at 5%

59.

45. 8.38 m per sec

[1

I

)

I

j

s (or s,

2:)

I

)

0

8

32

11. E I I

I I•

4

5. x > - 4

I'

65. El I I

1)

I I I I I•

-3

( -oo, 4 J

18

I I

-,

I I I

EI I

)

15.

I I I

[1

-1

0

4

(4, oo) 17. EI I I

I

I I I '

69.

0

( -oo, OJ

I I I•

19.

I

(1 -2

( - 2, oo)

2:

-2 1

I '

-7

( - 2 1, oo)

73. x > 5 75. x s 20 77. 83 or more

18

91. 5x - 100

99.

83. all numbers greater than 16

93. (5x - 100) - ( 125

[

j

I

8

95. - 1 < x < 2 I '

JO

103.

(-3,4)

101. I

E

j

I '

JO

5

(0, lOJ

I Ei I I I I -3

+ 4x), which simplifies to

97. - 1 < x s 2

0

(8, lOJ

I I I•

I I E i I

85. Itisneverlessthan - 13°F. 87. 32orgreater 89. 12min

I I I'

[- ~, oo)

j

I•

6

79. 80 or more 81. more than 3.8 in.

I

13.

1j

( -oo, 6 J

x - 225; x > 225

( -oo, - 3 )

I I

i2

67.

7. x S 4

)

5

0

3. (0, oo)

I

( 2, oo)

63. E I

71. x 2: ,

)

2

[Tz· 00) 1. >,< (or );

[1

61. I I I

( -oo, 32)

Section 2.9

I I I

0 I

( 4, oo)

65. width: 3 ft; length: 9 ft

Kevin: 2 1 yr old

1Q

(- ~. 00)

I I I

55. slower car: 40 mph; faster car: 60 mph

63. Bob: 7 yr old;

I I I

4

47. 7 2 hr 49. 5 hr 51. 14 hr 53. eastbound: 300 mph; 57. 75 hr 59. $2280 61. $97.54

57.

( -oo, OJ

25. 4 L

43. 9. 18 m per sec

I I I'

0

3

westbound: 450 mph

1j

55. El I

37. Arabian Mocha: 7 lb; Colombian Decaf: 3.5 lb

I

-, I

31. $1700 at 8%; $800 at 2% 33. 10 nickels 35. 49-cent stamps: 25; 2 1-cent stamps: 20

I I I•

I

( -oo, 1)

15. $ 17.50

13. $750

0

( -oo, ~)

9. Because the rate is given in

11. 45L

19. 133 L

9. EI I I I

1)

53. El I

I

45· 2= 2, or222m1.

39. 530 mi

I I•

2

3

miles per hour, the time must be in hours. So the distance is

17. 160 L

1j

49. El I I

0

-5

Section 2.8

45

)

( - 1, oo)

2

I

I I I

-1 0

51. E I I I 3. C

)

( - 3, oo)

( -oo,- 3 J

114. Bothmethodsgivethe

I

-3

-3

47. I

I•

[1

41. I I

20

43. EI I

I

(-oo, OJ

(20, 00)

101. 80%

111. 30

I 0

I

0

( -oo,- 3)

39.

)

0 2

-3

same solution set.

1. A

I I I I I I

-JO

( -oo, 6)

(:

99. 120%

11 11•

6

35. EI I I I

;;~~~,~~~;:: ,::8:L

97. 8%

113.

The second and third lines of the solution should have > symbols (instead of < )so that the last line is x > - 3. The solution set is ( - 3, oo).

65. (a) 30 min

91. 140.4 93. 700 95. 425

18

29. When each side of an inequality is divided by

by a negative number.

a negative number, the direction of the inequality symbol must be reversed.

81. x=22.5;y=25.5

:

-2 0

27. The inequality symbol must be reversed when multiplying or dividing

55. $30.00

73. 2~ cups

71. 2.0 in.

77. x=4 79. x=8

112. (a) 5x= 12

{T}

41.

61. $56.40 63. 50,000 fish

~-(•)a ,;~ ..

( 5, oo)

25. E I I ~ I I I I I I • ( -oo, - 11)

Section 2.7 I ; 100

A-5

0

1) 4

I•

105. I I Ei -2

(- 2, l J

1j 0

I

I •

A-6

Answers to Selected Exercises

107.

I [ 1I I

109.

I 1] I •

I

-1 0

I

(

0

I

)

I •

111. I I I [ I 0

113.

I) I

3

EI

I

7

[3, 7)

0

10

65.

-3

0

1[1

117.

6

-5

[-3, 6] 119.

J I I• 0

0

I

I]

I•

2

(~. 5] I

67. 88 or more

68. all numbers less than or equal to -3

0

Chapter 2 Mixed Review Exercises

-¥ - j 1( 1)

121.

I I I) •

-6

2

I•

-2

-~) ( _ !.!_ 6' 3

(-6, 2)

+I

I I I

I• 2

1(1

[-¥.o]

(1 I I

123.

I I I

1]

I

66.

[- 2.~]

24

I [1 I I I I I I 1] I •

I•

3

I

I

-2

[ -26, 6 ] -5

115.

2

[

I

I

4

3

I I I I] I •

I

(-oo, - 4)

6

-30 - 20 -10

I

-4

0

(-oo, - 5)

-26

)

64. EI

I I I I I I•

-5

( I, 3)

[- 1,6 ]

1)

63. EI I I

I I I•

7. 0

I I I ( I I I I )

124.

1 23 45 678

I 2 3 4 5 6 7 8

{4}

( 4 , oo)

125. EI I I )

1. {7 }

3. (-oo,2)

8. {all real numbers}

4. { - 9}

12. 24°, 66°

13. 13 hr

5. {70}

{143 }

6.

10. DiGiorno:

9. 4000 calories

11. 160 oz; $0.062

$ 10 14.6 million; Red Baron: $572.3 million 15. 44 m

I•

r=p;I

2.

14. fas ter train: 80 mph; slower train: 50 mph

16. 50 m or less

126. The graph would be the set

1 234567

of all real numbers.

(-oo, 4)

Chapter 2 Test [2. 1- 2.4] 1. {-6}

Chapter 2 Review Exercises 1. { 6}

2. { - 12}

7. {5}

8. {-4 }

3. {7} 9. {5 }

13. {all real numbers} 18. {- I }

17. 0

m

6.

5. {I I }

10. {- 12 }

14. {- 19}

6. { 17 }

11. {15}

15. {O}

12. {4}

16. {20}

19. -~ 20. Democrats: 67; Republicans: 51

mi2 ;

21. Hawaii: 6425

4.

Rhode Island: 12 12 mi2 22. Seven Falls:

300 ft; Twin Falls: 120 ft

23. oil: 4 oz; gasoline: 128 oz

24. 11, 13

25. shortest piece: 13 in.; middle-sized piece: 20 in.; longest piece: 39 in. 27. h = 11 28. s!l = 28

26. 80° 31. 2 cm

32. 42.2°; 92.8°

s!l 35. h = b

34. 100° ; 100°

42.

38. y = 2x-6

m -n {m 43. {

44.

47. 8 gold medals 51. 3.75 L

728 mi2 ; Kauai: 55 1 mi2

9. 50°

P- 2L [2.6) 11. (a) W = - 2

correct forms.) 16. {-29 }

(b) 18

13. 100°, 80°

17. 40%

5

12. y = 4x - 2

14. 75°, 75°

18. $264

~~ 21 00x < O

5

3

25.

3

48. 18 oz; $0.249

II

[

I

I I)

-3

II

0

[ -3,oo)

I

]

I•

27.

I ( 1 I I I I I 1] I • -2

(-oo, 4]

5

45. $3.06

I

20. 2300 mi

22. 4 hr

-2 0 or x > 0 and y < 0. If x < 0 and y > 0, then the point lies in quadrant II. If x > 0 and y < 0, negative

II . If X)'

then the point lies in quadrant IV.

13. between 2009 and 20 I0

15. 20 11: 9%; 20 12: 8%; decline: I% 23. yes

25. yes

27. no

29. 17

37. 8; 6; 3; (0, 8); (6, O); (3, 4 )

17. yes

33. - 4

21. no 35. - 7

17. (8, O); (0, - 8)

39. - 9; 4; 9; (-9, 0 ); (0, 4 ); (9, 8)

41. 12;1 2; 12; ( 12, 3 ); ( 12, 8);( 12, 0 ) (4, - 10) ; (0, - 10); (-4, - 10 )

19. yes

31. - 5

(0, 4 )

43. - 10; - 10;- 10;

21. (0, O); (0, 0 )

27. (0, 0); (0,0)

23. (2, O);

29. (4,0); noy-intercept

31. no x-intercept; (0, 2.5)

45. 2; 2; 2; (9, 2); (2, 2 ); (0, 2)

47. - 4; - 4; - 4; ( - 4, 4) ; (-4, O); (-4 , - 4 )

19. (4, O); (0, - 10 )

25. (6, 0 ); (0, - 2 )

33.

49. The ordered pair

(3, 4) represents the point 3 units to the right of the origin and 4 units up from the x-axis. The orde red pair (4, 3) represents the point 4 units to the right of the origin and 3 un its up fro m the x-axis. 51. (2, 4 ) ; I

53. (-5, 4); ll

55. (3, 0); no quadrant

59.-69.

4

71. - 3; 6; - 2; 4 ++ y

o -2

°-2 o x =

-

(0

2)~

5, 0

0 (

2

s

++ 0 ~ 63

73. - 3; 4; - 6; - 3

Y-

2x-5y = I0 +

39. •

ti

~7

I

43.

41.

57. (4, - 4); IV

-

rJ l(';..o}.

x

)

lr+7y = 14

~

f

65

49. 47 . •

2 4

1

,0

0 0

x

75. - 4; - 4; - 4; - 4

77. The points in each graph appear to lie on

-2x= Ox

- o.-

a straight li ne.

79. (a ) (5, 95 )

(b) (6, 110 ) 55. . .

- 5 53. • 0

5 0

-2 0

x

0

x

81. (a) (20 10, 608), (2011, 845), (2012, 1056), (201 3, 1228), (20 14, 1393), (20 15, 159 1), (20 16, 1860 )

( b ) (201 6, 1860)means thattheworld-

wide number of monthly active Facebook users was 1860 million at the

(c)

Y

j l 'E i;

~

~:1

1600 ... "' 1400 1200

~ :E,1000

...

· ' •

800

0

· .. . • .

~

59 . •

end o f the year 20 16.

•

•

61. ·Y-=

.

o. 2

X j

0

6,0

x

(d) The points lie approxi mately in a linear pattern. T he worldwide number

• ..

• · ..

of monthly active Facebook users was

In Exercises 63-69, descriptions may vary.

steadi ly increasing.

63. The graph is a line with x-intercept (-3, 0 ) and y-intercept (0, 9). 65. The graph is a vertical line with x-intercept ( 11 , 0 ). graph is a horizontal line with y-intercept (0, - 2 ) .

600

67. The

69. The graph has

01· '----'~~~-'+-

x- and y-intercepts (0, 0 ). It passes through the points (2, I) and (4, 2).

2010 2012 2014 2016

Year

71.

x= 3

73.

Y~

83. (a) 130; 11 7; 104; 9 1 (b) (20, 130), (40, 11 7), (60, 104), (80, 9 1) (c)

yes

y

minute; between I 17 and I 53 beats

- 200

~

i

.s

85. between 130 and 170 beats per pe r mi nute

ISO 100

so Age

Jl 12 1-1

49. y

=

23. •

I

25. ' • Ix S -y

y~~ '

'

' ' x

,

2.25x - 0.45

51. y = 0.2375x + 59.7

dX +

27. •

Y..

31.

s

S

29. •

y

_xi- 2j.:4

I 0

9

0

!;

0

1

~l.t

9

54. 5 55. F - 32 =5( C - O)

57. C = ~(F - 32)

xl

x>4

Year

59. 10°

I ~

O· I

0

012345

53. (0, 32);(100, 2 12)

11. (a) no

47. (a) ( 1, 33 10), (2, 3340),

(d) $3580 (x = 6)

···+···f··-+··+···1

56. F = ~ C + 32

+
3.

(d) no

(b) 9x +8y=- ll 1

29. (a) y = -3x +9 (b) 3x+9y =22 (b) x - 3y = - 4

11

25. (a) y = - 5x - 5

9

1.

line 3x - 4y = 12, which is not part of the graph because the symbol
- 6

58. 86°

33.

cs{

:1

60. - 40° 0-

x

C!::

2x +I

37. .... .y .

35. • 2

XI

0

0

x

2-0 ~

,-

x >:J=!+

S. (a) 15

7S. -8x6y 3 77. 9a6b4

c

7

2

I

67. 8

79. (a) negative (b) positive (c) positive (d ) negative 95 55 8S. 2 12x 12 87. 65x 10y1 5 89. x 21 91. 4w4x26y7 83. 81.

103. 6p7 22.

98 SS

6S.

97. (a) false (b) true (c) false 9S. -c-.18(d) true 99. Using power ru le (a), multiply the exponents. The base stays the same. Simplify as follows: ( 10 2)3 = I0 2·3 = 106. 101. I2x5

17. Magic Kingdom: 20.4 million visitors; Disneyla nd:

11 76 mi

18.0 million visitors

7

73. -

93. - r 18s 17

[4.4] 16. Memphis and Atlanta: 394 mi; M inneapolis and Houston:

63. Sq r

32x5

x3 71. 64

a3

J;3

69.

3 3

61. -5 5

4.

3. 0

61

900

11. 100,000x7y 14 I

22

l S. c

16. k4112

(e) J

S. I

(h) B

(g) I

(t) F

6. - I

7

8. 24

7. 0

729w 3x9 12. - 128a 1°b 15c4 13. --12y

17. y 12z3

x6

I

19. 2

18. 5 y

z

a6 y" 3 300x3 23. x 8 24. ~ 2S. /j 26. 6ab 27. I 22. 6 5x y q xl5 27y 18 343a6b 9 30. a8b12c16 31. 2 l 6z9 32. 8p6r3 28. - - 8 - 29. 4x8

21.

-3-

9 343 66 33. x y 34. - 1-5 3S. 6 x x I 39. 8p 10q 40. - 3 - 3 mnp

10 9 36. 5p q

37.

r 14 t -

2s

2

38.

Section 5.3

27. base: - 6x; exponent: 4 29. base: x; exponent: 4 31. 2; 5; 87 33. 5 8 3S. 4 12 37. 124 39. - 56r7 41. 42p 10

3. in scientific notation S. not in scientific notation; 5.6 X 106 7. not in scientific 3 10 1 9. not in scientific notation; 4 notation; 8

43. -30x9

11. (a) 6; 4; 6.3; 4

4S. - 30x 11

47. The product ru le does not apply. 49. The product rule does not apply. SL 4 6 S3. 120 SS. 73r 3

1. (a) C

(b) A

(c) B

(d ) D

X10-

X

(b) 5; 2; 5.7 1; -2

13. 5.876 X 109

9

20· ,-----2 r-s t 10

A-14

Answers to Selected Exercises 17. 7 x 10- 6 19. 2.03 x 10- 3 21. - 1.3 x 107

15. 8.23S x 104 23. - 6 x 10-3

25. ?S0,000

27. S,677,000,000,000

29. 1,000,000,000,000 31. 6.21 33. 0.00078 35. O.OOOOOOOOS 134 37. -0.004 39. -8 10,000 41. 0.000002 43. 4.2 x 1042 45. (a) 6 x 10 11

47. (a) l.S x 107

(b) 600,000,000,000

(b) IS,000,000 49. (a) -6 X 104 (b) -60,000 51. (a) 2.4 x 102 (b) 240 53. (a) 6.3 X 10- 2 (b) 0.063

55. (a) 3 x 10- 4

(b) 0.0003

59. (a) 1.3 x 10- 5

57. (a) - 4 X 10

(b) - 40 61. (a) S X 102 (b) SOO

(b) 0.00001 3

65. (a) 2 X 10- 7

63. (a) -3 x 106 (b) -3,000,000 67. 4.7EI

69. 2E7

71. IEI

73. 1.04

X 108

(b) 0.0000002

75. 9.2

X

10- 3

77. 6 x 109 79. l.S x 10 17 mi 81. $3064 83. $60,736 85. 3.S9 X 102, or 3S9 sec 87. $ 109.02 89. S.S80876 X 10 18; 3.348S2S6 x 1020 91. The Valdivia earthquake was 10 times as intense as the Southern Sumatra earthquake.

92. The Maule earthquake was

10 times as intense as the Gorkha earthquake. 93. The Valdivia earthquake was I00 times as intense as the Farkhar earthquake. 94. The Maule earthquake was I00,000 times as intense as the Mooreland earthquake. 95. The Gorkha earthquake was 10,000 times as intense as the Mooreland earthquake.

Section 5.4 I. 4;6

3. 9 5. 19; 19

- I; two

15. I, 8, S; three

31.

-Htu

35. 111114

-

5

17. 2111

11. I; one

13. - 19,

19. -,.S

25. Sp9 + 4p1

simplified. 23. -Sx5 1

7. 0 9. 6;one

21. It cannot be 27. 0 29. - 2y2

33. already simplified; 4; binomial 71113 - 31112; 4; trinomial 37. ..tA; 4; monomial

39. 7; O; monomial 41. - I 3ab; 2; monomial 43. The first term should be evaluated as -(-2) 2 = - 4, not 4, The correct answer is -6. 45. (a) 19

(b) -2

47. (a) 14

(b) - 19 7

49. (a) 36 2

69. - l lx2 - 3x - 3 71. a 4 - a 2 + I 2 73. Sm + 8m- IO 75. -6x2 - 12x+ l2 77. - 10 79.4b-Sc 81. 6x - ..ry - 7 83. c 4d - Sc2d2 + d 2 85. 8x2 + 8x + 6 87. 2r2 + 8x 89. 812 + 81 + 13 91. (a) 23y +St (b) 2S 0 , 67°, 88°

93. -?x - I

95. 0, -3, -4, -3, 0

97. 7, I, - 1, 1, 7 ~Y

1. (a) B

(b) D

7. 1Sy 11

9. 30a9

(c) A

(d) C 3. distributive 5. one 11. 1Spq2 13. - 181113112 15. 9y 10 17. -8x 10 19. 61112 + 4m 21. - 6p4 + 12p3 23. - 16z 2 - 24z 3 - 24z 4 25. 6y 3 + 4y 4 + l0y1 27. 28r5 - 32r 4 + 36r3 29. 6a 4 - l 2a 3b + I Sa2b 2 31. 31112; 2m11; -113; 2 Im5n2 + 14m4113 - 7m3n5

33. t2x3 + 26x2 + !Ox+ I 35. 72y 3 - 70y2 + 21y - 2 - m 3 - 8m 2 - 17111 - IS

37. 20m 4

- 4x4 + 4x3 - Sx2 + 8x - 3 3 41. Sx4 - 13x + 20x2 + ?x + S 43. 3x5 + 18x4 - 2x3 - 8x2 + 24x 45. first row: x2, 4x; second row: 3x, 12; Product: x 2 + ?x + 12 47. first row: 2x3, 6x 2, 4x; second row: x 2, 3x, 2; Product:

39. 6x6

-

3x5

2x3 + 1x2 + ?x + 2

49. (a) 2p; 3p; 6p2 (b) 2p; 7; 14p (c) -S ; 3p;

- ISp (d ) -S; 7; -3S (e) 6p2 - p - 3S 51. m 2 + 12m + 3S 53. 112 + 311 - 4 55. I2x2 + !Ox - 12 57. 8 1 - 12 59. 9x2 - 12x + 4 61. 10a2 + 37a + 7 63. 12 +Sm - 1Sm2 65. 20-7x-3x2 67. 312 + Ss1- 12s2 69. 8..ry-4x+6y - 3 71. 1Sx2 + ..ry - 6y 2 73. 6y 5 - 2 ly 4 - 4Sy3 75. -200r7 + 32r3 77. -3r 4 + 3r 79. (a) 3y2 + lOy + 7 (b) 8y + 16 81. x2 + l4x + 49 83. a 2 - 16 85. 4p2 - 20p + 2S 87. 2Sk 2 + 30kq + 9q2 89. m 3 - 1Sm2 + ?Sm - 12S 91. 8a3 + 12a2 + 6a + 1 93. -9a 3 + 33a2 + 12a 95. S61112 - 14111 - 21

97. 8 lr 4

-

2 16r3s + 2 16r 2s2

99. 6p8 + lSp 1 + l2p6 + 36p5 + 1Sp4 101. -24x8 - 28x1 + 32x6 + 20x5 103. 6p4 105. 14x + 49 107. 7TX2 - 9 109. 30x + 60

110. 30x + 60

=

600; { 18}

111. 10 ft by 60 ft

96rs3 + l6s4

-

5 2 25 zP CJ -

12CJ2

112. 140 ft

113. $4SO 114. $2870

(b) - 12

5

51. Sx2 - 2x 53. Sm2 + 3111 + 2 55. 6x2 - 15x + 6 57. 61113 + 1112 + 4111 - 14 59. 3y 3 - I Iy2 61. 4x4 - 4x2 + 4x 63. ISm3 - I3m2 + 8111 + I I 65. Sm2 - l 4m + 6 67. 4x3 + 2x2 + Sx

Section 5.5

Section 5.6 I. The student neglected to include the term 2ab. The correct answer is

+ 2ab + b2. 3. (a) 4x; 16x2 (b) 4x; 3; 24x (c) 3; 9 (d) 16x2 + 24x + 9 5. 1112 + 4111 + 4 7. r 2 - 6r + 9 9. x2 + 4..ry + 4y 2 11. 2Sp2 + 20pq + 4q2 13. 16x2 - 24x + 9 16 ' 15. 16a2 + 40ab + 2Sb2 17. 361112 - 48 511111 + 25n-

02

19. ~x2 +ix+~ 21. 2x2 + 24x + 72 23. 913 - 612 + t 25. 48t3 + 2412 + 31 27. - I6r2 + l6r - 4 29. k2 - 2S 31. 16 - 912 33. 2Sx2 - 4 35. 2Sy2 - 9x2 37. 100x2 - 9y2 39. 4x 4 - 2S 41. ~ - x 2 43. 8 Iy2 - ~ 45. 2Sq3 - q 47. - Sa2 + Sb6 49. 2k 2 - ~ 51. -x2 + 1 53. x 3 + 3x2 + 3x + I 55. 13 - 912 + 271 - 27 57. r 3 + 1Sr2 +?Sr+ 12S

~- 2x

101. 4, I, 0, I, 4

99. 0, 3, 4, 3, 0 ::) 4

- I

~

4

2l6r3

2

3

2

-

61. 2S6x4 - 2S6x3 + 96x2 - l 6x + I 2 63. 8 Ir t + 2 16r t - 96rt 3 + 1614 4 3 65. 2x + 6x + 6x2 + 2x 67. -414 - 3613 - 10812 - 1081 59. 8a3 + l 2a 2 + 6a + I

69. x4 - 2x2y2 + y 4 71. 9999 73. 39,999 75. 3994 77. i 1112 - 2112 79. 9a2 - 4 81. '1TX2 + 4'1Tx + 4'1T 83. x3 + 6x2 + 12r + 8 85. (a+ b) 2 86. a 2 87. 2ab

0

Y=~2 + 4

103. S; 17S

104. 9; 63

105. 6; 27

106. 2.S; 130

A-15

Answers to Selected Exercises 88. b2 89. a 2 + 2ab + b2 enti re large square. 93. l 22S

90. They both represent the area of the

•

94. They are equal.

Section 5.7 1. 1Ox2 + 8; 2; Sx2 + 4

3. The student wrote the second term of the 6x2 - 12x 6x2 12x quotient as - l 2x rather than -2x. = - = x2 - 2x 6 6 6 S. 6p4; 18p7 ; 2p2 + 6p5 7. 2m2 - m 9. 30x3 - lOx + S 11. 4m3 - 2m2 + 1 13. 4t4 - 2t2 + 21 lS. a4 - a +

~

a , 1 17. -3p- - 2 + - 19. 7r2 - 6 + -1 21. 4x3 - 3x2 + 2x p r 12 4x2 2 2 27. - + x + 23. - 9x + Sx + 1 2S. 2x + 8 + x 3 3x 26 29. 9r3 - 12r2 + 2r + - 3._ 31. - m2 + 3111 3 3r m

49. 4x2

7x + 3

-

SI. 4k 3 - k + 2

y+ l

SS. 3k-' + 2k - 2 + -6- S7. 2p2 - Sp + 4 + - ,6- k- 2 3p- + 1 -SS 6x + 19 + - x+3 13 , 1 , 6S. r2 + 2 + - 67. 3x- + 3x - 1 + - - 69. y- - 3y + 9 r2 - 4 x- l 4 9 71. a 2 + S 73. x2 - 4x + 2 + x 7S. x 3 + 3x2 - x + S x2 + 3

S9. x2 + 3x + 3

61. 3y2 - 2y + 2

63. 2x2

-

~a-

lO+_I!_ 79. x2 +~x-.!..+ -2 2a+6 3 3 3x-3 81. (x2 + x - 3) units 83. ( 48m2 + 96m + 24) units 77.

8S. (Sx2

4

-

7a + 2 S6. 6r3 + 8r2 - 17r + 6 3p3 + 2Sp2 + lSp S8. m2 - 7m - 18

and y

6n3 74. - 2111211 + mn + -

s

76. -6r5 s - 3r4 +

2. (-S) 11

3. -72x7

4. 10x 14

S. l9 5x 5

6. -47y 7 12.

13. - 1 14. -1 I

20. ST

3

I

21. 4 22. 36

28. 72r5

29. 4.8 X 107

32. - 4.82 X 106

23. x 2

64

17. -49

24. y 7

2S.

18. 25 r8

Sl

26.

8m9n 3

34. 78,300,000 (b) 800

(b) 4,000,000

79. x 2 + 3x - 4

80. m2 + 4m - 2

83. y 2 + 2y + 4

84. 1 00~ - IOx2 + I

243

---p;-

38. (a) S x lOo

40. (a) 2.S x 10- 2

4S. 201112 ; 2; monomial 46. p 3 - p 2 - 4p; 3; trinomial 47. - 8y5 - 7y4; S; binomial 48. -r3 - 2r + 7 49. 13x3y2 - Sxy 5 + 2 1x 2 SO. a 3 + 4a2 Sl. y 2 - lOy + 9 l Sk 2 + 18k

1 27. a3b5

3S. 0.000000897

41. 100,720,000,000,000,000 42. 38, 140,000 11 X 10 16 44. S.8S3 X 10

-

8S. 2y 2

-

-s

Sy + 4 + - - 3y2 + 1

-3

86. x 3 -2x2 +4+ - - 4x2 - 3

Chapter 5 Mixed Review Exercises 1. 2

216r6p 3 2. ~ 3. I44a2 - I

I

I

4. 16 S. 256

S 6. p - 3 + P 2

S 4 3x 14. - - + 2 S,ry 2y2

lS. 10p2

17. 49 - 28k + 4k2

1 18. ~y 12

-

3p - S

80 16. 3x2 + 9x + 2S + - x - 3

19. (a) 6x - 2

(b) 2x2 + x - 6

20. (a) 20x4 + 8x2 (b) 2Sx8 + 20x6 + 4x4

p

19. 64

43. 3.S0799SS2

S2. - l 3k4

6 Sa - 3 81. 4x - S 82. Sy - 10

-6-

30. 2.8988 X 10 10 31. 8.24 X 10-s

33. 24,000

39. (a) 4 x 106

(b) 0.02S

16. -2

37. (a) 8 x 102

36. - 0.00076 (b) S

lS. 2

I

7S. y 3 - 2y + 3

2__ 77. 2r + 7 78. 2a2 + 3a - 1 + - r2s5

Chapter 5 Review Exercises 1. 4 11

69. (a) Answers will vary. For example, let

=

2 7. - 3 8. 6k3 - 2 lk - 6 9. r 13 10. 4r2 + 20rs + 2Ss2 3m 11. y 2 +Sy+ 1 12. l0r2 +2 lr- 10 13. -y2 -4y+ 4

l lx + 14) hours

-

= I

Sy 2 73. - 3

-s 3 + --

S3. Sy3 + 2y -

-

2p4

as follows. (a+ b) 3 =(a+ b)(a + b) 2 =(a+ b)(a2 + 2ab + b2 ) = a 3 + 2a 2b + ab2 + a2b + 2ab2 + b 3 = a 3 + 3a2b + 3ab2 + b3 4 4 71. x6 + 6x4 + 12x2 + 8 72. :J7TX3 + 47TX2 + 41TX + :J 7T

a 41. 2y - S 74 47. 2a - 14 + - 2a+3

44

-

2. ( 1 + 2) 2 oF 12 + 22 , because 9 oF S. (b) Answers will vary. For example, let x = 1 and y = 2. ( 1 + 2) 3 oF 13 + 23, because 27 oF 9. 70. Find the third power of a binomial, such as (a + b ) 3,

39. x + 2

43. p - 4 + - - 4S. 6m - I p + 6

2a2

-

S9. 6k 2 - 9k - 6 60. 2a2 + Sab - 3b2 61. l 2k2 - 32kq - 3Sq2 62. s 3 - 3s2 + 3s - 1 63. a 2 + 8a + 16 64. 4r2 + 20rt + 2St2 6S. 36m2 - 2S 66. 2Sa2 - 36b2 67. r 3 + 6r2 + l2r + 8 x

.!3_ - ~ + ~ - !Q 3S. -4b2 + 3ab - ~

x ~ ~ ~ 37. 6x4y 2 - 4xy + 2.ry2 - x4 y

SS. a 3

S7. Sp 5

- 2 2

•

68. 2S13 - 30t2 + 9t

i

33.

S4. 10, I , - 2, I , 10

S3. 1, 4, S, 4, 1

92. 302 + 2(30) (S) + S2

91. 122S

Chapter 5 Test I 7 2 16 [S. l , S.2] 1. 625 2. 2 3. i2 4. -32 S. -;;;6 6. 9x3y5 7. 85 8. x 2y 6 9. (a) positive (b) positive (c) negati ve

(d) positive

(e) zero

(I) negative

[S.3) 10. (a) 4.S X 10 10

(b) 0.0000036 (c) 0.000 19 11. (a) I x 103; S.89 x 10 12 (b) S.89 X 10 15 mi [S.41 12. -7x2 + 8x; 2; binomial 13. 4n4 + l 3n3 - 1On2 ; 4; trinomial

14. 4, -2, -4, - 2, 4 lS. - 2y2 - 9y + 17 16. -2 la 3b2 + 7ab5 - Sa 2b2 17. 16r2

t J2

19

18. -12t2 + St + 8

+ 18x4 -

6x3 + 3x2 2 20. St - 24 21. 8x + 2.ry - 3y2 [S.6] 22. 2Sx2 - 20.xy + 4y2 23. 100v2 - 9w2 [S.S] 24. 2r3 + r2 - 16r + lS [S.6] 25. 12.r + 36

[S.S] 19.

-2

-

t2 -

-27x5

A-16

Answers to Selected Exercises

26. 9x2 + S4x + 8 1 [S.7] 27. 4y2 28. - 3.t:y2 + 2x3y2 + 4y2

-

s

111. commutative property

y

113. No, because the result is not a product. It is the difference of 2x(y - 4 )

3y + 2 + -

26 30. 3x2 + 6x + 11 + - x- 2

29. x - 2

Chapters R-5 Cumulative Review Exercises

and3(y-4).

112. 2x(y - 4) - 3(y - 4)

114. (y-4)(2x-3),or(2x-3)(y-4);yes

Section 6.2 1. a and b must have different signs, one positive and one negative. 3. C

5. a 2 + I 3a + 36 7. The greatest common factor must be

included in the factori zation. The correct factorization is x(x + 7)(x - 4) . 9. 1 and48, - I and -48, 2 and 24, -2 and -24, 3 and 16, -3 and - 16, 4 and 12, -4 and - 12, 6 and 8, -6 and -8; The pair with a sum of - 19 is -3 and - 16.

11. I and -24, - I and 24, 2 and - 12, -2 and 12,

3 and -8, -3 and 8, 4 and - 6, -4 and 6; The pair with a sum of-Sis 3 and -8.

13. 20; 12; table entries: 2, 2, 12, 4, 4, 9; 10 and 2;

(y+ IO)(y+ 2)

15. p+6

23. x + 11 25. y - 9 31. (111 + S)(111 - 4) 37. (z - 7)(z - 8)

33. (y - S)(y - 3) 39. (r - 6)(r + S)

43. prime 45. (x+ 16)(x-2)

[S.4] 33. 4, 1,0, 1,4

-4

.y

49. (x+y)(x+3y)

4

55. (111 + 611)(111 - 211)

21. y-S

29. (b + 3)(b + S) 35. prime

41. (a+ 4)(a - 12)

47. (r +2a)(r+a)

51. (t +2z)(t-3z)

53. (v-Sw)(v-6w)

57. (a - 6b)(a - 3b ) 59. 4(x + S)(x - 2)

63. 6z2 (z - 3)(z - I) 65. -20(x - 3)(x + 7) 67. S1112(1113 - Sm2 + 8) 61. 21(1 + I )(1 + 3)

0

69. x(x - 4y)(x - 3y) 73.

-a 3 (a

71. z 8 (z - 7y)(z + 3y)

+ 4b) (a - b)

75. mn(m - 611)(111 - 411)

77. yz(y + 3z)(y - 2z)

79. (a+ b)(x + 4)(x - 3)

81. (2p + q)(r - 9)(r - 3)

FACTORING AND APPLICATIONS

Section 6.1

Section 6.3

1. product; multiplying 15. xy 2

17. x+ 11 19. x-8

27. (y + 8)(y + 1)

3. 4

5. 4

7. 6

9. I

19. 6m3112 21. factored

17. 6

11. 8 13. 10x

3

23. not factored

1. (21+ ! )(St+2)

3. (3z - 2)(Sz - 3)

7. (a) 2, 12, 24, 11

(b) 3, 8 (Order is irrelevant.)

5. (2s -1)(4s + 31) (c) 3m, Sm

25. The correct factored form is l 8x3y 2 + 9Ay = 9xy( 2x2y + I ) . If a

(d) 21112 + 3111 + 8111 + 12

polynomial has two terms, then the product of the factors must have two

(f) (2m + 3)(m + 4) = 21112 + 8111 + 3111 + 12, and combining like terms

27. 31112

terms. 9xy(2x2y) = l 8x3y2 is j ust one term. 31. 2111114

33. y + 2

35. a - 2 43. 9111(31112 - I)

41. 31(21 + S)

(except I) 55. 8111113 ( I + 3111) 59.

-2x(2x2 -

Sx + 3)

63. a 3 (a 2 + 2b2

-

39. x(x - 4) 2 45. m (m - I ) 47. 8z2(2z2 + 3)

51. Sy6 ( 13y4 + 7)

49. -6x2(2x + I )

61.

53. no common factor

57. 13y2 (y6 + 2y 2

9p3q(4p3

3a2b2 + 4ab3 )

29. 2z4

37. 2 + 3xy

-

3)

gives the original trinomial 2m2 + I Im+ 12. obtain ( 4x - I )(4x - S) as the correct answer. 15. A

x+ 4,x- 1, orx - l,x+ 4 23.

form; (71 + 4)(8 + x)

73. in factored form 75. not in factored

3)

71. not in factored

77. The student should factor out -2, instead of 2, in the

13. B

common factor of 4, but those of the trinomial do not. The correct factored form is (4x - 3)(3x + 4).

-

11. The student

17. ( 4x + 4) cannot be a factor because its terms have a

65. (x + 2)(c - d)

69. (p - 4)(q2

9. B

stopped too soon. He needs to factor out the common factor 4x - I to

+ Sp2q3 + 9q)

67. (m + 211)(111 +11) form

(e) (2111 + 3)(111 + 4)

19. 2a +Sb 2z2

z - 3, 2z + I 25. (3a + 7)(a + I )

21. x 2 + 3x - 4;

- Sz-3;2z+ 1,z-3,or 27. (2y + 3)(y + 2)

29. (3m- l )(Sm+2)

31. (3s- 1)(4s+S)

33. (Sm - 4)(2111 - 3)

35. (4w - 1)(2w - 3)

second step to obtain x2 (x + 4) - 2(x + 4 ), which can be factored as

37. (4y + I )(Sy - II )

39. prime 41. 2(Sx + 3)(2x + I )

(x + 4)(x2 - 2).

43. 3(4x- 1)(2x-3)

45. prime 47. (3x+4)(x+4)

79. (S + 11)(m + 4)

83. (p + 4)(p + q)

81. (2y - 7)(3x + 4)

85. (a - 2)(a + b)

87. (z + 2)(7z - a)

89. (3r + 2y)(6r -x) 91. (w+ I)(w2 +9)

99. (4m-p2 )(4m2-p)

95. (x+S)( lOy+ I ) 97. (3-a)(4 -b) 101. (y + 3)(y + x)

103. (S - 2p)(m + 3)

107. (3r + 2y)(6r -1)

109. ( I + 2b)(a5

93. (a+2)(3a2 -2)

-

49. prime

51. (12x-S)(2x-3)

59. -q(Sm+2)(8m-3)

105. (z - 2)(2z - 3w)

63. (Sa+ 3b)(a - 2b)

3)

67. (3p + 4q) (4p - 3q)

53. 4x(8x+3)(x+ 1)

57. 3112 (S11 - 3)(n - 2)

55. -Sx(2x + 7)(x - 4)

61. y 2 (Sx-4)(3x+ 1)

65. (4s + St)(3s - t ) 69. (24y + 7x)(y - 2x)

Answers to Selected Exercises

A-17

71. 2(24a + b) (a - 2b) 73. (4x + 3y)(6x +Sy) 7S. x 2y 5 ( 10x- I)(x+4) 77. 4ab2(9a+ I)(a-3)

SUMMARY EXERCISES Recognizing and Applying Factoring Strategies

79. 111411(3111 + 211) (2111 + 11)

I. G 2. H 3. A 4. B S. E 6. I 7. C 8. F 9. I 10. E

81. - I(x + 7)(x- 3)

83. - 1(3x+4)(x- 1) 8S. - l (a+2b) (2a + b ) 87. (8x2 - 3) (3x2 + S) 89. ( 1Sx2 - Sy)(2r2 - 3y) 91. (m + I ) 3(Sq - 2)(Sq + I) 93. (r + 3) 3(3x + 2y) 2 9S. - 4, 4 97. - 11, -7, 7, 11

99. S · 7 100. - S( - 7)

101. The product of 3x - 4 and 2x - I is 6x2 - I Ix+ 4. 102. The product of 4 - 3x and I - 2x is 6x2

I Ix + 4.

-

103. The factors in Exercise I0 I are the opposites of the factors in Exercise 102.

104. (3 - 7t)(S - 21)

13. 6(y - 2) (y + I)

12. (a + S)(a + 9)

11. (a - 6 )(a + 2)

14. 7y 4 (y + 6)(y - 4)

18. (z + 7) (z - 6)

I7. (p- I l )(p - 6)

16. (m - 411)(111 + n)

IS. 6(a + 2b + 3c)

19. (Sz - 6)(2z +I)

20. 2(m - S)(m + 3) 21. I7xy(x2y + 3) 22. S(3y + I) 23. Sa3(a - 3)(a + 2) 24. (4k + 1)(2k- 3) 2S. (z - Sa) (z + 2a) 26. SO(z 2 - 2) 27. (x - S)(x - 4) 28. prime 31. 4(4x + S)

30. (3y- 1)(3y + S)

29. (311 - 2)(211- S)

32. (m + S)(m - 3) 33. (3y - 4) (2y + I) 3S. (6z+ l)(z + S) 36. ( 12x - l )(x+ 4)

34. (m + 9)(111 - 9) 37. (2k-3)2

Section 6.4

38. (Sp - I)(p + 3)

1. 1; 4;9; 16;2S;36;49; 64;S I; 100; 12 1; 144; 169; 196; 22S ; 2S6;2S9; 324; 361; 400 3. A, D S. The binomial 4x2 + 16 can be factored as

40. ( 4111 - 3)(2111 + I) 41. (3k - 2)(k + 2) 42. (2a - 3)(4a2 + 6a + 9) 43. 7k(2k + S) (k - 2)

4(x2 + 4 ). After any common factor is removed, a sum of squares (like x 2 + 4 here) cannot be factored. 7. (y + S) (y - S)

47. S111( I - 2111)

9. (x + 12)(x - 12)

lS. 4(m2 + 4)

11. prime 13. prime

17. (3r + 2)(3r - 2)

19. 4(3x + 2)(3x - 2)

21. ( 14p + IS)( 14p - IS) 2S. (9x + 7y)(9x - 7y)

23. (4r+Sa)(4r-Sa) 29. pri me 3S. x(x2 + 4)

27. 6(3x + y)(3x - y)

31. (2+x)(2-x) 33. (6+S1)(6-S1) 37. x 2(x + I )(x - I) 39. (p2 + 7)(p2 - 7) 41. (x2 + I )(x+ I)(x- 1)

43. (p2 + 16) (p + 4 )(p - 4) 4S. B, C 47. 10 49. 9 Sl. (w+ 1) 2 S3. (x-4)2 SS. prime S7. 2(x + 6) 2 S9. (2x + 3)2 61. (4x - S)2 63. 4r(3r + 4) 2 6S. (7x - 2y) 2 67. (Sx + 3y)2 69. 2(Sh - 2y) 2 71. z 2(2Sz 2 + Sz + I) 73. I; S; 27; 64; 12S;

39. 6(3111 + 2z)(3111 - 2z)

4S. (y2 + 4 )(y + 2)(y- 2)

44. (S + r) ( I -s)

SO. (y - S)(y + 7) Sl. prime S2. 9p8(3p + 7) (p - 4 ) S3. S1113(4m6 + 21112 + 3) S4. (2111 + S)(41112 - 10111 + 2S ) SS. (4r + 3m) 2 S6. (z - 6) 2 S7. (Sh + 7g) (3h - 2g) S9. (k - S)(k - 6)

S8. Sz (z - 7 )(z - 2)

60. 4(4p - Sm) (4p + Sm) 61. 3k(k - S)(k + I) 62. (y - 6k )(y + 2k) 63. ( !Op + 3)( 100p2 - 30p + 9) 64. (4r-7 )( 16r2 + 2Sr + 49) 6S. (2 + 111) (3+ p ) 66. (2111 - 311)(111 + Sn ) 67. (4z - 1) 2 70. ( IOa + 9y)( IOa - 9y)

(c) perfect square (d) perfect square (e) both of these (f) perfect cube 79. (a - I )(a2 + a + I) 81. (m + 2)(m2 - 2m + 4) 83. (y - 6)(y2 + 6y + 36) 8S. (k + IO)(k 2 - !Ok + 100) 87. (3x - 4) (9x 2 + 12x + 16)

78. 6y4 (3y + 4)(2y - S)

103. (St + 2s) (2St2 - IOts + 4s2)

llS. (6111 +

rn

119. (1 + i)2 12S. 411111

(x - 0.9) 2

I23.

113.

(P+ D(P - ~)

(x + i )(x2 -

ix+~)

I27. (m - p + 2)(m + p)

129. (2x + I + y)(2x + I - y)

80. 4(2k - 3) 2

79. prime

82. 12y 2(6yz 2 + I - 2y2z2)

81. (4 + m)(S + 311)

83. (4k - 3h)(2k + h) 84. (2a + S) (a - 6) 8S. 2(x + 4 )(x2 - 4x + 16) 86. 1Sa3b2(3b3 - 4a + Sa3b2) 87. (Sy - 6z)(2y + z) 88. (m - 2) 2 89. (Sa - b)(a + 3b) 90. Sm2(S111 - 3n)(Sm - 1311)

Section 6.5

(c) quadratic

6m - ~) 117. (x + O.S)(x - O.S) 121.

7S. (S111 - S11)2

77. 2(3a - I) (a + 2)

1. ax2 + bx + c 3. O; zero-factor S. The term with greatest degree is greater than two. (It is cubic.) 7. (a) linear (b) quadratic

lOS. (2x - Sy 2 )(4x2 + 10xy2 + 2Sy4 ) 107. (3m2 + 2n)(9m4 - 6m 2n + 4n2) 109. (Sk - 21113)(2Sk 2 + IOk1113 + 41116) 111. (x + y)(x2 - xy + y2)(x6 - x3y3 + y6)

71. prime 72. (2y + S)(2y- S)

74. S(2111 - 3)(111 +4)

73. Sz(4z- I)(z+2)

2 16; 343; S I2; 729; 1000 7S. C, D 77. (a) neither of these

S9. (Sx + 2)(2Sx2 - IOx + 4) 91. 6(p + I)(p2 - p + I) 93. S(x + 2)(x2 - 2x + 4) 9S. (y - 2x)(y2 + 2yx + 4x2) 97. 2(x - 2y)(x2 + 2ry + 4y2) 99. (2p + 9q) (4p2 - ISpq + Slq2) 101. (3a + 4b)(9a2 - 12ab + 16b2)

69. 3(6m - 1) 2

68. (a2 + 2S)(a + S)(a - S)

76. (2- q)(2 - 3p)

(b) perfect cube

46. prime 49. (z - 2)(z2 + 2z + 4)

48. (k + 4)(k - 4)

131. (3a - Sb+ 2c)(3a - Sb - 2c)

(d) linear

obtain 2x = 0 or 3x - 4 13. {3. n

9. Set each variable factor equal to 0 to

= 0. The solution set is {0, ~}-

1s. { - i.n 11.

II. { -S, 2}

{ -~. o} 19. {o.~} 2i. (6 )

23. {-2,-1}

2S. { 1, 2} 27. {-S, 3} 29. { - 1, 3}

31. { -2, -I}

33. { - 4 } 3s. { 10}

41.

{-fl

43.

SI. { -6, 0}

{i}

4S. {-3, 3}

S3. {0,7}

SS.

47.

{o.i}

37. {-2,

j}

39.

{-H}

{- ~. n 49. { - I I, I I} S7. {2,S}

S9. {- 4,n

A-18

Answers to Selected Exercises

61. {- 17, 4 }

63. { - 4. 12}

65. {-9.-2}

69. {-2. 0,4 } 11. { - s , o, 4 } 11.

{-~. o.fl

67.

{o.~. 4} 75. {-3, o, s }

73.

57. {- 2, - 1,

87.

{-~. 4}

S, 6, 7

{ -~.- 1 . n

89.

64; 144;4; 6

91.

-n

55. (6 )

56. {-3, 3}

n

59. length: 10 ft; width: 4 ft

62. length: 6 m; width: 4 m 63. - S, -4, -3 or

64. (a) 2S6 ft

(b) 1024 ft

2016: 197 1 thousand cars

(b) No time has elapsed, so the object hasn't fallen (been released) yet.

~. 0,

58. {-

60. S ft 61. 26 mi

{-~.~. s} 79. { -~.-3, 1} 81. {- 1,3} 83. {- 1,3 }

85. {3 )

54. {7}

53. {- 1,6 }

65. (a) 2013: 382 thousand cars;

(b) 20 13: The result is slightl y less than

388 thousand, the actual number; 2016: The result is less than 20 14 thousand, the actual number.

Section 6.6 Chapter 6 Mixed Review Exercises

1. Read; variable; equation; Solve; answer; Check; original II

+ 1)(x +

3. Sil= bh; Step 3: 4S = (2x

1); Step 4: x = 4 or x = -2;

1. D

3. (3k + S)(k

2. C

+ 2) 4. (z - x)(z - IOx )

Step 5: base: 9 units; height: S units; Step 6: 9 · S = 4S 5. Sil = LW; Step 3: 80 = (x + 8)(x - 8); Step 4: x = 12 or x = - 12; Step 5: length:

5. (y2 + 2S )(y + S )(y - S ) 6. 3m(2m + 3)(m - S)

20 un its; width: 4 units; Step 6: 20 · 4

=

10. 8abc(3b2c - 7ac2 + 9ab)

12 cm 9. base: 12 in.; height: S in.

ll. height: 13 in.; width: 10 in.

80

7. length: 14 cm; width:

13. length: IS in.; width: 12 in.

15. mirror: 7 ft ; painti ng: 9 ft

17. 20, 2 1 19. - 3, - 2or4, S

21. 0, 1, 2 or7, 8, 9

7,9

25. 7, 9, II

27. - 2,0, 2or 6,8, 10

23. -3, - 1 or

29. 12 cm

31. length : 20 in.; width: IS in.; diagonal: 2S in. 33. 12 mi 37. 11 2 ft (c) 3 sec

I

35. 8 ft

7. prime

8. 2a 3 (a + 2)(a - 6)

13. (7t + 4 ) 2

12. (2r +3q )(6r-S)

14. ( !Oa+3)( 100a2-30a +9) 17.

{-D

18. IS m, 36 m, 39 m

19. 6 m

16. {-S, 2 }

20. width: 10 m;

length: 17 m

I

(d) The negati ve solution, - I, does not make sense because 43. 3 sec

Chapter 6 Test [6. 1-6.4) 1. D

2. 6x(2x - S ) 3. m2n(2mn + 3m - Sn )

45. (a) IO (b) 297 million subscribers; The result obtained from the

4. (2x + y)(a - b)

model is slightly more than 296 million, the actual number for 20 I0.

7. (Sz - I )(2z - 3)

(c) 16 (d) 390 million subscribers; The result is less than 396 million, the actual number. 48. b2 49. a 2

(e) 20

(f) 443 million subscribers

50. a 2 + b2

= c2 ;

47. c 2

This is the equation of the

Pythagorean theorem.

10. prime

16. 3t (2t + 9)(t - 4) 2

-

1. 7(1 + 2)

2. 30z(2z2 + 1)

4. SOm 2n2 (2n - mn2 + 3) 6. (3y + 2x)(2y + 3) 9. (q + 9)(q - 3)

22. u. 6}

3. - 3x(x

2

-

2x - I)

23.

7. (x + 3)(x + 2)

14. (x+8) 2

17. (r - S)(r + Sr+ 2S ) 19. (x2 + 9)(x + 3)(x - 3)

2k + 4)

2y)(9x 2

+ 4y 2 )

{-n} 24.

8. (y - S)(y - 8)

Chapters R-6 Cumulative Review Exercises II 5 2. 18 3. 9 4. 2I [2. 1- 2.4) 5. {O }

15. (y - 3z) (y - Sz)

(1.7, 2.4) 7. (a) equation; { 6 }

5

17. p (p - 2q )(p + q )

18. -3r3 (r+ 3s)(r - Ss)

19. r and6r, 2rand 3r 20. Factorout z .

22. (3r - 1)(r + 4 ) 25. prime

27. - Sy(3y + 2)(2y - I )

23. (3r + 2)(2r - 3)

26. 4x3 (3x - 1)(2x - 1)

28. (2a - Sb )(7a + 4b) 30. rs(Sr + 6s) (2r + s)

32. D 33. (n + 7) (n - 7) 39. (3t - 7) 2 -

20kx + 16x2 )

-

51.

photos estimated to have been taken

-

{ -~.

xy + y 2 )

n

worldwide increased by I 3S billion photos per year.

IOm + 100)

{ -~. 1 } 46. {o.n 47. {-3.- 1} 48. , , ,4 ,

{ -~. S}

I3S means that the number of digital

(c)

37. prime

43. ( 10 - 3x2 )( 100 + 30x2 + 9X') 44. (x - y)(x + y)(x2 + xy + y 2 )(x 2

50.

[3. 1) 12. (a) negative; positive (b) negative; negative [3.2, 3.3) 13. (a) (-2, 0), (0, - 4 ) [3.3- 3.S] 14. (a) l 3S; A slope of (b) - 2

42. (7x - 4)(49x2 + 28x + 16)

52. {O, 8}

0.0S }

[2.6) 8. P = _ A_ 9. 110° and 70° [2.S] 10. gold: 9; silver: 8; I + rt bronze: 6 [R.2, 2.7] ll. 4 10; 340; 23%; 16%

34. (Sb + 11)(Sb - 11) 40. (m + IO)(m2

{

6.

(b) expression;~ m - I

31. B

35. (7y + Sw )(7y - Sw ) 36. 36(2p + q )(2p - q) 41. (Sk + 4x)(2Sk2

26. {-8.-H}

11. prime

14. (m + 3n )(m - 611 )

38. (r - 6) 2

25. { 10}

[6.6] 27. - 2, - I 28. 6 ft by 9 ft 29. 17 ft 30. $ 13,987 billion

(R. l ] 1. 2:I

29. ( 3m - Sn )(m + 811)

[6.S] 21. { - 3, 9 }

{o.9}

13. - 8p3 (p + 2)(p - S)

21. (2k - I )(k - 2)

15. (2x-7y) 2

2

12. 3x2 (x + 2)(x + 8)

24. (Sz+ 1)(2z- 1)

12. (3y + 8)(3y - 8)

5. (2y+ 3)(x - 4)

10. (r - 8)(r + 7)

16. (p + 12q) (p - lOq)

6. (2x + 3)(x- I )

8. 3(x + I )(x - S ) 9. prime

11. (2 - a)(6 + b)

18. 8(k + 2)(k2

Chapter 6 Review Exercises

5. (x + 3)(x- 8)

13. (9a+ ll b)(9a- l lb)

20. (3x + 2y)(3x -

49. (3, S)

15. {0, 7 }

39. 2S6 ft 41. (a) I sec (b) 2 sec and I 2 sec

t represents time, which cannot be negati ve in th is situation.

45.

9. (3m + 4 )( Sm - 4p)

ll. 6xyz(2xz 2 + 2y - Sx 2yz 3 )

[4. 1-4.3] 15. {(- 1,2)}

16. 0

19. _.!._ 20. __!__ [S.4] 21. - 4k 2 p2 m6

[S.6] 23. 9p2 + 12p + 4

[S. l , S.2] 17. 9 -

4k + 8

(b) ( 20 14, 800) 18. 2S6

[S.S] 22. 4Sx 2 + 3x - 18

[S.7] 24. 4x 3 + 6x2

[6.2-6.4] 25. (2a - l )(a + 4) 28. (Sr + 9t )(Sr - 9t)

16

-

3x + 10

26. (4t + 3v)(2t + v)

[6.S] 29.

{ -~, n

27. (2p - 3) 2

[6.6] 30. Sm, 12 m, 13 m

A-19

Answers to Selected Exercises

(x + y)2(x2 _ xy + y2)

RATIONAL EXPRESSIONS AND APPLICATIONS

7~

Section 7.1 1. 3; 3; - 5 3. B, D 5. B 7. - 3; - 3; - 3; - 6; 5 7

8

8

15. (a) undefined (b ) undefined

21. x # 0

29. m # - 3, m # 2

13. (a) -15 (b) undefined (b ) 0

23. y # 0

5

25. x # 6

27. x # -3

31. It is never undefi ned. 33. It is never 37. ~

2 x- 1 7 6 41. 3r2 43. 5 45. x + 47. 5 49. 7 51. m - n

3(2m + I ) 4

65. l 3x

7 z - 3 77. z + 5

x+ I a+b 81. a _ b

r +s

79. r _ s

x +3 x - 3

73. x + 2

75.

I

59. - 3-

61. x

69. x - 3 71. x + I

67. k - 3

+ 6 63.

3r - 2s

3m

5

57.

39.

2 53. t _

1

55.

x- 1 m+n

t 3

3 ~

x - 4

x2 + 1 85. - x -

83. - 2

1. B

3. The factor x should appear in the LCD the greatest number of

times it appears in any single denomi nator, not the total number of times. The correct LCD is 50x4. 5. 5; 5; one; 5; 2; 5; 50 11. x 5

23. (x + I )(x - I )

k-2

101. -(m + 1)

2

105. It is already in lowest terms. 107. - 2

47. k (k + 5) (k - 2)

55

109. -x

9k

z(z-3)(z-2)

- 2y(y -z) 2 y4 - z y

x +4

75. 7

x+4

'~'-(x- 3)' -x + 3

x - 3

-(2x - 3) - 2x + 3

113. - - - - , x+3

x+ 3

2x - 3

'

49. a(a + 6)(a - 3)

59. 60m2k3 61. ~ 32k4 6z - 18

57. - 45

20

· 55

65.

103. - I

Answers may vary in Exercises 111, 113, and 115.

-(x + 4 ) - x - 4

33. c-dord-c

35. m - 3or 3 - m 37.p - qorq - p 39.2(x+l )(x - 1) 41. 3 (x - 4 ) 2 43. 12p(p + 5) 2 45. 8(y + 2 )(y + I)

--~3-

lll.

27. 28m2 (3m - 5 )

25. 12p(p - 2)

31. 18(r-2)

14(z - 2)

z +3 I - 2r 95. - - 97. - - 99. - I

7. 60 9. 1800 17. 50m4 19. 15a 5b3 21. r 9t 3

15. 84r 5

13. 30p

51. (p + 3) ( p + 5) ( p - 6 ) 53. (k + 3) (k - 5 )(k + 7)(k + 8)

a +b

z

Section 7.3

29. 30(b - 2 )

b 2 + ba +a2 k 2 - 2k + 4 + p 2 89. x2 + 3x + 9 91. - - - - - 93. - - - -

87. I - p

2

19. (a) 0

35. numerator: x 2, 4x; denominator: x, 4

undefined.

3y(x - y)

M

(b) - I

17. (a) 0

(b ) 25

(x + y)2(x2 _ xy + y2) ,ill

3y(y - x)(x - y)

2 x + 10 5xy 77. - - 79. 10 4q

3

9. (a) iO (b) 15 11. (a) 0

-

67.

(t - r) (4r - t ) t3 - r

3

( r - 3) ( r

76.

18a - 36 2y(z - y)(y - z) ,or y4 _ z 3y

69.

36r(r + I ) 71. - - - - - - - 73

+ 2)( r + I)

- 4a

63.

+ 2b) + a 2b2 - ab3

ab(a

· 2a3b

77. identity property of multiplication

78. 7

79. I

80. identity property of multiplication

2x - 3

'

-(x+ 3 ) - x - 3

Section 7.4

-(3x - 1) -3x+ I 3x - I 3x - 1 115. ----5x - 6 ' 5x - 6 ' -( 5x - 6) ' - 5x + 6

1. E

117. (a) 0

expression must be subtracted. Using parentheses will help avoid this

(b) 1.6

yield 2x + I .

(c) 4. 1

119. x 2 + 3

also increases.

(d ) The number ofvehicleswaiting

121. Both yield 2x + 3.

123. Both yield x 2 + I .

122. Both

3. C

x- I

19. 4

Section 7.2 (b) D

t2

11. 4(x - y) 2

21.

22

: 7

:

3 29. - 2t4

13.

16q p 3 3

x+3

2

~ 12

15. ~ 17.

23. x - 2; 3; x - 2; 5;

1

41.

51. - I

- 4t(t + I ) t- I

3a

4k - I

7. - 3

ii

25.

m

r + 6s

67. - - 69. - - 71. m +5 r +s

53.

27. 5

x(x - 3)

3

9 2

(q - 3) 2 (q + 2) 2 q+ I

6m 2 + 23m - 2 (m + 2 )(m + l ) (m + 5)

-6

+3

25. x

27. y - 6

I

29. x _

3

I0 -7r - 3x- 2 61 37. - 39. ~ 41. t 14 28 7 - 6p

- k - 8

3P'2

49. k(k + 4)

x+4

51. - x+ 2

4y 2 - y + 5 3 55. - - - - - 57. (y + ! )Z(y- I ) t

- 2 2 61. - - , or - - 63. - 4 x- 5

5 -x

x +y -x -y 67. - - -,or - - 5x - 3y

6

3y - 5x

- m - n

69. - -, or - - 71. 3 73. (m _ n) 4p - 5 5 - 4p 2

x - 1 m +6

+9

-5 5 65. - -2 , or - 2- x- y y -x

55. (2x- l ) (x + 2 )

-

5x

45. ~ 47.

59. m - 2; 2 - m

45. -4 47.

5 7 . - - - - - - 59. - - 61. - - 63. (k+ 1)(2k- 1) 3k - 2 m +p m

I

4 y+ 4

15. - -

m

23. b

3z+5 33. 30 35. - - 15

43. - 2-

37. - 6-

m+4p

+

m - 1

21. m + I 11

31. - I x

7 19. r 2 + rp

m+ I

(k- 1) 2

2 9. c + d

40y

2

10

43. 9

53. _ m+ 2

p+ 2

~

2(x + 2) 35. x(x - I )

35

31. 4 33. - 8

5(x-4) 39. x2(x + 4) 49. p + 4

s

(d) A 3. i2 5.

(c) C

2 11 11. 3 13. -

x+5

124. Both yield x 2 + 2.

2

9. Each term in the numerator of the second

error. The correct answer is - -. 17. I

1. (a) B

5. B 7. G

y+ 3

65. - y+4

3-a-b

73. - - 2a - b

- 5q2 - 13q+7 - x 2 + 6x + ll 75. - - - - - - - (x + 3) (x - 3) (x + I ) 77. (3q - 2 )(q + 4)(2q - 3) 7x + 3 1 - 5x + l3 81. - - - 83. x +4 4x

85.

2x2

+6x

(x-?)(x-3)

79. y - 7

2x(x+3)

, or - - - - (x- 7)(x-3)

A-20

Answers to Selected Exercises

2a + 21 x- 8 8- x 1 87. - - - 89 - - - or - - - 91. - 3(a - 2) · 2(x - 3)' 2(3 - x) x- 2 9r + 2 93 · r (r + 2 )(r- 1) or

2(x2 + 3.iy + 4y 2 ) 9 S. (x + y)(x + y)(x + 3y)'

2(x2 + 3xy + 4y 2) (x

+ y) (x + 3y) 9k2

+ 6k + 26 5(3k + I)

99· (a)

(b) x # - 1

3.

~ S. Choice D

(d)

is correct, because every sign has been changed in the fraction. This . . . - I 13 5 S 13 5 means it was mulllphed by =I = 1. 7 . S tep I : 20, 6; tep 2 : 20 .;- 6; 39 zo13 · 56 , 50

31.

40 - 12p 85p

1

31

, or

9. -6

SS.

4( 10 -3p)

5y-2x 33. 3 + 4xy

85p

a-2 z-5 3S. - - 37. - 4 2a

(m - I )(m - 8)

x+y

+ xy +

29

29

y2

ab 4S. - b+a

2x-7 y+4 30 49. - - Sl. - - S3. - - - - 3x+ 1 y-8 (a + b)(a -b)

3m(m-3)

x2

S7 x - 3 • x - 5

S9 ·

~ 61.

3

T 63.

~+~

l 9r

6S

15

.

·

~ 2

106. (x

7

+ 3)(x + 1) 107. x +

1

I

-iOy 11. eq uation; {- 10}

. { 43 ~} 7. equation;

21. { - 15 }

13. {-6}

23. {-5}

9• expression; . 17. {7}

lS. {- 15 }

2S. {- 14}

29. { 12 } 31. {5 } 33. { l } 3S. { 2}

27. {9}

4. equation; { 9}

. 1 3. express10n; 2x2(x + ) 2 S. equation; { 39} 6. expression;

5k + 8 k(k- 4 )(k + 4)

. y+2 . t-5 7. express10n; - - 8. expression; ( ) y- 1 32t+ 1

p

. 9. expression; (p13 + ) 2 3 n.2}

12.

. { - 1, 512} 11. equation; 10. equat10n;

expression,~~

lS. expression; (m

13. expression;

3m + 5 + 3 )(m + 2 )(m +

m

{-H

SS.

{-~,3}

S7.

7S.

{fl

47. {-5}

{-~,3}

67. { -3}

77. {-6. D

49. 0

S9. {3}

si. {3}

61. {-4 }

s3. {-2. 12}

69. {xlx¥±n 11. {-6} 79. {6}

S. (a) the amount

81. Thisisanexpression,notan

(b) 5 + x

the quantity; x

=

x

31

83. F

=

ma

k

8S.

kF a=-;;;

5 +x 6

(c) -

=

13

3

+ 5; -6 13. x represents

+ ~x + ~x + ~x = 93; 36 lS. 18.809 min

17. 326.307 m per min 2S. 32 mph

8 24 _ x = +x 4 4 31. 18.5 mph

19. 3.37 1 hr 21.

equation. The student multiplied by the LCD, 14, instead of writing each coefficient with the LCD. The answer is 14t.

1

5 12 x+3 7. x represents the original numerator; (x + ) + = 7; i8 6 3 x+2 2 9. x represents the original numerator; x _ = I; 6 3 2

11. x represents the number; 6x

73. {o}

k+3 16. expression; ( k _ )

l)

5 t+ 2 19. expression; ( + ) 2 21 1

18. equation; 0

I

63. 0

14. equation; { 13}

1. into a headwind: (m - 5) mph ; with a tailwind: (m + 5 ) mph I

I

l~

Section 7.7

3. iO job per hr

37. x # - 2,x # 0

39. x # -3,x # 4,x # -2 41. x # - 9,x # l ,x # -2, x # 2 43.

109. 0

. y3 2. expression;~

. 10 1. expression; -

20. equation; { -7}

. c40x S. expression;

ll x + 21 108. ~

SUMMARY EXERCISES Simplifying Rational Expressions vs. Solving Rational Equations

Section 7.6

6s. {- 1}

0,

=

x= - 3.

17. equation; 0

4s.

lOS. If x

110. We know that -3 is not allowed because P and R a re undefined for

7

19. {-6}

15 104. 2x

(c) x # -3, x # - 1

72. -5

3. xyz

2

2S

I

66. 48 67. 48 68. Answers will vary. 69. 6 70. 3 71. -3

1. 12

h(B+b)

2a b

1

x2 + y 2 x2 + y 2 - m x +8 , or 39. - - 41. - - - 43. - 2- 2 m+ 2 -x+7 x -y (x+y)(x-y)

47.

91. s!l =

2

11. 50 13. lS. 17. - xy 6pq 3 m(m + 2) 2 8 a2 - 5 31 y2 + x2 19. 3(m _ 4 ) 21. :; 23. :; 2S. a 2 + 1 27. 50 29. xy(y _ x)

Step3:

r

the divisor R is equal to 0, and division by 0 is undefined.

Section 7.5 1. division

-

xz 93. a = , or a = - - L 9S. y = - ~ ~ x+z rs -rs 3y 97. t = , or t = 99. z = - - - , or rs-2s-3r - rs +2s+3r 5-9xy - 3y 2x- 1 -2x + 1 z= - - - 101. t = - - , or t = - - - 103. (a) x # -3 x+ 1 -x - 1 9.iy - 5

8000 + !Ox lOl. 49( 10 1 -x)

I

(b) 4

1

2S- ndL

15r 2 + !Ory - y2 97. - - - - - - - - (3r + 2y)(6r - y)(6r + y)

2

E - lr E 89. R = - - , or R = I

87. y = mx + b

27. 165 mph

29. 3 mph

23. 8 mph

Answers to Selected Exercises

O 47. I mL

.

I

7

41. 3 hr 43. 2!0 hr 45. 911 mm 49.

¥hr, or I ~ hr

(q - P) 2 xw + I IO 52. 13 53. - - 54. - - - 55. (x + 4)(x + 3), or pq xw- I

50. (a) The player mul tiplied: 5 · 3

5 + . - 2and d1.v1.ded by 2-that 1.s, h e averaged the times:

I

I

a

b

15.

=

(c) The player added the two times

(b) The player added: 5 + 3 = 8.

3

= 4.

x 2 + 7x + 12

60. 0

I - r-

56.

1

62. { 3}

61. {0}

= I to obtain x = - -.

65. d

a+ b

52. I0 hr; I0 hr; The same answer results.

b b - ac = - - , or d = - -

a

a

69. 2.020 hr 70. 10 mph I

7. inverse 9. inverse

/11

= kd

m+7 (m- 1)(111+ 1)

1.

13. (a) increases (b) decreases

19. direct; 10 21. (a) A= kb = ky2 23. 9 25. 2SO 27. 6 35. 15 in.2 37. $3S.20 39. 42~in.

17. direct; SO

15. inverse; 3 (b)

(c) F

29. 21 31. 125

= ka 33. ~

I

2

47. 20 lb 49. 52.8 17 in. 2

45. 122 amps

-t- I

6.

(d) x

41. 106 3 mph 43. IS ft

63.

59. {-4}

58. {- 16}

Ry

64. s

I = -

br -

=

I

rs s-r

3

20

c 66. t = - - 67. i5 68. i8 73.

71. 3TJ- hr 72. 2 hr

?f

Chapter 7 Mixed Review Exercises

11. Answers will vary; for example, number of movie tickets purchased and total price paid for the tickets.

57. 6

76. 32.97 in.

74. 2 75. 4 cm

Section 7.8 1. direct 3. direct 5. inverse

{35}

I

+r+t

111

a· b

.

51. Solve the equation - x + - x

A-21

(t + 2)(1 - 2)'

or

ll. {2}

10. {2 }

z+7 2 (z+ I )(z- 1)

1

2. 8p2 3. 6 4. 3 5.

x- 7 7. - - 8. {4} 2x + 3

I+ I (2 + 1) (2 - t) 12. v =at+ w

14. 2 hr

13. 3

18. 16

17. 4 ft

15. ISO km per hr 16. SO amps

9. { -2, 3}

22

51. 14 27 footcandles

Chapter 7 Test Chapter 7 Review Exercises 4

2.

1. (a) - 7 (b) - 16 4. (a) undefined

8.

/11

(b) 22

(a) 8

3. (a) undefined

(b) 2

I

(b) 2 5. x ¥- 3 6. y ¥- 0 7. k ¥- - 5,k ¥- -3

b a 3

¥- - 1, /11 ¥- 3 9.

3 13. x + y+4

II

(7.1 ] 1. (a) 6 13

11

10. - I

11.

-(2x + 3)

2p +Sq 12. - - 5p + q

2

15 .

4x - 9 4x-9 -(4x - 9) -4x+9 ----2x+3 ' 2x+3 '-(2x+3)' - 2x-3

16 ·

8 - 3x 8 - 3x -(8 - 3x) -8 + 3x 3 - 6x ' 3 - 6x ' -(3 - 6x)' - 3 + 6x 18. 2

19.

2

3m

20. 2 21. ii

29. 96

35

33. 56 34.

40 k 4

4b(b + 2) 15y 37. - - - 38. (b + 3)(b- l )(b + 2) so - IOy

28+ lly 42. y(7 + y)

7a + 6b 45 · (a - 2b)(a + 2b)

- 13p + 33 48. p(p - 2)(p - 3)

46

-2-3m 6 43.

-k2 -6k+3 · 3(k + 3)(k - 3)

~

-111 2 + 7111 + 2 17 • (2m + l )(m - S)(m - I )

25. D

19. (x - 5) (x - 3), or x 2

{-i. I }

[7.6] 21.

b

!Sa 35. IOa4 IS 39. x

50.

y-+3

-S4 _ x 18 6

36.

2 40. - p

3( 16 - x) 4x2 44. 47

Sz - 16 · z(z + 6)(z - 2)

4(y- 3)

a

49.

x2 - x - I +x + I ' or x - 3 3- x

2k [7.S] 18. -

22.

{-~. 5}

k dF- k F- , or D = d - F

[7.8] 29. 27

-

8x + 15

23. 0

24.

-2 - x 20. - - 4 +x

{-U

2

(7.7] 26. 2 9 hr

= -

28. -4

- 14 15. - ( - - ) Sy+2

(7.4] 14. 2

27. 3 mph

30. 27 days

31. m(m + 2) (m + 5)

30. 108y4

32. (x + 3)(x + I )(x + 4)

4k - 45 41. k(k - S)

21 240p2 -3 13. 42m - 84 64p

-

11. (2r + 3) (r + 2)(r - S)

-x2

_2(_p_+_2q_) 3z + I p + S y- 2 3a - I 23. - - 24. - - 25. - - 26. - - 27. p - 2q z+3 p+ l y-3 a+S 28. 2

x-5

16.

r+4 22 - • 3

5

3

-6

4. -3x2y 3

12.

Answers may vary in Exercises 15 and 16.

72 17. p

6x - S 6x - 5 -(6x - 5) -6x + S ) • - 2x , ---, ( - 3 2x+3 - 2x +3 2x+3 a- I 3k - 2 25 3a + 2 5. - - (7.2] 6. 27 7. 3k + 2 8. a + 4 a- l

3. (Answers may vary.)

9. - - (7.3] 10. l50p5 3-x

14. x2 + x + I

2. x ¥- -2,x ¥- 4

(b ) undefi ned

51.

6(3m + 2) 2111 - 5

Chapters R-7 Cumulative Review Exercises 71

5

[R. I] 1. 28 2. 24 [R.2] 3. - 24. 166 4. S.76 [R. l, 1.7] 5. 2 (2.3] 6. { 17} (2.9] 9.

2.sl/. [2.6 ] 7. b = h

I [1 I I I I I I I) -8

[-8, oo)

10.

2}

(2.7] 8. { -:; I I I

( I I) 4

0

(4, oo)

A-22

Answers to Selected Exercises

Section 8.2

[3. 1, 3.2) 11 . •

y

[5.4) 12.

2 I 0y

[5.4] 17.

of 2 and

+ 2k + l

14. 0

7(2z + I ) 24

[7.7] 35. l ~ hr

\/36 = 6 7. vlS

=

5. C

3. fal se; 2V7 represents the product

\122

9.

11.

\142

v'81, or 9

13.

4a 2 -

4ab +

I

x 16 7

16. m6

be simplified.

[6.4) 22. (4x + 1)(2x+ 1)(2x- l ) [6.6] 25. 6 m

26. -2 or - I

~: [7.6] 33.

{T}

33. - 4vlo 35. - 10V7

53. 24

\/36 · v2 =

6\12.

29. 4 Vs 31. It cannot 37. 9\/3 39. 25\12

,r,:

I

65. 5 67. 2 69. 6 v 5

25

71. 4

81. 3x4\12 83. 3c Vs 85. z

[7.8) 36. 2 ft

2

73. m

VZ

75. y

87. a6 Va

63. v'2 5 3 77. 6z 79. 20x

2

89. 8x3 Vx

\(i

103. 3-...yz

y2 x2y3 3 97. 99. 101. 2Vs x5 I0 13 105. 4VZ 107. 2~ 109. p 111. x 3 113. 4z2

115. 1a3 b

117. 21Vzt2

3 6

2

91. x y

34. { - 2, I }

V7 4

55. 6Vl0 57. 12Vs 59. 30Vs 61.

7

3r + 28 7 [7.4) 29. - - 30. ( ) 7r 15 q - 4

[7 .5] 32.

25. 3 vlo 27. 5 v3

41 . s vlo 43. 6\12 45. 9\12 47. 2Vi7 49. 12\12 51. 3\/6

p - l

n

~=

v'72 =

is 36. Simplify as follows: 21. 2 v'6 23. 3 Vs

b2

3 [5.7) 20. 6p2 + 7p + I + - -

-~,

\18

has a perfect square factor of 4. The largest perfect square factor of 72

2

24. {s,

[7. 1) 27. A 28. D

I

[5. l , 5.2) 15.

[5.6] 18.

[6.3) 21. (4t + 3v)(2t + v ) [6.5] 23. { -3, 5 }

V?.

15. 3 17. v'l3,= 19. 3\18 is not completely simplified because

[5.5) 19. 3y 3 + 8y2 + 12y - 5

[7.2] 31.

1. false; Y(-6) 2

1' 01

= -3x+ ;

[4. l-4.3) 13. {(- 1,3) } k2

•=-x'+ .. .

93. 9m n

95. -

m4

119. -

121. 6 cm

2

123. 6 in.

125. D

Section 8.3 1. radicand; index; like; 7; 7 3. (a) like; Both are square roots and

ROOTS AND RADICALS

have the same radicand, 6. is 3 and one is 2.

Section 8.1 1. 1; 4; 9; 16;25;36; 49; 64;81; LOO; 121 ; 144; 169; 196;225;256; 400; 625; 900; 2500 7. true

5. false; Zero has only one square root.

3. true

13. -3, 3

15. -8,8

5

17. - 13, 13

35. ~ 37.

-Ti-

25. l

5

19.-14.14 21. - 30,30

39. 0.8 41. - 0.2

29. LO

31. - 4

75. C

67. 9 and 10 77. c

85. 24 cm

=

69. 7 and 8

17 79. b

95. 9.434 97. 5

8 81. c = 11.705

=

89. 195 ft

87. 80 ft

99. 13

101.

VI3

103.

73. 4 and 5

83. a= 15.199

V34

105. v2

because 33 =27.

119. -6

121. 2

2

115. 9

6

123. 3 125. -5 127. 0.5

135. It is not a real number. 143. (a) 3 units, 4 units

117. -3

129. 3

137. -3

131. 5

139. 2

133. 6

2

5 is a true

145. (a) 2(a + b )(a+ b) I(

(c) 2 a + b

)(

)

I

I

2

(b) PWX: 2ab; PZY: 2ab; PXY: :zc

I

I

I

2

.

.

.

a + b = :zab + :zab + :zc ; When s1mphfied, we obtam

the equivalent equation a2 + b2

=

c2 .

19. 7\/3 21. l l Vs

23. 8\/3 25. -20\12

39. 3v2J 41. LL \/3 43. L9 V7

47. -2\12- 12\/3 49. 10..yz + 4 ~ 55. 7\[J;- 57. 3xv'6 59. ISx \/3 65.

6..;ypi 67. 21 U

69.

-\1'2

45. 12\/6+6Vs

51. 22\12 53.

\!'h

61. 2xv2 63. 13p\/3

-8pf/P 71. - 24z\V4z 73. 0

81. - 6a2 Vx

82. Combining like terms

and combining like radicals are essentially the same process-we use the distributive property to combine the numerical coefficients.

1. (a) I

(b) identity property of multiplication

3. The student

multiplying by a form of l. The correct process is • r::: • r::: v3 · v3

(c) Subtract 2ab from each side to obtain a 2 + b2 = c2 • I

13. 4\/3 15. 2\/6

4. \13

144. (a) (a + b) 2 , or a 2 + 2ab + b 2 (b) c2 + 2ab

I

s\/i7

did not use the identity property of multiplication, which requires

2

=

11.

17. These unlike radicals cannot be added using the distributive

Section 8.4

(b) If we let a = 3, b = 4, and c = 5, then

the Pythagorean theorem is satisfied because 3 + 4 statement.

3

141. - 4

2

4\12 - v2 = 4\12 - I v2 = (4 - I )v2 = 3\12.

- sV?

80. -6(p - 2q) 2 (a + b)

109. The student

67 =

113. 5

~. so both are cube roots of the

75. 42xVsZ 77. 6k 2h\/6 + 27hkv'6k 79. -6x2y

divided 27 by the index, 3, instead of taking the cube root. 111. I

9.

37. 24~

91. 11. l ft 93. 158.6 ft

107. I; 8; 27; 64; 125; 216; 343; 5 12; 729; 1000

1. 1\/3

property.

65. irrational;

71. -7 and -6

\l2ba =

27. -6\12 29. 4 \12 31. 3\/3 - 2Vs 33. 18 ~ 35.

59. rational; -8

61. irrational; - 17.32 1 63. It is not a real number. 34.641

33. - 16

43. It is not a real number.

57. irrational; 5.385

(0 like;

radicals, as follows:

47. 19 49. 19 51. ~ 53. 3x2 + 4

45. It is not a real number. 55. rational; 5

27. 7

(d) like; Both are square roots and have the

same radicand, 2ab. 5. Use the distributive property to subtract like

23. There is no real number that can be squared to obtain - 4. Therefore,

\/=4 is not a real number.

root and one is a cube root.

same radicand, 2x. (e) unlike; The rad icands are different-one is 3y and one is 6y.

9. a must be positive. 11. a must be negative.

(b) unlike; The radicands are different-one

(c) unlike; The indexes are different-one is a square

6Vs 5. - 5

7. Vs

2\/6 9. - 3

4\13

= --.

• r::: • r::: 3v'2 11. 2v3 13. v2 15.

10

3

A-23

Answers to Selected Exercises

17. - 3 V2 19. 2 1Vs 10 5 2 30 21. 3 v's 29. V3o 5 3 39. -1

53. 4 V3 27

51. 3 Vi4 4

47.

5

\!h x

3

SUMMARY EXERCISES Applying Operations with Radicals

25. - Vs 5

35. 8 Vi5 5

V3

33.

VJ

45. 55.

23 V2 . 2

5

V3

31.

0J 3

41. I 43.

6

V65

21.

Vi5

49

10

57 5 -v;;h • b

.

59.

37. V3o 2

10.

p\/3q

~ -

83. -

75. -

77. -

11

5

V1l5-

79. -

81.

3

2ef2

8 ~ 9Vl 7. - -n89. (a) - - sec 4 6

%Y zy

85.

7

€o -

'-Ym -

\Y4

3V2 4. 6V2

7. -3 - 2V2 8. 4v7 + 4Vs 9. - 33

Vt - VJ 11. 2xyz2 ~ t- 3

12. 4"0'3 13.

W

+ I

14.

\/6x ~

' r:: 2V1ls 22. These unlike . radicals . cannot be 20. 52 + 30 v J 21. - 9

Vi5 26. Vs 27. 20V'J 10 ViO 29. - 4- 30. 49 + 14Vx + x

23. 2v7 24. 12\/6 25.

combined.

8(4 + Vx) 28. 16 - x

Section 8.6

(b) 3. 182 sec

3. { 49 } 5. { 121 } 7. 0

1. squaring; squared; original

11.

Section 8.5 1. 13 3. 4 5. x - y

7. It is incorrect to combine -37 and -2vlS

because they are not like terms. This ex pression cannot be simplified further.

V6 + 2\/3 -

3. 2 -

15. ~ 16. 4 Vl 17. -2\1'2 18. 11 - 2V3o 19. 3~

q

xVhJ

73. - 2

v:

5. 73 - 12\/35 6.

0J 6

Wp 4Vm 3aVS,: x~ 61. - - 63. - - 65. - - 67. - - 69. - - y p m 5 6 71. B

Vi5

L -3Vi0 2. 5 -

\/35 11.

9. vlS -

-2v6 13. 30 + 2\/IO 15. 40

17. 57 + 23v'6 19. 7 1 - 160

21. 7 - 2v6 23. 37 + 120

{85}

13. {-45}

25. 0

23. {6}

43. { - I, 3}

31. 23

65. { - 1, 8}

V30 + vl5 + 6VS + 3\/IO 45. 61 - 3\i14t + 2V7/ - 7V2 (a) 5 - v3 (b) v6 + Vs (c) Vx +

17. 0

29. {7}

27. {5}

solution set is { 17}.

25. 187 - 20V2J 27. a+ 2Va + I 29. 49 - 14Vx + x 35. y- 10 37. 2\/3-2+3V2-v6

{-U

19. {8}

31. {6}

45. { 9}

55.{8}

47. { 8} 49. {9}

57. {2}

67. u . 2} 69. {-27,3}

61. {-5}

75. (a) 7 1 mph

(b) 60 mph

80. 6\/13 sq. units 81. h

13 units

63. {- 5}

71. 2 1 73. 8

79. s

=

41. {0,3}

51. { 2, 11}

59. {4,20}

43. 81 + I4V2J

I

33. When the left

35. {-2, I } 37. {1 2} 39. {5}

39. 15V2 - 15 41. 47. 3111 - 211 49.

9. {7 }

21. {I }

side is squared, the result shoul d be x - I , not -(x - I ). The correct

53.{9}

33. I

15.

(c) 54 mph

77. yes; 26 mi =

Vi3 units

82. 3\/13 sq. units 83. 6\/13 sq. units 84. They are both 6\/13.

51. There would still be a rad ical in the denominator. He should

4 - VJ

multiply by --::7::· 53. -2 + Vs 4- v3

57. 3 65.

v3

59.

-3 + 5 \13

61. -\/IO +

II

V6 + V2 + 3\13 + 3 2

vl5

63. 3 -

v3

-6V2 + 12 + ViO - 2Vs 67. 2

69.

-4\/3 -V2 + 10\/6 + 5 23 71.

73.

8(4 - Vx) 16 -x

81.

6 - ViO 2

75.

Chapter 8 Review Exercises

55. - 2 - v'TI

Vx-\/Y x-y

V2T + Vi4 +

\¥4

91. I

v6 + 2

-~ \/3+5 77. v7 - 2 79. - - 4

93. (a) 4 in. (b) 3.4 in.

95. 30 + 18x 96. They arenotlike terms.

97. 30 + 18Vs

98. They are not like radicals. 99. Make the first term 30x, so that 30x + I8x = 48x. Make the first term 30Vs, so that 30Vs + 18Vs = 48 Vs.

100. Both like terms and like radicals are

combined by adding their numerical coefficients. The variables in like terms are replaced by radicals in like radicals.

12. -30

7. 4 8. -6 13. 10

9. ~

10. 0.5

14. 3 15. -8 23.

VsJ

11.

16. -5

18. rational; 13 19. rational; - 25 21. 8 22. 6.633

2+V2 83. - 385. 2 - 3\1'4 87. 12 + 10~

89. - I + 3~ -

1. -7, 7 2. - 9,9 3. - 14, 14 4. - 11 , 11 6. -27, 27

5. - 15, 15 It is not a rea l numbe r.

17. irrational; 8.544

20. It is not a real number. 25. 4 v3

24. 48.3 cm

26. -3\/3 21. 4\/IO 28. 2~ 29. 5 '313 30. 5\/3 v6 V3 34. -5V3 35. 4 36. -Vs 2 31. 18 32. 16 6 33. - 76- 37. i5 38. 3V2 39. 8 40. 2V2 41. r9

42. x 5y 8 43. a7bJOy/;;b

44. I Ix3y5 45. y 2 46. 6x5 47. 8v'TI

48. 9V2 49. 2 1v3

50. 12\/3 51. 0 52. 30 53. 2\/3 + 3\/IO 54. 2V2 55. 6\i30 60. 66.

8

56.

5Vx

Vi0 61. v6 5

rVx -

67.

V'9 - 3-

57. -mVs 58. 1Ik2 vZn 59. Vs 62. V3o 15

68 68. - -

7 71. 22 - 16\/3 72. 2V21 4x

63. ViO 5

1

64. 8

65.

v'42 21

69. -vlS - 9 70. 3v'6 + 12

Vi4 +

12V2 -

4\13

73. - 2

A-24

Answers to Selected Exercises

-2 + 6Vz 74. x + 4 Vx + 4 75. - I + V2 7 6. - - - 17 2\/3 + 2 + 3Vz + W 3 + 2W I - 3V7 77. 2 78 . 3 79. 4 82. 0

80. 3+4\13 81. {25} 86. { - 2}

87. { -3,- 1}

91. { - I }

92. { 4}

89. { - 2}

34. -V3+Vs

[6.5] 35. {3,4}

(7.6) 38. { 19}

CD-AD 39. B = AC_ I

r~

(b) 347 species

14. { 8}

+

v'i3

2

-9 is also a solution, so her answer was not complete. The correct

13. {± II } 21. { ± 9}

Chapter 8 Test

(d ) A

2. (a) irrational (b) 11.9 16

(e) C

3. (a) B

2~

(f) A

43. {±2\/6} 49. { -2, 8}

12. -

- 13. 6V2 + 2 - 3Vl4 - v7 3 14. -5-v':h [8. 1J 15. (a) 6V2 in. (b) 8.485 in. 16. 50 ohms 5V14 [8.4) 17. - 7 21. -I + -Vz 2

\/fu

18. ~

[8.6) 22. 0

3

-'-Yi

19.

[8.5) 20.

- 12 - 3\/3 13

3

[I.I , 5.1, 8.1 ) 5. (a) 16 (g) It is not a real number.

(2.3] 9. { 3}

[2.9) 10. [ - 16, oo)

(d) - 16

7. 6

(e) 2

x+

0

=20

0 -2

[2.7) 12. (a) 9

x

x

4 -

-; 0

=2

[3.3-3.5) 16. (a) 16.6; The number of subscribers increased by an average of 16.6 million per year. (c) 429.1 million subscribers (x 18. {(x,y)l2x-y=6}

=

19. 0

(b) y

8)

y'5

y'5

. , or 23 36 5832

[5.7] 24. 412

-

l6.6x + 296.3

[5.1] 21. I2x 10y2 2

(5.4) 23. 3x3 + I Ix - 13

St+ 5 [6.2-6.4) 25. (m + 8)(111 + 4) 27. (9z + 4) 2 [7.2) 28. (t + 5)(1 + 3),

26. (6a + 5b)(2a - b)

{ 3}

57. 0, 2

59.

41. { ± 3\/3}

47. {

± 2 ~}

53. no real solution

{5 ± 2v'30}

{ - I ~ \/3 }

1

71. {

±:\/3} I

75. {-3.09, -0. 15 }

77. 2 sec 79. 9 in.

1. D 3. 25; (x + 5) 2 5. 100; (z - 10) 2 9.

>~

¥; (P - ~)2

11.

~; (x + ~)2

19. {- I ± \/6} 27. {- 1,

n

{9+ \/21}

39.

{-¥. I}

-

47. {

3 ±

6

21. {4 ± 2\/3}

23. {-3}

35.

41. { -

v'i3}

2

13. 0.04; (x - 0.2) 2 17. { 1, 3}

29. {- ~.

33.

7. I; (x + 1) 2

Vs}

15. 4;2; 4;4;x+2;5;{-2 ± x

[4.1-4.3) 17. { (3, -7)}

[4.4) 20. carfromChicago:

6 1 mph; car from Des Moines: 54 mph (5.2] 22.

=

Y7}

35. {±V3}

Section 9.2

14. • . [3.6] 15. •

.x

2

27. no real solution

(f) -2

(b) 70% [3.2) 13. •

33. {±8}

{-~.-n

81. 2%

[ 1.3) 8. 3

11. (5, oo)

25. { ± 4 \/3}

51. {8 ± 3 \/3}

73. {- 4.48, 0.20}

4

[I. I] 6. 54

45. { ±

67. {-i, 1} 69.

3. 5; 80% 4. 1.25; 125%

(b) 16 (c) - 16

Vi4}

19.

11. {3,5}

±3 3Vz} 63. {- 10 ± 4 •v r::} 3 65. {0, 4'}

61. { - I

Chapters R-8 Cumulative Review Exercises [R.2] 1. 0.04; 4% 2. 20; 0.15

{-3, 35}

55.

24. {'4. I } 25. {-4}

23. {3}

23. { ±

{-~,6}

17.

37. no real solution 39. { ± 3Vz}

~

11. -6x~

15. {±13}

29. {±fl 31. {±0.5}

(b) F

[8.2] 4. SVz 5. 6. 4\16 5 [8.3-8.5) 8. 31 9. 9v7 10. 11 +2\/3o

7. 2y~

3. C 5. D 7. According to the square root property,

solutionsetis{-9,9},or{±9} . 9. {-7,8}

not satisfy the original equation. The solution set is 0.

(c) D

[8.6) 40. { 16}

Section 9.1 1. B, C

15. (a) 57 species

16. 12 is not a solution. A check shows that it does

(8. 1) 1. -20, 20

36.{ -2,- 1} 37. { 12}

QUADRATIC EQUATIONS

V2

13. 0

(x+3)(x-l)

90. {4}

1. 2 Vi0 2. 5 3. - 5 4. 5yV2 5. 4 6 2 5 23 3s · 7. - II 8. 7-2W 9. -Vi0- 5v'i5 10. 166 +2v7 12. {7}

-2x - 14 30. - - - - -

(y + l )(y - 1)

[7.5J 31. -21 [S.31 32. 26\!3 [8.51 33. 21 - 5V2

Chapter 8 Mixed Review Exercises

11. II V3

y2

(7.4) 29.

84. { I} 85. {2}

83. {48 }

88. {-2,- 1}

or t 2 +St+ 15

31. no real solution

{-7 + V97} -

6

.

r; 37. { 1 ± v6}

~. 6} 43. { -4, 2} 45. {4 ± \/3}

49.

53. (a) { 3 ± 2\/6} 3

n

25. {O, 16}

{ -3 ± 2

Vs}

51.

(b) { -0.633, 2.633}

{2 ±

Vi4}

2

Answers to Selected Exercises

\/3}

55. (a) {-2 ±

59. I sec and 5 sec

61. 75 ft by I 00 ft

66. x2 + 8x 16 squares.

+ 8x + 16

67. x2

57. 3 sec and 5 sec

(b) { -3.732, - 0.268}

65. x2

63. 8 mi

68. It occurred when we added the

-7±Vs} 4

31. {

5 {- 83, - 6}

37.

{ -~,2} 38. ~'

n

- 5 ±Vs} 2

.

34.

41. {-

Section 9.3

32. {

36.

35. no real solutwn

{-H}

33.

A-25

{ 8±8V2} 3

{± 9IO}

39. {-4,n 4o. {-3, 5}

42. no real solution

1. E 3. A 5. D 7. a = 3, b = 4, c = -8 9. a = 8, b = -2, 13. a= 3, b = 7, c = 0

11. a= 3, b = - 4, c = -2

c = -3

15. a = l, b = l,c = - 12 17. a= 9,b=9,c=-26 19. 2a should be the denominator for -bas well. The correct formula is -b±Vb 2 -4ac . 21. { - 13, I } x= 2a

±vu}

27. { - 6 2

25. {2}

33. { ±r./6} 35. { ±

n

23.

{

6

±

47.

- I + -

2

~

y

~·

3

11. 3

-7

31. 0,7

(0, -6) 13.

4

67. r =

4

x 4

2

8

4

y =-x + x -

(4,0)

69. 3.5 ft 71. { 16, -8}; Only 16 ft

7T

H O 4

~

(3,4)

(-2, - 4 ) 23.

,_,

53. {-0.8, 0.3}

y

Y '= - X + I

21. .

2

4

x x

2-

3

2

y =x

2

V 7T2h2 + 7TS

6

3

O·

{-3,234} 57. {-I ±6 Yn} 59. no real solution 2 65. {-3 ± v'29} v 2 63. {- 1, 5} 61. { I ± -r::} ±

y - x 2 + 4x

17.

2 '] ·Y = x - 8x + 16

55.

- 7Th

(-3, 0)

(0,2) 15. •

(b) {-0.618, 1.618}

51. { - 1.8, 0.3}

6 -

=- x 2 + 2

y

;w} V61}

y = x + 3J'_y

3 -3 0 • 9.

-3 0

{ 12}

v'll} (b) {-2.158, 1. 158}

(a){ ' \Vs }

49. { -3.6, 0.6}

7.

5} - 1, :;

- 5 ± 39. no real solution 41. no real solution 43. { 2

45. (a) {

1. parabola 3. axis of symmetry; mirror 5. upward; downward

-

29. {- 1,0}

37. {

Section 9.4

+4.r +4

(0, I); (0, I);

(0, 1);(0, l );

(-2, 0); (0, 4);

no x-intercepts

(- 1, 0), ( !, 0)

(-2, 0)

25.

y

y;;-;'-:'47+3

27.

kY

29.

y

is a reasonable answer. 0 2

SUMMARY EXERCISES Applying Methods for Solving Quadratic Equations

{2 } 4. {± 97}

1. { ± 6} 2. { -3 ± Vs } 2 5. { I, 3}

6. { - 2, - I}

3. 5, 4

{I }

7. 4, I

8. { - 3±v'l7} 2

{I ± 2v'tO} 11. {-6} 12. {-5,7 I} Yl} 15. no real solution ± 2V6} 14. {' ± 4 7 3

3} 10. 9. {-3,I 5 13. { 7

16. no real solution 17. {- ~' 2} 18. {- ~' 19.

{ -~.

22. { -

5

n ±6

20. {-3,

Vi3}

2 25. { - 3 ±WI}

n

21. { , ±

23. {-4, 6}

5}

26. { -4

I ±2-\/3} 29. { - 2 ±3 28. { -

v'S}

27. {4, 5}

v'll} 30. { - 5 ±WI} 8

U

.+ -J TT

Cx

-6~~

;t°J

y =x - 4x-

(2, - I); (0, 3);

(- 1, 2); (0, 3);

(2, -6); (0, -2);

( I, 0), (3, 0)

no x-intercepts

(-0.4, 0), ( 4.4, 0)

31. 40 and 40 33. y = 5 !~5 x2

Chapter 9 Review Exercises 1. {-7, 4}

2. { - 9,-5}

5. {± 10}

6. {± 13}

4. n. 1}

3. {-5, n

7. {± 12}

8. { ± v'37}

v'tO} - \114} 13. no real solution 14. {-35} 12. 2 15. { -5, - I} 16. {-2 ± v'll} 17. {- I ± W} {

Vi}

=xz~

t - t t- J T

9. { ± 8V2}

I}

24. {- 3 ±

:Y.

10. {-7, 3}

11. {3 ±

- I +

18. { - 4 -+Vii} 2

19.

{- 7 2}

20. no real solution

21. 2.5 sec 22. 6, 8, I0 23. { I ± Vs } 25. {

24. { - ~' I}

v'tO} 26. {- 1 ±4 \/29} 27. { -3 ±WI} 2

2 ±2

A-26

Answers to Selected Exercises

Chapters R-9 Cumulative Review Exercises

28. no real solution 29.

30.

1~+5

[LS] 5. S 6. -~,or- I ~

(-4, 0)

11. L

32. mm~:s

[2.9) 13.

31. •

P

P -2W

=-

, or L = 2 - W

2

(1

I I I

m

2·

-2 1

I)

14.

1]

I I I

EI

0

I I I•

4

(-oo, 4]

(-2,oo)

y =x2+4x+ ~

-x2.;2r +

8. {2}

[2.7] 12. 44%

-2

0

-2

[2.1- 2.4) 7.

10. width: so ft; length: 94 ft

[2.6) 9. 100°, 80°

-4

(O,S)

4. 18.64

[R.I] 1. 14, or l 14 2. 48 [R.2] 3. 0.012

0 2

=

29

9

23

y

[3.2) 15.

( I, 4) ; (0, 3);

(-2, -2); (0, 2);

(- 1, 0 ),(3, 0)

(-3.4, 0), (-0.6, 0)

1111

16• •

:OS

0

0

x

=6

2r+3y

Chapter 9 Mixed Review Exercises i. {

-T· s}

2. {

-T, n 3· {-i. H 4• {-H 6. { -s±V17} 2

5. { - 1 ±V21} 2

8. { - I ± •v r,:j}

9.

{9 ± v41}

12. { ± 2Vz}

11. { -2 ± Vs}

I

-3 [3.4, 3.S) 18. 2x - y = -3

(4.1-4.3) 19. { (-3, 2)}

7. no real solution

10. { - I

2

[3.3) 17.

± 2.v 2} r,:

13. (a) { ± 3}

x4

bl6

24. - 2 [S. I , S.2] 23. c 9 [S.4] 25. 8x5 - 17x4 - x 2

[4.SJ 22.

(b) { ± 3}

[S.S] 26. 2x4 + x 3

19x2 + 2x + 20

-

[S.7] 27. 3x2 - 2x + I

(c) { ± 3} {d) Yes; Because there is only one solution set, we will always get the same resu lts, no matter which method of solution we use.

20. 0

[4.4) 21. 2010: 78 mi llion smartphones; 2016: 262 million smartphones

7

28. 2x + 2 + - - [6. 1) 29. 16x2 (x - 3y)

x-3

14. 0.7 sec and 10.4 sec rnrTTTTTrr 15. rrri..,. (3, - 1)

(6.3) 30. (2a+ J)(a - 3)

8

32. (Sm - 2) 2

[6.4) 31. (4x2 + 1)(2x+ 1)(2x-1)

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A Absolute value evaluating, 48 explanation of, 47 Addition associative property of, 79 commutative property of, 78 of decimals, 17-1 8 of fractions, 7-9 with grouping symbols, 29, 56--57, 77 identity element for, 80 identity property of, 80-8 1 inverse for, 81 of multivariable polynomials, 374 of negative numbers, 52 on a number line, 51-53 in order of operations, 29, 77 of polynomials, 372-373, 374 of radicals, 587-589 of rational expressions, 506-509 of real numbers, 51-54, 56--58 of signed numbers, 51-54 summary of properties of, 84 of terms, 380 word phrases for, 57-58 Addition property of equality, 104-109 of inequality, 183-1 85 Additive identity element, 80 Additive inverse explanation of, 47, 81 fi nding for real numbers, 46-47, 55 symbol for, 56 Algebraic expressions distinguishing from equations, 39-40 evaluating, 36--37 explanation of, 36 from word phrases, 37- 38, 9 1, 123 simplifying, 82, 88- 9 1 Angles complementary, 139-140 measure of, 139-140, 148-149 right, 139 straight, 139, 148-149 supplementary, 139-140 vertical, 148-149 Approximately equal to explanation of, 571 symbol for, 57 1 Area of a circle, 152, 632 rules for exponents for, 347-348 of a trapezoid, 146, 152 of a triangle, 152

Associative properties distinguishing from commutative, 79-80 explanation of, 79, 84 Average, 76 Axis of a coordinate system, 212 of a parabola, 375, 650 of symmetry, 375, 650 B Base of an exponential expression, 28, 342 in a percentage discussion, 162 Binomials explanation of, 371 factoring, 440 finding greater powers of, 390 finding product of the sum and difference of two terms, 388- 389 multiplication by FOIL method, 382-383 rationalizing denominator, 60 1-602 squares of, 387- 388 steps to multiply by FOIL method, 382 Boundary line explanation of, 263 graphing linear inequalities, 263-266 Braces explanation of, 29 as a grouping symbol, 29 for set notation, 38 Brackets as grouping symbol, 29-3 1 Break-even quantity, 322 Brin, Sergey, 366

c Cartesian coordinate system explanation of, 2 12 plotting points on, 213-214 Circle(s) area of, 152, 632 circumference of, 152 graph, 11 Circumference of a circle, 152 Coefficient(s), 88, 369 Common denominators, 7-8 Common factors, 3, 414 Commutative properties distinguishing from associative, 79- 80 explanation of, 78, 84 Complementary angles, 139-140 Completing the square method for solving quadratic equations, 632-638

Complex fractions explanation of, 5 14 simplifying, 515-520 steps to simplify , 515, 5 17 Complex numbers, 628 Components of an ordered pair, 268 Composite numbers, 2 Compound interest, 350 Concours d'Elegance, 5 14 Conditional equation, 12 1-1 22 Conjugates explanation of, 390, 601 multiplying, 601 -602 of radicals, 601-602 Consecutive integers even, 138, 462 explanation of, 137, 46 1 odd, 138-1 39,462 solving problems involving, 137-139 Consistent system, 291-292 Constant(s), 36 Constant of variation, 547 Contradiction, 122 Coordinate(s) of a point, 44, 212 Coordinate system Cartesian, 212 explanation of, 212 origin of, 212 quadrants of, 212 rectangular, 212 Cost, unit, 158 Counting numbers. See Natu ral numbers Cross products of a proportion, 159-160 Cube(s) difference of, 444-445, 446 of a number, 28 perfect, 444, 584 sum of, 445-446 volume of, 587 Cube root(s) application of, 573 explanation of, 573 rationalizing denominators with, 595-596 symbol for, 573 D Data, interpreting, 48, 59 Data sets, 256-257 Decimal(s) adding, 17- 18 approximating irrational numbers with, 571 dividing, 19-20 equivalents, 21

1-1

1-2

Index

Decimal(s) (continued) explanation of, 16 linear equations with, 128- 129 multi plying, 18, 20 place value in, 16 repeating, 2 1, 43 rounding of, 19-20 solving linear systems with, 302 subtracting, 17-1 8 terminati ng, 21, 43 writing as percents, 2 1-22, 162 writing fractions as, 20-2 1 writing percents as, 21-22, 162 written as fractions, 17 Deci mal places, 18 Degree explanation of (angle measure unit), 139 of a polynomial, 37 1 of a term, 37 1 Denominator(s) adding fractions with different, 8-9 adding fractions with same, 7 common, 7-8 with cube roots, 595-596 explanation of, 1, 3-4 least common, 8, 499- 501 in rational expressions, 482, 501 -502 rationalizing, 592-596, 601-602 Dependent equations explanation of, 291-292 substitution method for solving, 300-301 Depreciation, 230 Descartes, Rene, 2 12 Descending powers, 370 Difference ofcubes,444-445,446 explanation of, 9, 54 of squares, 439-44 I, 446 Direct variation explanation of, 547 as a power, 549 Discriminant of quadratic formula, 644 Distance, rate, and time relationship as example of a fu nction, 270 formulas for, 174 solving problems, 175- 176, 318- 320, 538-539 Distance formula application of, 573 explanation of, 174, 572 for falling objects, 452, 460 Distributive properties addition property of equality and, 109 explanation of, 82-84 Dividend, 6, I 9, 68 Divisibility tests for numbers, 76, 4 14 Division of decimals, 19-20 explanation of, 68 of fractions, 6- 7 long, 394-399

multiplication property of equality extended to, 113 in order of operations, 29, 77 of polynomials, 393-399 by powers of ten, 20 of radicals, 58 1-582 of rational expressions, 494-496 of real numbers, 67-73 with scientific notation, 363 of signed numbers, 68 word phrases for, 72 by zero, 69 Divisor(s) explanation of, 6, 19, 68 factors as, 4 I 4 Domain explanation of, 269 finding for functions, 272 identifying for relations, 269 Double negative rule, 47 Double solution, 456-457, 644

E Earthquake, intensity of, 368 Elements of a set, 38 Eli mination method explanation of, 305 for solving equivalent equations, 3 10 for solving linear systems, 305-3 10, 312 steps to solve by, 306-307 Elimination number, I 56 Empty set explanation of, 122 symbols for, 122 Equality addition property of, I 04-109 multiplication property of, 112-116 squaring property of, 607- 6 11 symbol for, 3 1, 38 Equation(s) conditional, I 2 1-1 22 defini ng functions, 269- 270 dependent,29 1-292 distinguishing from expressions, 39-40, 523- 524 equivalent, 104, 310 explanation of, 38, 104 extraneous solution of, 608 from sentences, 72-73 of a horizontal line, 224-225 independent, 29 1-292 linear in one variable. See Linear equations in one variable linear in two variables. See Linear equations in two variables literal, 149-15 1 with no solutions, 608-609 quadratic. See Quadratic equations simplify ing before solving, 109, 115- 116 slope of, 236-237

solution set of, 104, 196 solutions of, 38, 104 solving with radicals, 607-612 solving with rational expressions, 524-531 square root property of, 627-630 translating sentences into, 72-73 of a vertical line, 224-225 Equilibrium demand, 297 Equilibrium supply, 297 Equivalent equations, 104, 3 10 Equivalent forms for a rational expression, 487-488 Even consecutive integers, 138, 462 Exponential expressions base of, 28, 342 evaluating, 28 explanation of, 28, 342 Exponents explanation of, 28, 342 integer, 350-357 negative, 35 1-354, 356 in order of operations, 29, 77 positive, 354, 356 power rules for, 344-347, 356 product rule for, 342-344, 346- 347, 356 quotient rule for, 354-356 and scientific notation, 361-364 scientific notation and, 361, 362 summary of rules for, 346, 356 zero, 35 1, 356 Expressions algebraic. See Algebraic expressions distinguishing from equations, 39-40, 523-524 equality and inequality symbols in, 33 exponential, 28, 342 from word phrases, 9 1 quadratic, 452 radical. See Radical expressions rational. See Rational expressions simplifying, 82, 88-9 1 terms of, 88- 89, 369 Extraneous solution, 608 Extremes of a proportion, 158

F Factor(s) common, 3, 4 14 distinguishing between terms and, 89 explanation of, 2, 414 greatest common, 3, 4 14-418 of integers, 67 of numbers, 2, 4 14 prime, 2 Factor tree, 2 Factoring binomials, 440 difference of cubes, 444-445 , 446 difference of squares, 439-441

Index

explanation of, 4 14 with four terms by grouping, 4 19--420 greatest common factor, 4 16--418 by grouping, 4 18--420 guidelines for, 426 perfect square trinomials, 441--443, 446 solving quadratic equations, 626-630 special techniques, 439--446 sum of cubes, 445--446 summary for polynomials, 449 summary of special techniques, 446 trinomials, 423--428, 430--436 zero-factor property, 452--458, 626 Fibonacci, 568 First-degree equation. See Linear equations Fixed cost, 253 FOIL method explanation of, 382 factoring binomials, 440 factoring trinomials, 424, 433--436 inner product of, 382 outer product of, 382 Formulas, explanation of, 146 Fourth root, 573 Fraction(s) adding, 7-9, 10 applications of, 10 complex, 514-520 denominator of, 1 dividing, 6-7 equivalents, 21 explanation of, 1 improper, 1, 4 least common denominator of, 8, 499-501 linear equations with, 125-127 linear systems with, 301-302 lowest terms of, 3 mixed numbers, 4, 8, 10 multiplying, 5- 6 numerator of, l operations on, 5-10 proper, 1 quadratic equations with, 645 reciprocals of, 6, 67, 81 simplifying, 3 subtracting, 9-10 writing as percents, 23 writing decimals as, 17 writing percents as, 22-23 written as decimals, 20-21 Fraction bar explanation of, 1, 68 as grouping symbol, 29-3 1 Functions domain of, 272 equations defining, 270-27 1 explanation of, 269 input-output, 270 linear, 277 notation, 273

range of, 272 as relations, 268-269 vertical line test for, 27 1 Fundamental property of rational expressions, 484 f(x) notation, 273

1-3

H

Horizontal lines equation of, 224-225 explanation of, 224 graphing, 224-225 slope of, 235-236 Hypotenuse of a right triangle, 463

G

Galileo Galilei, 452, 460 Garfield, James A., 578 Geometry formulas, explanation of, 146. See also specific formulas Googol, 366 Graph(s) circle, 11 explanation of, 2 19 line, 208-209, 246--247 of linear equations, 2 19-226 of parabolas, 375 pie charts, 11 region of, 262- 266 Graphing explanation of, 2 19 horizontal lines, 224-225 inequalities, 182-183 linear equations in two variables, 2 19-226,288-292 linear inequalities in one variable, 182-183 linear inequalities in two variables, 262- 266 numbers, 44 ordered pairs, 2 12- 2 13 parabolas, 375, 649-653 polynomials, 375 quadratic equations, 649- 654 rational numbers, 44 solving linear inequalities by, 326- 328 vertical lines, 224-225 Greater powers of binomials, 390 "(is) Greater than" explanation of, 46 orderi ng of real numbers, 46 symbol for, 3 1, 182 "(is) Greater than or equal to," 3 1, 182 Greatest common factor explanation of, 3, 4 14 factoring out, 4 16--418 finding, 4 14--416 of numbers, 4 14 steps to fi nd, 4 15 for variable terms, 4 15--416 Grouping factoring by, 4 18--420 factoring by fo ur terms, 4 19--420 factoring trinomials by, 430--432 Grouping symbols addition with, 56--57 explanation of, 29 subtraction with, 56--57

Identity, 122 Identity element for addition, 80 for multiplication, 80 Identity properties, 80-8 1 Improper fractions converting between mixed numbers and, 4 explanation of, 1 Incidence rate, 49 1 Inconsistent system explanation of, 291- 292 substitution method for solving, 300 Independentequations, 291-292 Index of a radical, 573 Inequalities addition property of, 183-185 applied problems using, 189- 190 explanation of, 31 , 182 graphing, 182-183 linear in one variable. See Linear inequalities in one variable linear in two variables. See Linear inequalities in two variables multiplication property of, 185-187 solving linear, 182-192 symbols of, 31 - 33, 45--46, 182 three-part, 190-192 Infinity symbol, 182 Inner product, 382 Input of an ordered pair, 270 Input-output machine, 270 Integers consecutive, 137- 139, 461 consecutive even, 138, 462 consecutive odd, 138- 139, 462 explanation of, 42--43 as exponents, 350-357 factors of, 67 Intensity of an earthquake, 368 Intercepts graphing, 221-223 of a linear equation, 221-223, 245-249 of a parabola, 650-654 point-slope form, 253-254 of a quadratic equation, 650-654 slope-intercept form, 245-249 Interest compound, 350 formula for, 170 simple, 153, 172-173

1-4

Index

Interval notation, 182-183, 192 Interval on a number line, 182-183 Inverse additive, 46-47, 55-56, 81 multiplicative, 67, 8 1 Inverse properties, 81, 84 Inverse variation explanation of, 550 as a power, 550 Irrational numbers decimal approximation of, 57 1 explanation of, 44, 570 L

Leading term, 369 Least common denominators (LCDs) explanation of, 8, 499 fi nding, 8, 499-50 1 rational expressions and, 524 steps to find, 499 Legs of a right triangle, 463 Leonardo da Pisa, 568 "(is) Less than" explanation of, 46 ordering of real numbers, 46 symbol for, 3 1, 182 "(is) Less than or equal to," 3 1, 182 Light-year, 409 Like radicals, 587 Like terms combini ng, 89, 369- 370 explanation of, 89, 369 Line(s) graphing, 208- 209, 246-247 horizontal. See Hori zontal lines number. See Number line parallel, 238-240 perpendicular, 238-240 slope of, 23 1-240 of symmetry, 375 vertical. See Vertical lines Line graphs explanation of, 208 interpreting, 208-209 Linear equations in one variable applications of, 132- 140, 169- 176 with decimal coefficients, 128-129 explanation of, 104 with fractions, 125-127 geometric applications of, 146-148 with infi nitely many solutions, 12 1-1 22 with no solutions, 122 solving, 117-123, 125-129 steps to solve, 117 Linear equations in two variables explanation of, 209 graphing, 2 19- 226, 288- 292 intercepts of, 22 1-223 point-slope form of, 253- 254, 255 slope-intercept form of, 245-249, 255 slope of, 236-237

solution of, 209 standard form of, 209, 253, 254-255 summary of forms of, 255 systems of. See Systems of linear equations use to model, 225- 226, 256- 257 Linear fu nctions, 277 Linear inequalities in one variable graph of, 182- 183 solution of, 187- 188 steps to solve, 187 Linear inequalities in two variables boundary line of, 263-266 explanation of, 262 graphing, 262-266 steps to graph, 265 system of. See System of linear inequalities Linear systems. See Systems of linear equations Literal equations, 149-15 1 Long division, 394-399 Lowest terms of a fraction, 3 of a rational expression, 484-487 writing radical expressions with quotients in, 603 M

Magic number, 156 Mathematical model, 42 Means of a proportion, 158 Measure of an angle, 139-140, 148-149 Minuend, 54 Mixed numbers applications with, 10 converting between improper fractions and,4 explanation of, 4, 43 Mixture problems formula for, 170 gasoline-oil , 136 involving percent, 3 16-3 17 steps to solve, 170-172 Model(s) mathematical, 42 quadratic, 464-465 using a linear equation to, 225- 226, 256-257 Money denomination problems, 173-174 Monomials dividi ng polynomials by, 393-394 explanation of, 37 1 multiplying, 380 Motion problems. See Distance, rate, and time relationship Multiplication associative property of, 79 of binomials, 382- 383 commutative property of, 78 of conj ugates, 601-602

of decimals, 18, 20 distributive property, 82-84, 109 FOIL method of, 382-383 of fractions, 5-6 identity element for, 80 identity property of, 80--81 inverse property for, 81 of a monomial, 380 in order of operations, 29, 77 of a polynomial, 380-382 by powers of ten, 20 of radicals, 579-58 1 of rational expressions, 492-494, 496 of real numbers, 65-67, 70-72 with scientific notation, 363 of signed numbers, 65-67 of sum and difference of two terms, 388-389 summary of properties of, 84 of terms, 380 word phrases for, 7 1-72 by zero, 65 Multiplication property of equality, 112-116 of inequality, 185-187 of zero, 65 Multiplicative identity element, 80 Multiplicative inverse, 67, 81 Multivariable polynomials adding, 374 subtracting, 374 N

Nanometer, 364 Natu ral numbers explanation of, 1, 42 negative of, 42-43 opposite of, 42-43 Negati ve exponents changing to positive, 354, 356 explanation of, 352, 356 scientific notation and, 362 Negative infinity symbol, 182 Negative numbers adding, 5 1- 52 dividing, 68 explanation of, 42-43 as exponents, 35 1- 354, 356 multiplying, 66-67 subtracting, 55-56 symbol for, 56 Negative slope, 235 Negative square roots, 568 "(is) Not equal to," 31 Notation function, 273 interval, 182-1 83, 192 scientific, 36 1-364 set-builder, 43, 192, 291 standard, 363 subscript, 233

Index nth root, 573 Null set explanation of, 122 symbols for, 122 Number(s) absolute value of, 47-48 composite, 2 cube of, 28 divisibility tests for, 76, 414 factors of, 2, 41 4 fractions, 1 graph of, 44 greatest common factor of, 4 14-418 integers, 42-43 irrational, 44, 570-57 1 mixed, 4, 8, 10, 43 natural, 1, 42 negative. See Negative numbers opposite of. See Opposite(s) ordering of, 46 perfect square, 570 positive, 43 prime, 2, 4 15 prime factors of, 2 rational, 43-44 real. See Real nu mbers reciprocal of, 6, 67 signed . See Signed numbers solving problems about, 537 square of, 28, 568 square roots of, 568 whole, 1, 42 Number line addition on, 51-53 explanation of, 42 graphing a number on, 44 graphing intervals on, 182-183 subtraction on, 54 Numerator(s) explanation of, 1, 3-4 in rational expressions, 482 Numerical coefficient, 88, 369 Numerical expressions evaluating, 70-72 from word phrases, 57- 58, 7 1-72

0 Odd consecutive integers, 138- 139, 462 One, properties of, 3 Opposite(s) as additive inverse, 46-47 , 55, 81 as negative number, 42-43, 52 quotient of, 487 Order of a radical, 573 Order of operations, 29, 56-57, 77 Ordered pairs completing, 2 10-2 11 components of, 268 explanation of, 209 function as set of, 269 graphing, 2 12-213

input of, 270 output of, 270 plotting, 212-213 as solutions, 2 10 table of values, 2 11-212 writing solutions as, 209 Ordering of real numbers, 46 Origin boundary line through, 265-266 explanation of, 2 12 line passing through, 223 Outer product, 382 Output of an ordered pair, 270

p Pairs, ordered. See Ordered pairs Parabola(s) axis of, 375, 650 axis of symmetry of, 375, 650 explanation of, 375 graph of, 375 graphing, 650-653 intercepts of, 650-654 line of symmetry of, 375 steps to graph, 652 vertex of, 375, 649 Parallel lines explanation of, 238 slopes of, 239 Parentheses, 29 Percents and percentages applied problem involving, 24 applied problems involving, 163 equi valents, 2 1 explanation of, 2 1, 162 mixture problems involving, 3 16--3 17 writing as decimals, 2 1-22, 162 writing as fractions, 22 writing decimals as, 22, 162 writing fractions as, 23 Perfectcubes,444, 573 , 584 Perfect fourth powers, 573 Perfect square trinomials creating, 632-633 explanation of, 441 factoring of, 441-443 Perfect squares, 440, 570 Perimeter explanation of, 15, 147 ofarectangle, 147, 152 ofa triangle, 147-148, 152 Perpendicular lines explanation of, 238 slopes of, 239 Pi (rr) approximation of, 152 explanation of, 45, 570 Pie charts, 11 Pisa, Leonardo da, 568 Place value in decimals, 16 Plane, 212

1-5

Plotting ordered pairs, 212-2 13 Plotting points, 2 12-213, 650 Plus or minus symbol, 627 Point-slope form, 247-249, 253-254 Polynomial(s) adding,372- 373, 374 binomials. See Binomials classifying, 371 coeffi cients of, 369 degree of, 37 1 of degree two, 375 in descending powers, 370 dividing, 393-399 evaluating, 372 factoring sum mary, 449 graphing, 375 graphing equations defined by, 375 leading term of, 369 monomials. See Monomials multiplying, 380-382 multivariable, 374 numerical coefficients of, 369 operations on, 372-374, 380-382, 393-399 prime, 426 special factoring techniques, 439-446 subtracting, 373- 374 terms of, 369 trinomials. See Trinomials in x, 370 Positive exponents changing from negative to, 354, 356 scientific notation and, 362 Positive infinity symbol, 182 Positive numbers, 43 Positive slope, 235 Positive square roots, 568 Power(s) descending, 370 direct variation as, 549 explanation of, 28, 342 inverse variation as, 550 Power rules for exponents, 344-347, 356 Powers of ten division by, 20 explanation of, 16 multiplication by, 20 Price per unit, 158 Prime factors of a number, 2 Prime nu mbers, 2, 41 5 Prime polynomials, 426 Principal square root, 568, 608 Product explanation of, 2, 65 simplifying for radical expressions, 599- 601 of the sum and difference of two terms, 388- 389 translating words and phrases, 7 1 Product rule for exponents, 342- 344, 346- 347, 356 for radicals, 579-58 1 Proper fractions, 1

1-6

Index

Properties of equality, 104-109, 112- 116, 607- 608 of inequality, 183- 187 of one, 3 of radicals, 584 of real numbers, 65, 78-84 Proportions applications of, 16 1 cross products of, 159- 160 explanation of, 158 extremes of, 158 means of, 158 solving, 158- 160 terms of, 158 Pyramids, volume of, 153 Pythagorean formula explanation of, 463, 57 1 proof of, 578 Q

Quadrants, 212 Quadratic equations applications of, 460--465, 638 completing the square method for solving, 632-638 with double solution, 644 explanation of, 452, 626 with fractions, 645 graphing of, 649- 654 intercepts, 650- 654 with no real solution, 628- 629 with one distinct solution, 644-645 quadratic formula method for solving, 641-645 square root property for solving, 626-630 standard form of, 452, 626, 641 steps to solve, 454 steps to solve an applied problem, 460 summary of solution methods, 648 zero-factor property for solving, 452--458, 626 Quadratic expressions, 452 Quadratic formula discri minant of, 644 explanation of, 642 for solving quadratic equations, 64 1- 645 Quadratic models, 464--465 Quotient(s) explanation of, 6, 19, 68 of opposites, 487 translating words and phrases, 72 writing in lowest terms, 603 Quotient rule for exponents, 354- 356 for radicals, 58 1- 582 R

Radical(s) adding, 587- 589 conjugates of, 601-602

dividing, 58 1- 582 explanation of, 568 index of, 573 like, 587 multiplying, 579-581 operations on, 579- 582, 587- 589 order of, 573 product rule for, 579- 58 1 properties of, 584 quotient rule for, 58 1-582 simplified form of, 579, 582, 593 simplifying, 579-5 84, 589- 590, 593- 596 simplifying higher roots, 584 solving equations with, 607- 612 subtracting, 587- 589 unlike, 588 with variables, 582-5 83 Radical equations solving, 607-612 steps to solve, 609 Radical expressions conditions for simplifying, 593 explanation of, 568 guidelines for simplifying, 599 simplifying, 589-590 simplifying products of, 599- 601 squaring of, 569 using conjugates to rationalize denominators of, 601-602 writing quotients in lowest terms, 603 Radicand, 568 Range explanation of, 269 finding for functions, 272 identifying for relations, 269 Rate of work, 539-542 Ratio explanation of, 157 from word phrases, 157-1 58 Rational expressions adding, 506-509 applications of, 537- 542 with denominator zero, 483 dividing, 494--496 equivalent forms for, 487--488 evaluating, 482 explanation of, 482 fundamental property of, 484 in lowest terms, 484--487 multiplying, 492--494, 496 with numerator zero, 482 operations on, 492--496, 506-5 11 solving equations with, 524-531 with specified denominator, 501 - 502 steps for division of, 496 steps for multiplication of, 496 subtracting, 509-51 1 summary of operations on, 535- 536 undefined values for, 483 Rational numbers explanation of, 43 graphing, 44

Rationalizing denominators, 592- 596, 601- 602 Real numbers. See also Number(s) absolute value of, 47--48 adding, 5 1-54, 56-58 additive inverse of, 46--47 dividing, 67- 73 explanation of, 45 inequality symbols used with, 45--46 multi plying, 65-67, 70-73 opposites of, 46--47 order of operations of, 56- 57 ordering of, 46 properties of, 65, 78- 84 reciprocals of, 67 sets of, 45 subtracting, 54-59 summary of operations on, 29, 77 Reciprocals explanation of, 6 of fractions, 6, 67, 81 Rectangles, perimeter of, 147, 152 Rectangular box, volume of, 152 Rectangular coordinate system, 2 12 Region of a graph, 262-266 Relations domain of, 269 explanation of, 269 functions as, 269-270 range of, 269 Repeating decimals, 2 1, 43 Richter, Charles F., 368 Richter scale, 368 Right angles, 139 Right triangles hypotenuse of, 463 legs of, 463 Pythagorean theorem for, 463 Rise, 232 Roots cube,573, 595- 596 fourth, 573 nth , 573 principal, 568, 608 square, 568- 571 Rounding, of decimals, 19-20 Run, 232

5 Scatter diagrams, 2 14 Scientific notation application of, 361, 364 in calculations, 363-364 dividing with, 363 explanation of, 361 and exponents, 361, 362 multiplying with, 363 steps to write a number in, 362 Second-degree equation. See Quadratic equations

Index Sentences equality and inequality symbols in, 33 translating into equations, 72-73 Set(s) elements of, 38 empty, 122 explanation of, 38 null, 122 of real numbers, 45 Set braces, 38 Set-builder notation, 43, 192, 29 1 Sight distance, 615 Signed numbers adding, 51-54 dividing, 68 explanation of, 43 interpreting data with, 59 multiplying, 65- 67 subtracting, 55- 56 Similar triangles, 167 Simple interest formula for, 153, 172 solving problems, 173 Simplified form of a radical, 579, 582, 593 Simplifying algebraic expressions, 82, 88-91 equations before solving, 109, 115-116 fractions, 3 radicals, 579-584, 593-596 Six-step method for solving applied problems, 132 Slope explanation of, 232 formula for, 232, 233 from an equation, 236--237 of horizontal lines, 235-236 negative, 235 of parallel lines, 238-239 of perpendicular lines, 238- 239 positive, 235 undefined,235-236 of vertical lines, 235- 236 Slope-intercept form, 245- 249 Solution(s) double, 456--457, 644 of equations, 38, 104 extraneous, 608 writing as ordered pairs, 209 Solution set(s) of an equation, l 04, 196 of a system of linear equations, 288 of a system of linear inequalities, 326--328 Sphere(s) surface area of, 632 volume of, 153, 587 Square(s) of binomials, 387-388 completing, 632- 638 difference of, 439--441 of a number, 28, 568

perfect, 440, 570 sum of, 440 Square root(s) of a, 569 approximation of, 571 explanation of, 568 finding, 569 identifying types of, 570 negative, 568 of a number, 568 positive, 568 principal, 568, 608 symbol for, 568 Square root property for solving quadratic equations, 626- 630 Squaring of radical expressions, 569 Squaring property of equality, 607- 611 Standard form of a linear equation, 209, 253, 254- 255 of a quadratic equation, 452, 626 Standard notation, 363 Straight angles, 139, 148-149 Subscript notation, 233 Substitution method explanation of, 297 for solving dependent equations, 300-30 1 for solving inconsistent systems, 300 for solving linear systems, 297-303 , 312 steps to solve by, 299 Subtraction addition property of equality extended to, 106 of decimals, 17-18 explanation of, 55 of fractions, 9- 10 with grouping symbols, 56--57 of a multivariable polynomial, 374 on a number line, 54 in order of operations, 29, 77 of polynomials, 373- 374 of radicals, 587-589 of rational expressions, 509-5 11 of real numbers, 54-59 of signed numbers, 55- 56 symbol for, 56 word phrases for, 57- 58 Subtrahend, 54 Sum of cubes, 445--446 explanation of, 7 of squares, 440 Supplementary angles, 139- 140 Supply and demand, 297 Surface area of a sphere, 632 Symbol(s) for additive inverse, 56 for approximately equal to, 57 1 for cube root, 573 for empty sets, 122 for equality, 31 , 38 for "(is) greater than," 31, 182 grouping. See Grouping symbols

1-7

for inequality, 31-33, 45--46, 182 for "(is) less than," 3 1, 182 for negative infinity, 182 for negative numbers, 56 fornull set, 122 plus or minus, 627 for positive infinity, 182 for square roots, 568 for subtraction, 56 word conversions to, 32, 57-58 Symmetry axis of a parabola, 375 System of linear inequalities graphical method for solving, 326-328 solution set of, 326--328 steps to solve, 326 Systems of linear equations alternative method for solving, 308- 309 applications of, 314-320 choosing a method to solve, 312-313 consistent, 291 - 292 with decimals, 302 elimination method for solving, 305-3 10,3 12 explanation of, 288 with fractions, 301-302 graphing method for solving, 288-292 inconsistent, 291-292 with no solution, 291-292 solution of, 288 solution set of, 288 solving by graphing, 288-292 steps for solving by elimination, 306 steps for solving by substitution, 299 steps to solve by graphing, 290 substitution method for solving, 297-303 summary of outcomes, 292

T Table of data, interpreting, 48 Table of values, 2 11- 212 Terminating decimals, 2 1, 43 Terms adding, 380 combining, 89, 369-370 degree of, 371 distinguishing between factors and, 89 explanation of, 88- 89 of an expression, 88- 89, 369 like, 89, 369 multiplying, 380 numerical coefficient of, 88, 369 of a polynomial, 369 of a proportion, 158 unlike, 89, 370 Tests for di visibility, 76, 414 Three-part inequalities, 190--192 Traffic intensity, 491 Translating sentences into equations, 72- 73 word phrases into algebraic expressions, 37- 38,9 1, 123

1-8

Index

Trapezoids, area of, 146, 152 Triangle( s) area of, 152 perimeterof, 147- 148, 152 right, 463 similar, 167 Trinomials explanation of, 37 1 factoring of, 423-428, 430-436 perfect square, 441-443, 446, 632-633

u Undefined rational expressions, 483 Undefined slope, 236 Union of solution sets, 196 Unit cost, 158 Unit pricing, 158 Unlike radicals, 588 Unlike terms, 89, 370

v Variable cost, 253 Variables explanation of, 36 formulas to evaluate, 146- 148 radicals with, 582- 583 solving for specified, 149-151 , 530-531

Variation constant of, 547 direct, 547- 550 inverse, 550-551 Velocity problem, 638 Vertex of parabolas explanation of, 375, 649 solving for, 650-654 Vertical angles, 148-149 Vertical line test for functions, 27 1 Vertical lines equation of, 224-225 explanation of, 224 graphing, 224- 225 graphing linear inequalities with, 265 slope of, 235-236 Volume of a cube, 587 explanation of, 152 of a pyramid, 153 of a rectangular box, 152 of a sphere, 153, 587

w Whole numbers, 1, 42 Word phrases for addition, 57-58 to algebraic expressions, 37- 38, 91, 123 for division, 72 to expressions, 91

for multiplication, 7 1- 72 to numerical expressions, 57- 58, 7 1- 72 to ratios, 157- 158 for subtraction, 57- 58 Word statements to equations, 72-73 Words to symbols conversions, 32, 57-58 Work rate problems, 539- 542

x x-axis, 212 x-intercept explanation of, 221 of a parabola, 650-654

y y-axis, 212 y-intercept explanation of, 221 of a parabola, 650-653 slope-intercept form and, 245- 247

z Zero division by, 69 multiplication by, 65 Zero denominator in a rational expression, 483 Zero exponent, 351 , 356 Zero-factor property, 452-458, 626

Triangles and Angles Right Triangle Triangle has one 90° (right) angle.

~ a

Right Angle Measure is 90°.

c

JO° b

Pythagorean Theorem (for right triangles)

L

a2 + b2 = c2

B

Isosceles Triangle Two sides are equal.

Straight Angle Measure is 180°.

180°

~

AB = BC

Equilateral Triangle All sides are equal.

Complementary Angles The sum of the measures of two complementary angles is 90°.

AB = BC = CA

fCD/

~

CD 0

Angles and are complementary.

Sum of the Angles of Any Triangle A+ B

+ C = 180°

Supplementary Angles The sum of the measures of two supplementary angles is 180°.

B

A~C Similar Triangles Corresponding angles are equal. Corresponding sides are proportional. A

= D , B = E, C = F

AB DE

AC DF

BC EF

Vertical Angles Vertical angles have equal measures.

E

Angles

0

and

*

0

are supplementary.

3

4

F

CD = Angle 0 Ang le 0 = Angle 0 Angle

Geometry Formulas Figure

Formulas

Square

Perimeter: Area:

Rectangle

= 4s

P

:il =

Perimeter: Area:

Illustration

s2

= 2L + 2 W

P

:il = L W

w L

Triangle

Perimeter: Area:

Parallelogram

:il

Perimeter: Area:

:il

=a+b+c

P

1 2

= - bh

= 2a + 2b

P

b

= bh b

Trapezoid

Perimeter: Area:

=a + b + c+ B

P 1 2

:il = - h(b

b

+ B) B

Circle

Diameter:

d

= 2r

Circumference: Area:

:il = 7Tr 2

C = 27Tr

c = 7Td

Geometry Formulas Figure

Formulas

Illustration

Cube

Volume:

V

Rectangular Solid

Volume:

V

= e3 Surface area: S = 6e2

= LWH

H

Surface area: S = 2HW + 2LW + 2LH

Right Circular Cylinder

Volume:

V

=

7Tr2 h

Surface area: S = 27rrh + 27rr2 (Includes both circular bases)

I I

I

hi I I

__ _,__ _ I

Cone

Volume:

V

=

. _r_:::

1

- 7Tr 2h

3

Surface area: S = 7Tr~ + 7rr2 (Includes circular base)

Right Pyramid

Volume:

1

V=-Bh 3

B = area of the base

Sphere

4 3 7Tr 3

V

=-

Surface area:

S

Volume:

=

47rr 2

,--,

-------

,.

Other Formulas Distance: d = rt ( r = rate or speed, t = time) Percent: p = br (p = percentage, b = base, r = rate) 9 5 Temperature: F = - C + 32 C = - (F - 32) 5 9 Simple Interest: I = prt (p = principal or amount invested, r

= rate or percent, t = time in years)