probability

🚀The assignment is divided into two parts, please complete it separately. Thefirst part is HW4 Base on Chapter 6. I have attached the textbook and the answers, so please be sure to complete the whole process All work must be shown.Don’t just copy the answers. The second part is Select Problem Set 4, only three questions also base on chapter 6. I have attached the screenshot of the questions of Select Problem Set 4.

🚀 The handwriting needs to be clear. Please send me the PDF format

HW4 based on Chapter 6.

Section 6.1 Exercises: 2, 4, 23, 26

Section 6.2 Exercises: 8, 18, 26, 34, 42, 44

Section 6.3 Exercises: 2, 23, 25, 34

Section 6.4 Exercises: 4, 10, 14, 20, 46

Section 6.5 Exercises: 11, 30, 34, 46

Section 6.6 Exercises: 3, 12, 18, 20

Instructions:1. The problems must be written out neatly on loose-leaf sheets and must be done in numerical order.2. All problems must be completed.3. All work must be shown. Just presenting a numerical answer without having some work to back it up will not get any credit. This will result in deductions in HW score.4. HW sets will be collected on the dates indicated. The entire Chapter’s work is collected on a single day. Completion will constitute half your HW score. Scan the papers in which you have written your solutions to form a SINGLE pdf file and upload to blackboard. Instructions on how to name your file:

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twelfth edition
Finite
Mathematics
& ITS APPLICATIONS
Larry J. Goldstein
Goldstein Educational Technologies
David I. Schneider
University of Maryland
Martha J. Siegel
Towson University
Steven M. Hair
The Pennsylvania State University
330 Hudson Street, NY, NY 10013
Director, Portfolio Management: Deirdre Lynch
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Cover Design: Cenveo
Cover Image: Doug Chinnery/Getty Images
Copyright © 2018, 2014, 2010 by Pearson Education, Inc. All Rights Reserved. Printed in the United States of
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Library of Congress Cataloging-in-Publication Data
Names: Goldstein, Larry Joel. | Schneider, David I. | Siegel, Martha J. |
   Hair, Steven M.
Title: Finite mathematics & its applications.
Other titles: Finite mathematics and its applications
Description: Twelfth edition / Larry J. Goldstein, Goldstein Educational
   Technologies, David I. Schneider, University of Maryland, Martha J.
   Siegel, Towson State University, Steven M. Hair, Pennsylvania State
   University. | Boston: Pearson Education, [2018] | Includes indexes.
Identifiers: LCCN 2016030690 | ISBN 9780134437767 (hardcover) | ISBN
  0134437764 (hardcover)
Subjects: LCSH: Mathematics—Textbooks.
Classification: LCC QA39.3 .G65 2018 | DDC 511/.1—dc23
LC record available at https://lccn.loc.gov/2016030690
1
16
Student Edition ISBN-13: 978-0-134-43776-7
Student Edition ISBN-10:
0-134-43776-4
Contents
The book divides naturally into four parts. The first part consists of linear mathematics: linear
equations, matrices, and linear programming (Chapters 1–4); the second part is devoted to
probability and statistics (Chapters 5–7); the third part covers topics utilizing the ideas of the
other parts (Chapters 8 and 9); and the fourth part explores key topics from discrete
mathematics that are sometimes covered in the modern finite mathematics curriculum
(Chapters 10–12).
Preface
vii
PART ONE
1
Linear Equations and Straight Lines
1.1
1.2
1.3
1.4
2
The Slope of a Straight Line
1
8
The Intersection Point of a Pair of Lines
20
The Method of Least Squares 25
Chapter Summary and Chapter Review Exercises
Chapter Project: Break-Even Analysis 38
34
Matrices
2.1
2.2
2.3
2.4
2.5
2.6
3
Coordinate Systems and Graphs
1
39
Systems of Linear Equations with Unique Solutions
General Systems of Linear Equations
Arithmetic Operations on Matrices
The Inverse of a Square Matrix
39
50
59
73
The Gauss–Jordan Method for Calculating Inverses
Input–Output Analysis 84
Chapter Summary and Chapter Review Exercises
Chapter Project: Population Dynamics 95
80
90
Linear Programming, A Geometric Approach
3.1
3.2
3.3
3.4
Linear Inequalities
97
97
A Linear Programming Problem
105
Fundamental Theorem of Linear Programming
111
Linear Programming 121
Chapter Summary and Chapter Review Exercises
Chapter Project: Shadow Prices 135
132
iii
iv
CONTENTS
4
The Simplex Method
4.1
4.2
4.3
4.4
136
Slack Variables and the Simplex Tableau
136
The Simplex Method I: Maximum Problems
144
The Simplex Method II: Nonstandard and Minimum Problems
154
Sensitivity Analysis and Matrix Formulations of Linear Programming
Problems 161
4.5 Duality
168
Chapter Summary and Chapter Review Exercises
Chapter Project: Shadow Prices Revisited 183
178
PART TWO
5
Sets and Counting
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
6
184
A Fundamental Principle of Counting
Venn Diagrams and Counting
The Multiplication Principle
197
203
Permutations and Combinations
Further Counting Techniques
The Binomial Theorem
191
209
216
222
Multinomial Coefficients and Partitions 226
Chapter Summary and Chapter Review Exercises
Chapter Project: Pascal’s Triangle 237
232
Probability
6.1
6.2
6.3
6.4
6.5
6.6
6.7
7
Sets
184
240
Experiments, Outcomes, Sample Spaces, and Events
Assignment of Probabilities
246
Calculating Probabilities of Events
257
Conditional Probability and Independence
Tree Diagrams
265
275
Bayes’ Theorem, Natural Frequencies
282
Simulation 288
Chapter Summary and Chapter Review Exercises
Chapter Project: Two Paradoxes 298
Probability and Statistics
7.1
7.2
7.3
7.4
7.5
7.6
Visual Representations of Data
The Mean
300
307
317
325
The Variance and Standard Deviation
The Normal Distribution
346
294
300
Frequency and Probability Distributions
Binomial Trials
240
336
CONTENTS
v
7.7 Normal Approximation to the Binomial Distribution
359
Chapter Summary and Chapter Review Exercises 363
Chapter Project: An Unexpected Expected Value 368
PART THREE
8
Markov Processes
369
8.1 The Transition Matrix 369
8.2 Regular Stochastic Matrices 381
8.3 Absorbing Stochastic Matrices 389
Chapter Summary and Chapter Review Exercises 399
Chapter Project: Doubly Stochastic Matrices 401
9
The Theory of Games
404
9.1 Games and Strategies 404
9.2 Mixed Strategies 410
9.3 Determining Optimal Mixed Strategies
417
Chapter Summary and Chapter Review Exercises 426
Chapter Project: Simulating the Outcomes of Mixed-Strategy
Games 428
PART FOUR
10
The Mathematics of Finance
10.1
10.2
10.3
10.4
10.5
11
Interest
430
430
Annuities
440
Amortization of Loans
449
Personal Financial Decisions
458
A Unifying Equation 474
Chapter Summary and Chapter Review Exercises
Chapter Project: Two Items of Interest 489
485
Logic
11.1
11.2
11.3
11.4
11.5
11.6
11.7
491
Introduction to Logic
Truth Tables
Implication
491
495
504
Logical Implication and Equivalence
Valid Argument
510
518
Predicate Calculus
525
Logic Circuits 533
Chapter Summary and Chapter Review Exercises
Chapter Project: A Logic Puzzle 542
537
vi
CONTENTS
12
Difference Equations and Mathematical Models (Online*)
12.1
12.2
12.3
12.4
12.5
Introduction to Difference Equations D1
Difference Equations and Interest D8
Graphing Difference Equations D13
Mathematics of Personal Finance D22
Modeling with Difference Equations D26
Chapter Summary and Chapter Review Exercises D30
Chapter Project: Connections to Markov Processes D33
Appendix A
Areas Under the Standard Normal Curve A-1
Appendix B
Using the TI-84 Plus Graphing Calculator
Appendix C
Spreadsheet Fundamentals
Appendix D
Wolfram|Alpha
Learning Objectives
Selected Answers
(Online*) A-11
SA-1
Index of Applications
Index
A-10
IA-1
I-1
*www.pearsonhighered.com/mathstatsresources
A-6
A-2
Preface
T
his work is the twelfth edition of our text for the finite mathematics course taught to
first- and second-year college students, especially those majoring in business and the
social and biological sciences. Finite mathematics courses exhibit tremendous diversity
with respect to both content and approach. Therefore, in developing this book, we
incorporated a wide range of topics from which an instructor may design a curriculum,
as well as a high degree of flexibility in the order in which the topics may be presented.
For the mathematics of finance, we even allow for flexibility in the approach of the presentation.
The Series
This text is part of a highly successful series consisting of three texts: Finite Mathematics
& Its Applications, Calculus & Its Applications, and Calculus & Its Applications, Brief
Version. All three titles are available for purchase in a variety of formats, including as an
eBook within the MyMathLab online course.
twelfth edition
Finite
Mathematics
fourteenth edition
fourteenth edition
Calculus
Calculus
& ITS APPLICATIONS
& ITS APPLICATIONS
BRIEF VERSION
& ITS APPLICATIONS
Goldstein
Schneider
Goldstein
Goldstein
Siegel
Lay
Lay
Hair
Schneider
Schneider
Asmar
Asmar
Topics Included
This edition has more material than can be covered in most one-semester courses. Therefore, the instructor can structure the course to the students’ needs and interests. The
book divides naturally into four parts:
• Part One (Chapters 1–4) consists of linear mathematics: linear equations, matrices,
and linear programming.
• Part Two (Chapters 5–7) is devoted to counting, probability, and statistics.
• Part Three (Chapters 8 and 9) covers topics utilizing the ideas of the other parts.
• Part Four (Chapters 10–12) explores key topics from discrete mathematics that are
sometimes included in the modern finite mathematics curriculum.
Minimal Prerequisites
Because of great variation in student preparation, we keep formal prerequisites to a minimum. We assume only a first year of high school algebra, and we review, as needed,
those topics that are typically weak spots for students.
vii
viii
PREFACE
New to This Edition
We welcome to this edition a new co-author, Steven Hair from Penn State University.
Steve has brought a fresh eye to the content and to the MyMathLab course that accompanies the text.
We are grateful for the many helpful suggestions made by reviewers and users of the
text. We incorporated many of these into this new edition. We also analyzed aggregated
student usage and performance data from MyMathLab for the previous edition of this
text. The results of this analysis helped improve the quality and quantity of exercises
that matter the most to instructors and students. Additionally, we made the following
improvements in this edition:
• Help-Text Added. We added blue “help text” next to steps within worked-out
examples to point out key algebraic and numerical transitions.
• Updated Technology. We changed the graphing calculator screen captures to the
more current TI-84 Plus CE format.The discussions of Excel now refer to Excel
2013 and Excel 2016.
• Additional Exercises and Updated Data. We have added or updated 440 exercises and have updated the real-world data appearing in the examples and exercises.
The book now contains 3580 exercises and 370 worked-out examples.
• Technology Solutions. We added technology-based solutions to more examples to
provide flexibility for instructors who incorporate technology. For instance, the section on the method of least-squares (1.4) now relies more on technology and less on
complicated calculations. In Section 7.6, several examples now demonstrate how to
compute the area under a normal curve using a graphing calculator, in addition to
the table-based method. In the finance chapter, many TI-84 Plus TVM Solver screen
captures accompany examples to confirm answers. Instructors have the option of
using TVM Solver for financial calculations instead of complicated formulas.
• Linear Inequalities Section Relocated. We moved this section from 1.2 (in the 11e)
to the beginning of the linear programming chapter (Ch. 3) in this edition. The
move places the topic in the chapter where it is used. Also, the move allows us to use
conventional names (such as slope-intercept form) in the section.
• Improved Coverage of Counting Material. In Chapter 5, we added several definitions and discussions to aid student comprehension of counting problems. We
moved the definition of factorials to 5.4 and rewrote the permutation and combination formulas in 5.5 in terms of factorials. In 5.6, the complement rule for counting
is now formally defined, and we have added a discussion of when addition, subtraction, and multiplication is appropriate for solving counting problems.
• Section Added to the End of the Finance Chapter. Titled “A Unifying Equation,” this new section shows that the basic financial concepts can be described by a
difference equation of the form yn = a # yn – 1 + b, y0 given, and that many of the
calculations from the chapter can be obtained by solving this difference equation.
Examples and exercises show that this difference equation also can be used to solve
problems in the physical, biological, and social sciences. This section can be taught as
a standalone section without covering the preceding sections of the finance chapter.
• Revision of Logic Material. We substantially revised Chapter 11 on logic to better
meet student needs.We moved the definition of logical equivalence and De ­Morgan’s
laws from 11.4 to 11.2. By stating key ideas related to truth tables and implications
in terms of logical equivalence, students will be better equipped to understand these
concepts. To remove confusion between the inclusive and exclusive “or” statements,
we removed the word “either” from inclusive “or” statements in English. In 11.4, we
added the definition of the inverse of an implication. This is a key concept in the
topic of implications and logical arguments. To help students understand when a
logical argument is invalid, we expanded 11.5 to include more discussion of invalid
arguments. Additionally, we added the fallacies of the inverse and converse, and two
new examples where arguments are proven to be invalid.
PREFACE
ix
• Difference Equation Chapter Moved Online. We moved former Chapter 11 online
(relabeling it Chapter 12 in the process). The chapter is available directly to students
at www.pearsonhighered.com/mathstatsresources and within MyMathLab. All
­support materials for the chapter appear online within MyMathLab. Note: The new
section at the end of the finance chapter contains the fundamental concepts from the
difference equation chapter.
New to MyMathLab
Many improvements have been made to the overall functionality of MyMathLab
(MML) since the previous edition. However, beyond that, we have also invested in
increasing and improving the content specific to this text.
• Instructors now have more exercises than ever to choose from in assigning homework. There are approximately 2540 assignable exercises in MML.
• We heard from users that the Annotated Instructor Edition for the previous edition
required too much flipping of pages to find answers, so MML now contains a downloadable Instructor Answers document—with all answers in one place. (This augments the downloadable Instructor Solutions Manual, which contains all solutions.)
• Interactive Figures are now in HTML format (no plug-in required) and are supported by assignable exercises and tutorial videos.
• An Integrated Review version of the MML course contains pre-made quizzes to
assess the prerequisite skills needed for each chapter, plus personalized remediation
for any gaps in skills that are identified.
• New Setup & Solve exercises require students to show how they set up a problem as
well as the solution, better mirroring what is required of students on tests.
• StatCrunch, a fully functional statistics package, is provided to support the statistics
content in the course.
• MathTalk and StatTalk videos highlight applications of the content of the course to
business. The videos are supported by assignable exercises.
• Study skills modules help students with the life skills that can make the difference
between passing and failing.
• 110 new tutorial videos by Brian Rickard (University of Arkansas) were added to
support student learning.
• Tutorial videos involving graphing calculators are now included within MML exercises to augment videos showing “by hand” methods. If you require graphing calculator usage for the course, your students will find these videos very helpful. (If you
do not use calculators, you can hide these videos from students.)
• Graphing Calculator and Excel Spreadsheet Manuals, specific to this course, are
now downloadable from MML.
Trusted Features
Though this edition has been improved in a variety of ways to reflect changing student
needs, we have maintained the popular overall approach that has helped students be
successful over the years.
Relevant and Varied Applications
We provide realistic applications that illustrate the uses of finite mathematics in other
disciplines and everyday life. The variety of applications is evident in the Index of Applications at the end of the text. Wherever possible, we attempt to use applications to motivate the mathematics. For example, the concept of linear programming is introduced in
Chapter 3 via a discussion of production options for a factory with labor limitations.
Plentiful Examples
The twelfth edition includes 370 worked examples. Furthermore, we include computational details to enhance comprehension by students whose basic skills are weak.
x
PREFACE
­ nowing that students often refer back to examples for help, we built in fidelity between
K
exercises and examples. In addition, students are given Now Try exercise references
immediately following most examples to encourage them to check their understanding
of the given example.
Exercises to Meet All Student Needs
The 3580 exercises comprise about one-quarter of the book—the most important part of
the text, in our opinion. The exercises at the ends of the sections are typically arranged
in the order in which the text proceeds, so that homework assignments may be made
easily after only part of a section is discussed. Interesting applications and more challenging problems tend to be located near the ends of the exercise sets. Exercises have
odd-even pairing, when appropriate. Chapter Review Exercises are designed to prepare
students for end-of-chapter tests. Answers to the odd-numbered exercises, and all Chapter Review Exercises, are included at the back of the book.
Check Your Understanding Problems
The Check Your Understanding problems are a popular and useful feature of the
book. They are carefully selected exercises located at the end of each section, just
before the exercise set. Complete solutions follow the exercise set. These problems prepare students for the exercise sets beyond just covering simple examples. They give
students a chance to think about the skills they are about to apply and reflect on what
they’ve learned.
Use of Technology
We incorporated technology usage into the text in ways that provide you with flexibility,
knowing that the course can vary quite a bit based on how technology is incorporated.
Our basic approach in the text is to assume minimal use of technology and clearly label
the opportunities to make it a greater part of the course. Many of the sections contain
Incorporating Technology features that show how to use Texas Instruments graphing
calculators, Excel spreadsheets, and Wolfram|Alpha. In addition, the text contains
appendixes on the use of these technologies. Each type of technology is clearly labeled
with an icon:
(Graphing Calculator),
(Spreadsheet),
(Wolfram|Alpha)
In our discussions of graphing calculators, we specifically refer to the TI-84 Plus
models, since these are the most popular graphing calculators. New to this edition,
screen shots display the new color versions of the TI-84. Spreadsheets refer to Microsoft Excel 2016. The web application discussed is Wolfram|Alpha, which is an exceptionally fine and versatile product that is available online or on mobile devices for free
or at low cost. We feel that Wolfram|Alpha is a powerful tool for learning and exploring
mathematics, which is why we chose to include activities that use it. We hope that by
modeling appropriate use of this technology, students will come to appreciate the application for its true worth.
End-of-Chapter Study Aids
Near the end of each chapter is a set of problems entitled Fundamental Concept Check
Exercises that help students recall key ideas of the chapter and focus on the relevance of
these concepts as well as prepare for exams. Each chapter also contains a two-column
grid giving a section-by-section summary of key terms and concepts with examples.
Finally, each chapter has Chapter Review Exercises that provide more practice and
preparation for chapter-level exams.
PREFACE
xi
Chapter Projects
Each chapter ends with an extended project that can be used as an in-class or out-ofclass group project or special assignment. These projects develop interesting applications or enhance key concepts of the chapters.
Technology and Supplements
MyMathLab® Online Course (access code required)
Built around Pearson’s best-selling content, MyMathLab is an online homework, tutorial, and
assessment program designed to work with this text to engage students and improve results.
MyMathLab can be successfully implemented in any classroom environment—lab-based, hybrid,
fully online, or traditional. By addressing instructor and student needs, MyMathLab
improves student learning.
Used by more than 37 million students worldwide, MyMathLab delivers consistent, measurable gains in student learning outcomes, retention, and subsequent course success. Visit www.
mymathlab.com/results to learn more.
Preparedness
One of the biggest challenges in Finite Mathematics courses is making sure students are adequately prepared with the prerequisite skills needed to successfully complete their course work.
Pearson offers a variety of content and course options to support students with just-in-time remediation and key-concept review.
•
Integrated Review Courses can be used for just-in-time prerequisite review. These courses
provide additional content on review topics, along with pre-made, assignable skill-check quizzes, personalized homework assignments, and videos integrated throughout the course.
Motivation
Students are motivated to succeed when they’re engaged in the learning experience and understand the relevance and power of mathematics. MyMathLab’s online homework offers students immediate feedback and tutorial assistance that motivates them to do more, which
means they retain more knowledge and improve their test scores.
• Exercises with immediate feedback—over 2540 assignable exercises—are based on the textbook exercises, and regenerate algorithmically to give students unlimited opportunity for
practice and mastery. MyMathLab provides helpful feedback when students enter incorrect
answers and includes optional learning aids including Help Me Solve This, View an Example,
videos, and an eText.
xii
PREFACE
• Setup and Solve Exercises ask students to first describe how they will set up and approach the problem. This reinforces students’ conceptual understanding of the process
they are applying and promotes long-term retention of the skill.
• MathTalk and StatTalk videos connect the math to the real world (particularly business). The videos include assignable exercises to gauge students’ understanding of video
content.
• Learning Catalytics™ is a student response tool that uses students’ smartphones, tablets, or laptops to engage
them in more interactive tasks and
thinking. Learning Catalytics fosters
student engagement and peer-to-peer
learning with real-time analytics.
Learning and Teaching Tools
• Interactive Figures illustrate key concepts and allow manipulation for use as teaching and
learning tools. MyMathLab includes assignable exercises that require use of figures and
instructional videos that explain the concept behind each figure.
• Instructional videos—238 example-based videos—are available as learning aids within
exercises and for self-study. The Guide to Video-Based Assignments makes it easy to assign
videos for homework by showing which MyMathLab exercises correspond to each video.
PREFACE
xiii
• Graphing Calculator videos are available to augment “by hand” methods, allowing you to
match the help that students receive to how graphing calculators are used in the course. Videos
are available within select exercises and in the Multimedia Library.
• Complete eText is available to students through their MyMathLab courses for the lifetime
of the edition, giving students unlimited access to the eText within any course using that edition of the textbook.
• StatCrunch, a fully functional statistics package, is provided to support the statistics content
in the course.
• Skills for Success Modules help students with the life skills that can make the difference
between passing and failing. Topics include “Time Management” and “Stress Management.”
• Excel Spreadsheet Manual, specifically written for this course.
• Graphing Calculator Manual, specifically written for this course.
• PowerPoint Presentations are available for download for each section of the book.
• Accessibility and achievement go hand in hand. MyMathLab is compatible with the JAWS
screen reader, and enables multiple-choice and free-response problem types to be read and
interacted with via keyboard controls and math notation input. MyMathLab also works with
screen enlargers, including ZoomText, MAGic, and SuperNova. And, all MyMathLab videos
have closed-captioning. More information is available at http://mymathlab.com/accessibility.
• A comprehensive gradebook with enhanced reporting functionality allows you to efficiently
manage your course.
• The Reporting Dashboard provides insight to view, analyze, and report learning outcomes.
Student performance data is presented at the class, section, and program levels in an accessible, visual manner so you’ll have the information you need to keep your students on track.
• Item Analysis tracks class-wide understanding of particular exercises so you can refine
your class lectures or adjust the course/department syllabus. Just-in-time teaching has
never been easier!
MyMathLab comes from an experienced partner with educational expertise and an eye on the
future. Whether you are just getting started with MyMathLab, or have a question along the way,
we’re here to help you learn about our technologies and how to incorporate them into your course.
To learn more about how MyMathLab helps students succeed, visit www.mymathlab.com or contact your Pearson rep.
MathXL® is the homework and assessment engine that runs MyMathLab. (MyMathLab is
MathXL plus a learning management system.) MathXL access codes are also an option.
Student Solutions Manual
ISBN-10: 0-134-46344-7 | ISBN-13: 978-0-134-46344-5
Contains fully worked-out solutions to odd-numbered exercises. Available in print and
downloadable from within MyMathLab.
Instructor Answers / Instructor Solutions Manual (downloadable)
ISBN-10: 0-134-46343-9 | ISBN-13: 978-0-134-46343-8
The Instructor Answers document contains a list of answers to all student edition exercises. The
Instructor Solutions Manual contains solutions to all student edition exercises. Downloadable
from the Pearson Instructor Resource Center www.pearsonhighered.com/irc, or from within
MyMathLab.
xiv
PREFACE
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Acknowledgments
While writing this book, we have received assistance from many people, and our heartfelt
thanks go out to them all. Especially, we should like to thank the following reviewers,
who took the time and energy to share their ideas, preferences, and often their enthusiasm, with us during this revision:
Jeff Dodd, Jacksonville State University
Timothy M. Doyle, University of Illinois at Chicago
Sami M. Hamid, University of North Florida
R. Warren Lemerich, Laramie County Community College
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Arthur J. Rosenthal, Salem State University
Mary E. Rudis, Great Bay Community College
Richard Smatt, Mount Washington College
Paul J. Welsh, Pima Community College
The following faculty members provided direction on the development of the
MyMathLab course for this edition:
Mark A. Crawford, Jr., Waubonsee Community College
Cymra Haskell, University of Southern California
Ryan Andrew Hass, Oregon State University
Melissa Hedlund, Christopher Newport University
R. Warren Lemerich, Laramie County Community College
Sara Talley Lenhart, Christopher Newport University
Enyinda Onunwor, Stark State College
Lynda Zenati, Robert Morris University
We wish to thank the many people at Pearson who have contributed to the success
of this book. We appreciate the efforts of the production, design, manufacturing, marketing, and sales departments. We are grateful to Lisa Collette for her thorough proofreading and John Morin and Rhea Meyerholtz for their careful and thorough checking
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book on schedule. The authors wish to extend special thanks to editor Jeff Weidenaar.
If you have any comments or suggestions, we would like to hear from you. We hope
you enjoy using this book as much as we have enjoyed writing it.
Larry J. Goldstein
larrygoldstein@predictiveanalyticsshop.com
David I. Schneider
dis@math.umd.edu
Martha J. Siegel
msiegel@towson.edu
Steven M. Hair
smh384@psu.edu
chapter
1
Linear Equations
and Straight Lines
1.1
1.2
1.3
1.4
Coordinate Systems and Graphs
The Slope of a Straight Line
The Intersection Point of a Pair of Lines
The Method of Least Squares
M
any applications considered later in this text involve linear equations and their geometric counterparts—straight lines. So let us begin by studying the basic facts
about these two important notions.
Coordinate Systems and Graphs
1.1
Often, we can display numerical data by using a Cartesian coordinate system on either
a line or a plane. We construct a Cartesian coordinate system on a line by choosing an
arbitrary point O (the origin) on the line and a unit of distance along the line. We then
assign to each point on the line a number that reflects its directed distance from the origin. Positive numbers refer to points on the right of the origin, negative numbers to
points on the left. In Fig. 1, we have drawn a Cartesian coordinate system on the line
and have labeled a number of points with their corresponding numbers. Each point on
the line corresponds to a number (positive, negative, or zero).
y
y-axis
232
(a, b)
b
22
origin
O
x
a
1
2
21
0
15
8
1
2
3
Figure 1
x-axis
Figure 2
In a similar fashion, we can construct a Cartesian coordinate system to numerically
locate points on a plane. Each point of the plane is identified by a pair of numbers (a, b).
See Fig. 2. To reach the point (a, b), begin at the origin, move a units in the x direction
(to the right if a is positive, to the left if a is negative), and then move b units in the y
1
2
chapter 1 Linear Equations and Straight Lines
direction (up if b is positive, down if b is negative). The numbers a and b are called,
respectively, the x- and y-coordinates of the point.
EXAMPLE 1
Plotting Points Plot the following points:
(a) (2, 1)     (b) ( -1, 3)     (c) ( -2, -1)     (d) (0, -3)
SOLUTION
y
(21, 3)
(2, 1)
x
(22, 21)
(0, 23)
Now Try Exercise 1
An equation in x and y is satisfied by the point (a, b) if the equation is true when x
is replaced by a and y is replaced by b. This collection of points is usually a curve of
some sort and is called the graph of the equation.
EXAMPLE 2
Solution of an Equation Are the following points on the graph of the equation
8x – 4y = 4?
(a) (3, 5)     (b) (5, 17)
SOLUTION
(a)
8x – 4y = 4
Given equation
8#3 – 4#5 = 4
?
x = 3, y = 5
?
24 – 20 = 4
Multiply.
4=4
Subtract.
Since the equation is satisfied, the point (3, 5) is on the graph of the equation.
(b)
8x – 4y = 4
Given equation
8 # 5 – 4 # 17 = 4
?
x = 5, y = 17
?
40 – 68 = 4
Multiply.
-28 = 4
Subtract.
?
The equation is not satisfied, so the point (5, 17) is not on the graph of the
Now Try Exercises 11 and 13
­equation.
Linear Equations
A linear equation is an equation whose graph is a straight line. Figure 3 shows four
examples of linear equations, along with their graphs and some points on their graphs.
y
y
5
y
5
5
(0, 3)
25
x53
(0, 6)
(0, 0)
x
5
10
(6, 3)
(3, 0)
25
y
(3, 0)
x
5
25
25
y53
Figure 3 Four linear equations and their graphs
x
10
210
25
y 5 12 x
x
5
25
210
y 5 22 x 1 6
1.1 Coordinate Systems and Graphs
3
Intercepts
The intercepts of a line are the points where the line crosses the x- and y-axes. These
points have 0 for at least one of their coordinates. For the graph of y = -2x + 6 in
Fig. 3, the x-intercept is the point (3, 0) and the y-intercept is the point (0, 6).* The
y-intercept of a line having an equation of the form y = mx + b is the point (0, b),
since setting x equal to 0 gives y the value b. The x-intercept is the point having the
solution of the equation 0 = mx + b as the first coordinate and 0 as the second
­coordinate.
Table 1 shows how to draw the graphs of the four types of linear equations
shown in Fig. 3. The equations y = b and y = mx are actually special cases of
y = mx + b.
Table 1 Graphs of Linear Equations
Equation
Description of Graph
How to Draw Graph
x=a
Vertical line through the point
(a, 0)
Plot (a, 0) and draw the vertical line
through the point.
y=b
Horizontal line through the
point (0, b)
Plot (0, b) and draw the horizontal line
through the point.
y = mx
Line through the origin
Draw the line through the origin and
any other point on the graph.
y = mx + b;
m ≠ 0, b ≠ 0
Line having two different
intercepts
Draw the line through any two points
(often the two intercepts) of the line.
General Form of a Linear Equation
be written in the general form
Any equation whose graph is a straight line can
cx + dy = e
where c, d, and e are constants and c and d are not both zero.
An equation in general form having d ≠ 0 (that is, an equation in which y
appears) can be solved for y. The resulting equation will have the form of one of the
last three equations in Table 1. An equation in which y does not appear can be
solved for x and the resulting equation will have the form of the first equation in
Table 1.
EXAMPLE 3
SOLUTION
Graph of an Equation Write the equation x – 2y = 4 in one of the forms shown in
Table 1 and draw its graph.
Since y appears in the equation, solve for y.
x – 2y = 4
Given equation
-2y = -x + 4
y=
1
2x
-2
Subtract x from both sides.
Divide both sides by -2.
Since the equation y = 12 x – 2 has the form of the last equation in Table 1, it can be
graphed by finding its two intercepts and drawing the straight line through them.
*Intercepts are sometimes defined as numbers, such as x-intercept 3 and y-intercept 6. In this text, we define
them as pairs of numbers, such as (3, 0) and (0, 6).
4
chapter 1 Linear Equations and Straight Lines
The y-intercept is the point (0, -2) since setting x equal to 0 gives y the value -2.
The x-intercept is found by setting y equal to 0 and solving for x.
y
5
(4, 0)
(0, 22)
Given equation
-2
Set y equal to 0.
0=
x
25
y = 12 x – 2
2=
5
1
2x
1
2x
Add 2 to both sides.
Multiply both sides by 2. Rewrite.
x=4
25
Figure 4 Graph of x – 2y = 4
EXAMPLE 4
SOLUTION
Therefore, the x-intercept is the point (4, 0).
The graph in Fig. 4 was obtained by plotting the intercepts (4, 0) and (0, -2) and
Now Try Exercise 27
drawing the straight line through them.
Graph of an Equation Write the equation -2x + 3y = 0 in one of the forms shown in
Table 1 and draw its graph.
Since y appears in the equation, solve for y.
y
-2x + 3y = 0
5
3y = 2x
(6, 4)
y=
x
(0, 0)
210
10
25
Figure 5 Graph of -2x + 3y = 0
2
3x
Given equation
Add 2x to both sides.
Divide both sides by 3.
Because the graph of the equation y = 23x passes through the origin, the point (0, 0)
is both the x-intercept and the y-intercept of the graph. In order to draw the graph, we
must locate another point on the graph. Let’s choose x = 6. Then y = 23 # 6 = 4. Therefore, the point (6, 4) is on the graph. The graph in Fig. 5 was obtained by plotting the
points (0, 0) and (6, 4) and drawing the straight line through them.
Now Try Exercise 19
The next example gives an application of linear equations.
EXAMPLE 5
Linear Depreciation For tax purposes, businesses must keep track of the current v­ alues
of each of their assets. A common mathematical model is to assume that the current
value y is related to the age x of the asset by a linear equation. A moving company buys
a 40-foot van with a useful lifetime of 5 years. After x months of use, the value y, in
­dollars, of the van is estimated by the linear equation
y = 25,000 – 400x.
(a)
(b)
(c)
(d)
SOLUTION
Draw the graph of this linear equation.
What is the value of the van after 5 years?
When will the value of the van be $15,000?
What economic interpretation can be given to the y-intercept of the graph?
(a) The y-intercept is (0, 25,000). To find the x-intercept, set y = 0 and solve for x.
0 = 25,000 – 400x
y
dollars
(0, 25,000)
(62.5, 0)
Figure 6
x (months)
Set y = 0.
400x = 25,000
Add 400x to both sides.
x = 62.5
Divide both sides by 400.
The x-intercept is (62.5, 0). The graph of the linear equation is sketched in Fig. 6.
Note how the value decreases as the age of the van increases. The value of the
van reaches 0 after 62.5 months. Note also that we have sketched only the portion of the graph that has physical meaning—namely, the portion for x between
0 and 62.5.
1.1 Coordinate Systems and Graphs
5
(b) After 5 years (or 60 months), the value of the van is
y = 25,000 – 400(60) = 25,000 – 24,000 = 1000.
Since the useful life of the van is 5 years, this value represents the salvage value of
the van.
(c) Set the value of y to 15,000, and solve for x.
15,000 = 25,000 – 400x
400x + 15,000 = 25,000
400x = 10,000
x = 25
Set y = 15,000.
Add 400x to both sides.
Subtract 15,000 from both sides.
Divide both sides by 400.
The value of the van will be $15,000 after 25 months.
(d) The y-intercept corresponds to the value of the van at x = 0 months—that is, the
Now Try Exercise 41
initial value of the van, $25,000.
INCORPORATING
TECHNOLOGY
Appendix B contains instructions for TI-84 Plus calculators. (For the specifics of
other calculators, consult the guidebook for the calculator.) The appendix shows
how to obtain the graph of a linear equation of the form y = mx + b, find coordinates
of points on the line, and determine intercepts. Vertical lines can be drawn with the
Vertical command from the draw menu. To draw the vertical line x = k, go to the
­ ertical, type in the value of k,
home screen, press 2nd [draw] 4 to display the word V
and press ENTER .
Appendix D contains an introduction to Wolfram | Alpha.
Straight lines can be drawn with instructions of the following forms:
plot ax + by = c; plot y = ax + b; plot x = a
If a phrase of the form for x from x1 to x2 is appended to the instruction, only the portion
of the line having x-values from x1 to x2 will be drawn.
An equation of the form ax + by = c, with b ≠ 0, can be converted to the form
y = mx + b with the instruction solve ax + by = c for y.
The intercepts of an equation can be found with an instruction of the form intercepts
[equation]. An expression in x can be evaluated at x = a with an instruction of the form
evaluate [expression] at x = a. For instance, the instruction
evaluate 2500 − 400x at x = 5
gives the result 500.
Check Your Understanding 1.1
Solutions can be found following the section exercises.
2. Is the point (4, -7) on the graph of the linear equation
2x – 3y = 1? Is the point (5, 3)?
1. Plot the point (500, 200).
EXERCISES 1.1
In Exercises 1–8, plot the given point.
1. (2, 3)
2. ( -1, 4)
5. ( -2, 1)
6. ( -1, – 52 )
3. (0, -2)
4. (2, 0)
7. ( -20, 40)
8. (25, 30)
6
chapter 1 Linear Equations and Straight Lines
y
38. Which of the following equations is graphed in Fig. 9?
(a) x + y = 3   (b) y = x – 1   (c) 2y = x + 3
Q
y
1
x
1
(5, 4)
P
(1, 2)
Figure 7
x
9. What are the coordinates of the point Q in Fig. 7?
10. What are the coordinates of the point P in Fig. 7?
In Exercises 11–14, determine whether the point is on the graph of
the equation -2x + 13 y = -1.
11. (1, 3)
12. (2, 6)
13.
( 12, 3 )
( 13, -1 )
14.
In Exercises 15–18, each linear equation is in the form y = mx + b.
Identify m and b.
15. y = 5x + 8
16. y = -2x – 6
17. y = 3
18. y = 23 x
In Exercises 19–22, write each linear equation in the form
y = mx + b or x = a.
19. 14x + 7y = 21
20. x – y = 3
21. 3x = 5
22. – 12 x + 23 y = 10
In Exercises 23–26, find the x-intercept and the y-intercept of each line.
23. y = -4x + 8
24. y = 5
25. x = 7
26. y = -8x
In Exercises 27–34, graph the given linear equation.
5
2
27. y = 13 x – 1
28. y = 2x
29. y =
30. x = 0
31. 3x + 4y = 24
32. x + y = 3
33. x =
– 52
34.
1
2x
–
1
3y
= -1
35. Which of the following equations describe the same line as the
equation 2x + 3y = 6?
(a) 4x + 6y = 12
(b) y = – 23 x + 2
(c) x = 3 – 32 y
2
(d) 6 – 2x – y = 0 (e) y = 2 – 3 x
(f) x + y = 1
36. Which of the following equations describe the same line as the
equation 12 x – 5y = 1?
(a) 2x – 15 y = 1
(b) x = 5y + 2
(c) 2 – 5x + 10y = 0
(d) y = .1(x – 2)
(e) 10y – x = -2
(f) 1 + .5x = 2 + 5y
37. Each of the lines L1, L2, and L3 in Fig. 8 is the graph of one
of the equations (a), (b), and (c). Match each of the equations
with its corresponding line.
(a) x + y = 3
(b) 2x – y = -2
(c) x = 3y + 3
y
(0, 3)
L1
(0, 2)
(3, 0)
(21, 0)
(0, 21)
Figure 8
L2
x
L3
Figure 9
39. Heating Water The temperature of water in a heating tea kettle rises according to the equation y = 30x + 72, where y is
the temperature (in degrees Fahrenheit) x minutes after the
kettle was put on the burner.
(a) What physical interpretation can be given to the y-­intercept
of the graph?
(b) What will the temperature of the water be after 3 minutes?
(c) After how many minutes will the water be at its boiling
point of 212°?
40. Life Expectancy The average life expectancy y of a person
born x years after 1960 can be approximated by the linear
equation y = 16x + 70.
(a) What interpretation can be given to the y-intercept of the
graph?
(b) In what year did people born that year have an average
life expectancy of 75 years?
(c) What is the average life expectancy of people born in
1999?
41. Cigarette Consumption The worldwide consumption of cigarettes has been increasing steadily in recent years. The number
of trillions of cigarettes, y, purchased x years after 1960, is
estimated by the linear equation y = .075x + 2.5.
(a) Draw the graph of this linear equation.
(b) What interpretation can be given to the y-intercept of the
graph?
(c) When were there 4 trillion cigarettes sold?
(d) If this trend continues, how many cigarettes will be sold
in the year 2024?
42. Ecotourism Income In a certain developing country, ecotourism income has been increasing in recent years. The income y
(in thousands of dollars) x years after 2000 can be modeled
by y = 1.15x + 14.
(a) Draw the graph of this linear equation.
(b) What interpretation can be given to the y-intercept of
this graph?
(c) When was there $20,000 in ecotourism income?
(d) If this trend continues, how much ecotourism income will
there be in 2022?
43. Insurance Rates Yearly car insurance rates have been increasing steadily in the last few years. The rate y (in dollars) for a
small car x years after 1999 can be modeled by y = 23x + 756.
(a) Draw the graph of this linear equation.
(b) What interpretation can be given to the y-intercept of
this graph?
(c) What was the yearly rate in 2007?
(d) If this trend continues, when will the yearly rate be
$1308?
1.1 Coordinate Systems and Graphs
44. Simple Interest If $1000 is deposited at 3% simple interest, the balance y after x years will be given by the equation
y = 30x + 1000.
(a) Draw the graph of this linear equation.
(b) Find the balance after two years.
(c) When will the balance reach $1180?
45. College Freshmen The percentage, y, of college freshmen
who entered college intending to major in general biology
increased steadily from the year 2000 to the year 2014 and can
be approximated by the linear equation y = .2x + 4.1 where x
represents the number of years since 2000. Thus, x = 0 represents 2000, x = 1 represents 2001, and so on. (Source: The
American Freshman: National Norms.)
(a) What interpretation can be given to the y-intercept of the
graph of the equation?
(b) In 2014, approximately what percent of college freshmen
intended to major in general biology?
(c) In what year did approximately 5.5% of college freshmen
intend to major in general biology?
46. College Freshmen The percentage, y, of college freshmen who
smoke cigarettes decreased steadily from the year 2004 to the
year 2014 and can be approximated by the linear equation
y = -.46x + 6.32 where x represents the number of years
since 2004. Thus, x = 0 represents 2004, x = 1 represents
2005, and so on. (Source: The American Freshman: National
Norms.)
(a) What interpretation can be given to the y-intercept of the
graph of the equation?
(b) In 2014, approximately what percent of college freshmen
smoked?
(c) In what year did approximately 2.6% of college freshmen
smoke?
47. College Tuition Average tuition (including room and board)
for all institutions of higher learning in year x can be
approximated by y = 461x + 16,800 dollars, where x = 0
corresponds to 2004, x = 1 corresponds to 2005, and so on.
(Source: U.S. National Center of Education Statistics.)
(a) Approximately what was the average tuition in 2011?
(b) Assuming that the formula continues to hold, when will
the average tuition exceed $25,000?
48. Bachelor’s Degrees The number of bachelor’s degrees conferred in mathematics and statistics in year x can be approximated by y = 667x + 12,403, where x = 0 corresponds to
2003, x = 1 corresponds to 2004, and so on. (Source: U.S.
National Center of Education Statistics.)
(a) Approximately how many bachelor’s degrees in mathematics and statistics were awarded in 2007?
(b) Assuming that the model continues to hold, approximately when will the number of bachelor’s degrees in
mathematics and statistics awarded exceed 25,000?
49. Find an equation of the line having x-intercept (16, 0) and
y-intercept (0, 8).
50. Find an equation of the line having x-intercept (.6, 0) and
y-intercept (0, .9).
7
51. Find an equation of the line having y-intercept (0, 5) and
x-intercept (4, 0).
52. Find an equation of the line having x-intercept (5, 0) and parallel to the y-axis.
53. What is the equation of the x-axis?
54. Can a line other than the x-axis have more than one
x-intercept?
55. What is the general form of the equation of a line that is parallel to the y-axis?
56. What is the general form of the equation of a line that is parallel to the x-axis?
In Exercises 57–60, find a general form of the given equation.
57. y = 2x + 3
59. y =
– 23x
-5
58. y = 3x – 4
60. y = 4x –
5
6
61. Show that the straight line with x-intercept (a, 0) and y-intercept
(0, b), where a and b are not zero, has bx + ay = ab as a general
form of its equation.
62. Use the result of Exercise 61 to find a general form of the
equation of the line having x-intercept (5, 0) and y-intercept
(0, 6).
In Exercises 63–70, give the equation of a line having the stated
property. Note: There are many answers to each exercise.
63. x-intercept (9, 0)
64. y-intercept (0, 10)
65. passes through the point ( -2, 5)
66. passes through the point (3, -3)
67. crosses the positive part of the y-axis
68. passes through the origin
69. crosses the negative part of the x-axis
70. crosses the positive part of the x-axis
71. The lines with equations y = 23x – 2 and y = -4x + c have
the same x-intercept. What is the value of c?
72. The lines with equations 6x – 3y = 9 and y = 4x + b have
the same y-intercept. What is the value of b?
TECHNOLOGY EXERCISES
In Exercises 73–76, (a) graph the line, (b) use the utility to determine the two intercepts, (c) use the utility to find the y-coordinate
of the point on the line with x-coordinate 2.
73. y = -3x + 6
74. y = .25x – 2
75. 3y – 2x = 9
76. 2y + 5x = 8
In Exercises 77 and 78, determine an appropriate window, and
graph the line.
77. 2y + x = 100
78. x – 3y = 60
Solutions to Check Your Understanding 1.1
1. Because the numbers are large, make each hatchmark correspond to 100. Then the point (500, 200) is found by starting at
the origin, moving 500 units to the right and 200 units up
(Fig. 10 on the next page).
8
chapter 1 Linear Equations and Straight Lines
2.
y
2x – 3y = 1
?
2(4) – 3( -7) = 1
?
29 = 1
(500, 200)
x = 4, y = -7
False
Since the equation is not satisfied, (4, -7) is not on the graph.
100
x
2x – 3y = 1
100
?
2(5) – 3(3) = 1
1=1
Figure 10
1.2
Given equation
Given equation
x = 5, y = 3
True
Since the equation is satisfied, (5, 3) is on the graph.
The Slope of a Straight Line
In this section, we consider only lines whose equations can be written in the form
y = mx + b. Geometrically, this means that we will consider only nonvertical lines.
Slope is not defined for vertical lines.
DEFINITION Given a nonvertical line L with equation y = mx + b, the number m is
called the slope of L. That is, the slope is the coefficient of x in the equation of the
line. The equation is called the slope–intercept form of the equation of the line.
EXAMPLE 1
SOLUTION
Finding the Slope of a Line from its Equation Find the slopes of the lines having the
following equations:
(a) y = 2x + 1   (b) y = – 34 x + 2   (c) y = 3   (d) -8x + 2y = 4
(a)
(b)
(c)
(d)
m = 2.
m = – 34 .
When we write the equation in the form y = 0 # x + 3, we see that m = 0.
First, write the equation in slope–intercept form.
-8x + 2y = 4
Given equation
2y = 8x + 4
y = 4x + 2
Add 8x to both sides.
Divide both sides by 2.
Thus, m = 4.
Now Try Exercise 1
The definition of the slope is given in terms of an equation of the line. There is an
alternative equivalent definition of slope.
DEFINITION Alternative Definition of Slope Let L be a line passing through the
points (x1, y1) and (x2, y2), where x1 ≠ x2. Then, the slope of L is given by the formula
m=
y2 – y1
.
x2 – x1
(1)
That is, the slope is the difference in the y-coordinates divided by the difference in the
x-coordinates, with both differences formed in the same order. Note: x1 is pronounced
“x sub 1.”
Before proving this definition equivalent to the first one given, let us show how it
can be used.
1.2 The Slope of a Straight Line
EXAMPLE 2
SOLUTION
9
Finding the Slope of a Line from Two Points Find the slope of the line passing through
the points (1, 3) and (4, 6).
We have
m=
[difference in y@coordinates] 6 – 3 3
=
= = 1.
[difference in x@coordinates] 4 – 1 3
Thus, m = 1. Note that if we reverse the order of the points and use formula (1) to compute the slope, then we get
3-6
-3
=
= 1,
1-4
-3
m=
which is the same answer. The order of the points is immaterial. The important concern is to make sure that the differences in the x- and y-coordinates are formed in the
Now Try Exercise 7
same order.
The slope of a line does not depend on which pair of points we choose as (x1, y1)
and (x2, y2). Consider the line y = 4x – 3 and two points (1, 1) and (3, 9), which are on
the line. Using these two points, we calculate the slope to be
m=
9-1 8
= = 4.
3-1 2
Now, let us choose two other points on the line—say, (2, 5) and ( -1, -7)—and use these
points to determine m. We obtain
m=
-7 – 5
-12
=
= 4.
-1 – 2
-3
The two pairs of points give the same slope.
Justification of Formula (1) Since (x1, y1) and (x2, y2) are both on the line, both
points satisfy the equation of the line, which has the form y = mx + b. Thus,
y2 = mx2 + b
y1 = mx1 + b.
Subtracting these two equations gives
y2 – y1 = mx2 – mx1 = m(x2 – x1).
Dividing by x2 – x1, we have
m=
y2 – y1
,
x2 – x1
which is formula (1). So the two definitions of slope lead to the same number.
Let us now study four of the most important properties of the slope of a straight
line. We begin with the steepness property, since it provides us with a geometric interpretation for the number m.
Steepness Property Let the line L have slope m. If we start at any point on the line
and move 1 unit to the right, then we must move m units vertically in order to return
to the line (Fig. 1 on the next page). (Of course, if m is positive, then we move up; and
if m is negative, we move down.)
10 chapter 1 Linear Equations and Straight Lines
y
y
m
y
1
1
x
Figure 1
EXAMPLE 3
1
m
x
x
m negative
m positive
m50
Steepness Property of a Line Illustrate the steepness property for each of the lines.
(b) y = – 34 x + 2
(c) y = 3
(a) y = 2x + 1
SOLUTION
(a) Here, m = 2. So starting from any point on the line, proceeding 1 unit to the right,
we must go 2 units up to return to the line (Fig. 2).
(b) Here, m = – 34 . So starting from any point on the line, proceeding 1 unit to the
right, we must go 34 unit down to return to the line (Fig. 3).
(c) Here, m = 0. So going 1 unit to the right requires going 0 units vertically to return
to the line (Fig. 4).
y
y
y
y 5 2x 1 1
y 5 2 34 x 1 2
2
1
1
1
y53
2 34
x
x
x
Figure 3
Figure 2
Figure 4
Now Try Exercise 59
In the next example, we introduce a new method for graphing a linear equation.
This method relies on the steepness property and is often more efficient than finding
two points on the line (e.g., the two intercepts).
EXAMPLE 4
SOLUTION
Using the Steepness Property to Graph a Line Use the steepness property to draw the
graph of y = 12 x + 32 .
The y-intercept is ( 0, 32 ) , as we read from the equation. We can find another point on the
line by using the steepness property. Start at ( 0, 32 ) . Go 1 unit to the right. Since the slope
is 12 , we must move vertically 12 unit to return to the line. But this locates a second point
on the line. So we draw the line through the two points. The entire procedure is illustrated in Fig. 5.
y
y
(1, 2)
(0, 32 )
1
(1, 2)
(0, 32 )
1
2
x
x
Figure 5
Now Try Exercise 13
1.2 The Slope of a Straight Line
11
Actually, to use the steepness property to graph an equation, all that is needed is
the slope plus any point (not necessarily the y-intercept).
EXAMPLE 5
Using the Steepness Property to Graph a Line Graph the line of slope -1, which
passes through the point (2, 2).
SOLUTION
y
1
21
(2, 2)
(3, 1)
x
Start at (2, 2), move 1 unit to the right and then -1 unit vertically—that is, 1 unit down.
The line through (2, 2) and the resulting point is the desired line. (See Fig. 6.)
Slope measures the steepness of a line. That is, the slope of a line tells whether it is
rising or falling, and how fast. Specifically, lines of positive slope rise as we move from
left to right. Lines of negative slope fall, and lines of zero slope stay level. The larger the
magnitude of the slope, the steeper the ascent or descent will be. These facts are directly
implied by the steepness property. (See Fig. 7.)
y
Figure 6
y
m52
m 5 22
m51
m5
m 5 21
1
2
m 5 2 12
x
x
Figure 7
Justification of the Steepness Property Consider a line with equation y = mx + b,
y
(x1 1 1, y2)
(x1, y1)
1
x
y 5 mx 1 b
and let (x1, y1) be any point on the line. If we start from this point and move 1 unit to
the right, the first coordinate of the new point will be x1 + 1, since the x-coordinate is
increased by 1. Now, go far enough vertically to return to the line. Denote the y-coordinate
of this new point by y2. (See Fig. 8.) We must show that to get y2, we add m to y1. That
is, y2 = y1 + m. By equation (1), we can compute m as
m=
[difference in y@coordinates] y2 – y1
=
= y2 – y1.
[difference in x@coordinates]
1
In other words, y2 = y1 + m, which is what we desired to show.
Figure 8
Often, the slopes of the straight lines that occur in applications have interesting and
significant interpretations. An application in the field of economics is illustrated in the
next example.
EXAMPLE 6
SOLUTION
Slope of the Cost Line A manufacturer finds that the cost y of producing x units of a
certain commodity is given by the equation y = 2x + 5000. What interpretation can be
given to the slope of the graph of this equation?
Suppose that the firm is producing at a certain level and increases production by 1 unit.
That is, x is increased by 1 unit. By the steepness property, the value of y then increases
by 2, which is the slope of the line whose equation is y = 2x + 5000. Thus, each additional unit of production costs $2. The graph of y = 2x + 5000 is called a cost curve. It
relates the size of production to total cost. The graph is a straight line, and economists
call its slope the marginal cost of production. The y-coordinate of the y-intercept is
called the fixed cost. In this case, the fixed cost is $5000, and it includes costs such as
rent and insurance, which are incurred even if no units are produced.
Now Try Exercise 35
12 chapter 1 Linear Equations and Straight Lines
In applied problems having time as a variable, the letter t is often used in place of
the letter x. If so, straight lines have equations of the form y = mt + b and are graphed
on a ty-coordinate system.
EXAMPLE 7
SOLUTION
Straight-Line Depreciation The federal government allows businesses an income tax
deduction for the decrease in value (or depreciation) of capital assets (such as buildings
and equipment). One method of calculating the depreciation is to take equal amounts
over the expected lifetime of the asset. This method is called straight-line depreciation.
Suppose that, for tax purposes, the value V of a piece of equipment t years after purchase is figured according to the equation V = -100,000t + 700,000 and the expected
life of the piece of equipment is 5 years.
(a) How much did the piece of equipment originally cost?
(b) What is the annual deduction for depreciation?
(c) What is the salvage value of the piece of equipment? (That is, what is the value of
the piece of equipment after 5 years?)
(a) The original cost is the value of V at t = 0, namely
V = -100,000(0) + 700,000 = 700,000.
That is, the piece of equipment originally cost $700,000.
(b) By the steepness property, each increase of 1 in t causes a decrease in V of 100,000.
That is, the value is decreasing by $100,000 per year. So the depreciation deduction
is $100,000 each year.
(c) After 5 years, the value of V is given by
V = -100,000(5) + 700,000 = 200,000.
The salvage value is $200,000.
We have seen in Example 5 how to sketch a straight line when given its slope and
one point on it. Let us now see how to find the equation of the line from this data.
Point-Slope Equation The equation of the straight line passing through (x1, y1) and
having slope m is given by y – y1 = m(x – x1).
EXAMPLE 8
SOLUTION
Finding the Equation of a Line from Its Slope and a Point on the Line Find the slope–
intercept equation of the line that passes through (2, 3) and has slope 12 .
Here, x1 = 2, y1 = 3, and m = 12 . So the point–slope equation is
y – 3 = 12 (x – 2)
y – 3 = 12 x – 1
y=
EXAMPLE 9
SOLUTION
1
2x
+2
Perform multiplication on right side.
Add 3 to both sides.
Now Try Exercise 49
Finding the Equation of a Line Find the slope–intercept equation of the line through
the points (3, 1) and (6, 0).
We can compute the slope from equation (1).
m=
y2 – y1
1-0
1
=
= – .
x2 – x1 3 – 6
3
Now, we can determine the equation from the point–slope equation with (x1, y1) = (3, 1)
and m = – 13 .
1.2 The Slope of a Straight Line
13
y – 1 = – 13 (x – 3) Point-slope equation
y – 1 = – 13 x + 1
y=
– 13 x
+2
Perform multiplication on right side.
Add 1 to both sides.
[Question: What would the equation be if we had chosen (x1, y1) = (6, 0)?]
Now Try Exercise 55
EXAMPLE 10
SOLUTION
Sales Generated by Advertising For each dollar of monthly advertising expenditure,
a store experiences a 6-dollar increase in sales. Even without advertising, the store has
$30,000 in sales per month. Let x be the number of dollars of advertising expenditure
per month, and let y be the number of dollars in sales per month.
(a) Find the equation of the line that expresses the relationship between x and y.
(b) If the store spends $10,000 in advertising, what will be the sales for the month?
(c) How much would the store have to spend on advertising to attain $150,000 in sales
for the month?
(a) The steepness property tells us that the line has slope m = 6. Since x = 0 (no advertising expenditure) yields y = $30,000, the y-intercept of the line is (0, 30,000).
Therefore, the slope–intercept equation of the line is
y = 6x + 30,000.
(b) If x = 10,000, then y = 6(10,000) + 30,000 = 90,000. Therefore, the sales for the
month will be $90,000.
(c) We are given that y = 150,000, and we must find the value of x for which
150,000 = 6x + 30,000.
Solving for x, we obtain 6x = 120,000, and hence, x = $20,000. To attain $150,000
Now Try Exercise 45
in sales, the store should invest $20,000 in advertising.
Verification of the Point–Slope Equation Let (x, y) be any point on the line passing
through the point (x1, y1) and having slope m. Then, by equation (1), we have
m=
y – y1
.
x – x1
Multiplying through by x – x1 gives
y – y1 = m(x – x1).
(2)
Thus, every point (x, y) on the line satisfies equation (2). So (2) gives the equation of the
line passing through (x1, y1) and having slope m.
Perpendicular and Parallel Lines
The next property of slope relates the slopes of two perpendicular lines.
Perpendicular Property When two nonvertical lines are perpendicular, their slopes
are negative reciprocals of one another. That is, if two lines with nonzero slopes m
and n are perpendicular to one another, then
1
m= – .
n
Conversely, if two lines have slopes that are negative reciprocals of one another, they
are perpendicular.
14 chapter 1 Linear Equations and Straight Lines
A proof of the perpendicular property is outlined in Exercise 88. Let us show how it
can be used to help find equations of lines.
EXAMPLE 11
SOLUTION
Perpendicular Lines Find an equation of the line perpendicular to the graph of
y = 2x – 5 and passing through (1, 2).
The slope of the graph of y = 2x – 5 is 2. By the perpendicular property, the slope of a
line perpendicular to it is – 12 . If a line has slope – 12 and passes through (1, 2), it has the
point-slope equation
y – 2 = – 12 (x – 1) or y = – 12 x + 52 .

Now Try Exercise 21
The final property of slope gives the relationship between slopes of parallel lines. A
proof is outlined in Exercise 87.
Parallel Property Parallel lines have the same slope. Conversely, if two different
lines have the same slope, they are parallel.
EXAMPLE 12
SOLUTION
Parallel Lines Find an equation of the line through (2, 0) and parallel to the line whose
equation is y = 13 x – 11.
The slope of the line having equation y = 13 x – 11 is 13 . Therefore, any line parallel to it
also has slope 13 . Thus, the desired line passes through (2, 0) and has slope 13 , so its equation is
y – 0 = 13 (x – 2) or y = 13 x – 23.
Now Try Exercise 23
INCORPORATING
TECHNOLOGY
A graphing calculator can find the equation of the line through two points. Refer to
the graphing calculator discussion in the Incorporating Technology feature of
­ ection 1.4 and find the equation of the least-squares fit to the two points.
S
Excel can find the equation of the line through two points. Refer to the Excel
discussion in the Incorporating Technology feature of Section 1.4 and find the
equation of the least-squares fit to the two points.
The following instructions produce the equation of the line described.
line through (a, b) and (c, d )
line through (a, b) with slope m
line through (a, b) perpendicular to y = mx + b
line through (a, b) parallel to y = mx + b
Check Your Understanding 1.2
Suppose that the revenue y from selling x units of a certain commodity is given by the formula y = 4x. (Revenue is the amount of
money received from the sale of the commodity.)
1. What interpretation can be given to the slope of the graph of
this equation?
Solutions can be found following the section exercises.
2. The cost curve discussed in Example 6 intersects the revenue
curve at the point (2500, 10,000). What economic interpretation can be given to the value of the x-coordinate of the
intersection point?
1.2 The Slope of a Straight Line
EXERCISES 1.2
In Exercises 1–6, find the slope of the line having the given equation.
1. y = 23 x + 7
2. y = -4
3. y – 3 = 5(x + 4)
4. 7x + 5y = 10
5.
x y
+ =6
5 4
6.
20.
y
L
x y
– =1
7 8
(1, 2)
(21, 12 )
In Exercises 7–10, plot each pair of points, draw the straight line
through them, and find its slope.
x
7. (3, 4), (7, 9)
8. ( -2, 1), (3,-3)
y
21.
9. (0, 0), (5, 4)
y 5 24x 1 10
10. (4, 17), ( -2, 17)
11. What is the slope of any line parallel to the y-axis?
12. Why doesn’t it make sense to talk about the slope of the line
between the two points (2, 3) and (2, -1)?
(2, 2)
L
In Exercises 13–16, graph the given linear equation by beginning at
the y-intercept, and moving 1 unit to the right and m units in the
y-direction.
13. y = -2x + 1
14. y = 4x – 2
15. y = 3x
16. y = -2
x
L perpendicular to y 5 24x 1 10
22.
y
In Exercises 17–24, find the equation of line L.
17.
(5, 3)
y
(2, 3)
L
y 5 13 x
1
x
22
L parallel to y 5 13 x
x
L
y
23.
18.
y
L
(3, 1)
1
y 5 2x 1 2
x
1
2
L
x
L parallel to y 5 2x 1 2
19.
24.
y
y
L
x
(2, 21)
(1, 2)
(2, 0)
L
x
y 5 212 x
L perpendicular to y 5 212 x
15
16 chapter 1 Linear Equations and Straight Lines
In Exercises 25–28, give the slope–intercept form of the equation
of the line.
35. Cost Curve A manufacturer has fixed costs (such as rent and
insurance) of $2000 per month. The cost of producing each
unit of goods is $4. Give the linear equation for the cost of
producing x units per month.
y
25.
5
x
5
25
25
y
26.
5
x
5
25
25
y
27.
5
x
5
25
36. Demand Curve The price p that must be set in order to sell q
items is given by the equation p = -3q + 1200.
(a) Find and interpret the p-intercept of the graph of the
equation.
(b) Find and interpret the q-intercept of the graph of the
equation.
(c) Find and interpret the slope of the graph of the equation.
(d) What price must be set in order to sell 350 items?
(e) What quantity will be sold if the price is $300?
(f) Draw the graph of the equation.
37. Boiling Point of Water At sea level, water boils at a temperature of 212°F. As the altitude increases, the boiling point
of water decreases. For instance, at an altitude of 5000 feet,
water boils at about 202.8°F.
(a) Find a linear equation giving the boiling point of water in
terms of altitude.
(b) At what temperature does water boil at the top of Mt.
Everest (altitude 29,029 feet)?
38. Cricket Chirps Biologists have found that the number of chirps
that crickets of a certain species make per minute is related to
the temperature. The relationship is very close to linear. At
68°F, those crickets chirp about 124 times a minute. At 80°F,
they chirp about 172 times a minute.
(a) Find the linear equation relating Fahrenheit temperature
F and the number of chirps c.
(b) If you count chirps for only 15 seconds, how can you
quickly estimate the temperature?
39. Cost Equation Suppose that the cost of making 20 cell phones
is $6800 and the cost of making 50 cell phones is $9500.
(a) Find the cost equation.
(b) What is the fixed cost?
(c) What is the marginal cost of production?
(d) Draw the graph of the equation.
25
y
28.
34. Find the equation of the line passing through the point (1, 4)
and having y-intercept (0, 4).
5
Exercises 40–42 are related.
x
5
25
25
29. Find the equation of the line passing through the point (2, 3)
and parallel to the x-axis.
30. Find the equation of the line passing through the point (2, 3)
and parallel to the y-axis.
31. Find the y-intercept of the line passing through the point
(5, 6) and having slope 35 .
32. Find the y-intercept of the line passing through the points
( -1, 3) and (4, 6).
33. Find the equation of the line passing through (0, 4) and having undefined slope.
40. Cost Equation Suppose that the total cost y of making x coats
is given by the formula y = 40x + 2400.
(a) What is the cost of making 100 coats?
(b) How many coats can be made for $3600?
(c) Find and interpret the y-intercept of the graph of the
equation.
(d) Find and interpret the slope of the graph of the equation.
41. Revenue Equation Suppose that the total revenue y from the
sale of x coats is given by the formula y = 100x.
(a) What is the revenue if 300 coats are sold?
(b) How many coats must be sold to have a revenue of $6000?
(c) Find and interpret the y-intercept of the graph of the
equation.
(d) Find and interpret the slope of the graph of the equation.
42. Profit Equation Consider a coat factory with the cost and revenue equations given in Exercises 40 and 41.
(a) Find the equation giving the profit y resulting from making and selling x coats.
17
1.2 The Slope of a Straight Line
(b) Find and interpret the y-intercept of the graph of the
profit equation.
(c) Find and interpret the x-intercept of the graph of the
profit equation.
(d) Find and interpret the slope of the graph of the profit
equation.
(e) How much profit will be made if 80 coats are sold?
(f) How many coats must be sold to have a profit of $6000?
(g) Draw the graph of the equation found in part (a).
43. Heating Oil An apartment complex has a storage tank to
hold its heating oil. The tank was filled on January 1, but no
more deliveries of oil will be made until sometime in March.
Let t denote the number of days after January 1, and let y
denote the number of gallons of fuel oil in the tank. Current records show that y and t will be related by the equation
y = 30,000 – 400t.
(a) Graph the equation y = 30,000 – 400t.
(b) How much oil will be in the tank on February 1?
(c) How much oil will be in the tank on February 15?
(d) Determine the y-intercept of the graph. Explain its significance.
(e) Determine the t-intercept of the graph. Explain its significance.
44. Cash Reserves A corporation receives payment for a large
contract on July 1, bringing its cash reserves to $2.3 million.
Let y denote its cash reserves (in millions) t days after July 1.
The corporation’s accountants estimate that y and t will be
related by the equation y = 2.3 – .15t.
(a) Graph the equation y = 2.3 – .15t.
(b) How much cash does the corporation have on the morning of July 16?
(c) Determine the y-intercept of the graph. Explain its
significance.
(d) Determine the t-intercept of the graph. Explain its
significance.
(e) Determine the cash reserves on July 4.
(f) When will the cash reserves be $.8 million?
45. Weekly Pay A furniture salesperson earns $220 a week plus
10% commission on her sales. Let x denote her sales and y her
income for a week.
(a) Express y in terms of x.
(b) Determine her week’s income if she sells $2000 in merchandise that week.
(c) How much must she sell in a week in order to earn $540?
46. Weekly Pay A salesperson’s weekly pay depends on the volume of sales. If she sells x units of goods, then her pay is
y = 5x + 60 dollars. Give an interpretation to the slope and
the y-intercept of this straight line.
In Exercises 47–58, find an equation for each of the following lines.
47. Slope is – 12 ; y-intercept is (0, 0).
48. Slope is 3; y-intercept is (0, -1).
49. Slope is – 13 ; (6, -2) on line.
50. Slope is 1; (1, 2) on line.
51. Slope is 12 ; (2, -3) on line.
52. Slope is -7; (5, 0) on line.
53. Slope is – 25 ; (0, 5) on line.
54. Slope is 0; (7, 4) on line.
55. (5, -3) and ( -1, 3) on line.
56. (2, 1) and (4, 2) on line.
57. (2, -1) and (3, -1) on line.
58. (0, 0) and (1, -2) on line.
In each of Exercises 59–62, we specify a line by giving the slope
and one point on the line. We give the first coordinate of some
points on the line. Without deriving an equation of the line, find the
second coordinate of each of the points.
59. Slope is 2, (1, 3) on line; (2, ); (0, ); ( -1, ).
60. Slope is -3, (2, 2) on line; (3, ); (4, ); (1, ).
61. Slope is – 14 , ( -1, -1) on line; (0, ); (1, ); ( -2, ).
62. Slope is 13 , ( -5, 2) on line; ( -4, ); ( -3, ); ( -2, ).
63. Each of the lines (A), (B), (C), and (D) in Fig. 9 is the graph
of one of the linear equations (a), (b), (c), and (d). Match
each line with its equation.
y
y
2
2
22
2
x
22
22
2
x
22
sAd
sBd
y
y
2
22
2
2
x
22
22
2
x
22
sCd
sDd
Figure 9
(a) x + y = 1
(c) x + y = -1
(b) x – y = 1
(d) x – y = -1
64. The table that follows gives several points on the line
Y1 = mx + b. Find m and b.
X
Y1
4.8
4.9
5.0
5.1
5.2
5.3
5.4
3.6
4.8
6.0
7.2
8.4
9.6
10.8
Y1 5 10.8
18 chapter 1 Linear Equations and Straight Lines
In Exercises 65–70, give an equation of a line with the stated property. Note: There are many answers to each exercise.
65. rises as you move from left to right
66. falls as you move from left to right
67. has slope 0
68. slope not defined
69. parallel to the line 2x + 3y = 4
70. perpendicular to the line 5x + 6y = 7
71. Temperature Conversion Celsius and Fahrenheit temperatures
are related by a linear equation. Use the fact that 0°C = 32°F
and 100°C = 212°F to find an equation.
72. Dating of Artifacts An archaeologist dates a bone fragment
discovered at a depth of 4 feet as approximately 1500 b.c. and
dates a pottery shard at a depth of 8 feet as approximately
2100 b.c. Assuming that there is a linear relationship between
depths and dates at this archeological site, find the equation
that relates depth to date. How deep should the archaeologist
dig to look for relics from 3000 b.c.?
73. College Tuition The average college tuition and fees at fouryear public colleges increased from $3735 in 2001 to $8312
in 2013. (See Fig. 10.) Assuming that average tuition and fees
increased linearly with respect to time, find the equation that
relates the average tuition and fees, y, to the number of years
after 2001, x. What were the average tuition and fees in 2009?
(Source: National Center for Education Statistics, Digest of
Education Statistics.)
$12000
8
$9000
6
$6000
4
$3000
2
2001
2013
Figure 10 College Tuition
77. Bachelor’s Degrees in Business According to the U.S.
National Center of Education Statistics, 263,515 bachelor’s degrees in business were awarded in 2001 and 360,823
were awarded in 2013. If the number of bachelor’s degrees
in business continues to grow linearly, how many bachelor’s degrees in business will be awarded in 2020? (Source:
National Center for Education Statistics, Digest of Education Statistics.)
78. Pizza Stores According to Pizza Marketing Quarterly, the
number of U.S. Domino’s Pizza stores grew from 4818 in 2001
to 4986 in 2013. If the number of stores continues to grow
linearly, when will there be 5100 stores?
79. Super Bowl Commercials The average cost of a 30-second
advertising slot during the Super Bowl increased linearly from
$3.5 million in 2012 to $4.5 million in 2015. Find the equation
that relates the cost (in millions of dollars) of a 30-second
slot, y, to the number of years after 2012, x. What was the
average cost in 2014?
80. Straight-Line Depreciation A multi-function laser printer purchased for $3000 depreciates to a salvage value of $500 after
4 years. Find a linear equation that gives the depreciated value
of the multi-function laser printer after x years.
81. Supply Curve Suppose that 5 million tons of apples will be
supplied at a price of $3000 per ton and 6 million tons of
apples will be supplied at a price of $3400 per ton. Find the
equation for the supply curve and draw its graph. Let the
units for q be millions of tons and the units for p be thousands of dollars.
0
$0
number of home health aide jobs increases linearly during
that time, find the equation that relates the number of jobs,
y, to the number of years after 2014, x. Use the equation
to estimate the number of home health aide jobs in 2018.
(Source: Bureau of Labor Statistics, Occupational Projections
Data.)
2000
2013
Figure 11 College
Enrollments (in millions)
74. College Enrollments Two-year college enrollments increased
from 5.9 million in 2000 to 7.0 million in 2013. (See Fig. 11.)
Assuming that enrollments increased linearly with respect
to time, find the equation that relates the enrollment, y, to
the number of years after 2000, x. When was the enrollment
6.5 million? (Source: National Center for Education Statistics,
Digest of Education Statistics.)
75. Gas Mileage A certain car gets 25 miles per gallon when the
tires are properly inflated. For every pound of pressure that
the tires are underinflated, the gas mileage decreases by 12 mile
per gallon. Find the equation that relates miles per gallon, y, to
the amount that the tires are underinflated, x. Use the equation to calculate the gas mileage when the tires are underinflated by 8 pounds of pressure.
76. Home Health Aid Jobs According to the U.S. Department of
Labor, home health aide jobs are expected to increase from
913,500 in 2014 to 1,261,900 in 2024. Assuming that the
82. Demand Curve Suppose that 5 million tons of apples will be
demanded at a price of $3000 per ton and 4.5 million tons of
apples will be demanded at a price of $3100 per ton. Find the
equation for the demand curve and draw its graph. Let the
units for q be millions of tons and the units for p be thousands of dollars.
83. Show that the points (1, 3), (2, 4), and (3, -1) are not on the
same line.
84. For what value of k will the three points (1, 5), (2, 7), and
(3, k) be on the same line?
85. Find the value of a for which the line through the points (a, 1)
and (2, -3.1) is parallel to the line through the points ( -1, 0)
and (3.8, 2.4)
86. Rework Exercise 85, where the word parallel is replaced by
the word perpendicular.
87. Prove the parallel property. [Hint: If y = mx + b and
y = m′x + b′ are the equations of two lines, then the two
lines have a point in common if and only if the equation
mx + b = m′x + b′ has a solution for x.]
88. Prove the perpendicular property. [Hint: Without loss of
generality, assume that both lines pass through the origin.
Use the point–slope formula, the Pythagorean theorem,
and Fig. 12.]
1.2 The Slope of a Straight Line
are $1,000,000. If the product sells for $130 per unit, how
many units must the company produce and sell in order to
attain an annual profit of $2,000,000?
y
l2
l1
a
b
m1
x
m2
Figure 12
89. Temperature Conversion Figure 13 gives the conversion of
temperatures from Celsius to Fahrenheit. What is the Fahrenheit equivalent of 30°C?
212
Fahrenheit
32
0
19
100
Celsius
Figure 13
90. Shipping Costs Figure 14 gives the cost of shipping a package
from coast to coast. What is the cost of shipping a 20-pound
package?
$
93. Demand and Revenue Suppose that the quantity q of a certain brand of mountain bike sold each week depends on price
according to the equation q = 800 – 4p. What is the total
weekly revenue if a bike sells for $150?
94. Demand and Revenue Suppose that the number n of singleuse cameras sold each month varies with the price, according
to the equation n = 2200 – 25p . What is the monthly revenue
if the price of each camera is $8?
95. Setting a Price During 2015, a manufacturer produced 50,000
items that sold for $100 each. The manufacturer had fixed
costs of $600,000 and made a profit before income taxes of
$400,000. In 2016, rent and insurance combined increased
by $200,000. Assuming that the quantity produced and all
other costs were unchanged, what should the 2016 price be if
the manufacturer is to make the same $400,000 profit before
income taxes?
96. Setting a Price Rework Exercise 95 with a 2015 fixed cost of
$800,000 and a profit before income taxes of $300,000.
TECHNOLOGY EXERCISES
  97. Graph the three lines y = 2x – 3, y = 2x, and y = 2x + 3
together, and then identify each line without using trace.
  98.  Graph the two lines y = .5x + 1 and y = -2x + 9 in the
standard window [-10, 10] by [ -10, 10]. Do they appear
perpendicular? If not, use ZSquare to obtain true aspect,
and look at the graphs.
38
Cost
5
0
60
Weight
in pounds
Figure 14
91. Costs and Revenue A T-shirt company has fixed costs of
$25,000 per year. Each T-shirt costs $8.00 to produce and sells
for $12.50. How many T-shirts must the company produce
and sell each year in order to make a profit of $65,000?
92. Costs and Revenue A company produces a single product for
which variable costs are $100 per unit and annual fixed costs
  99.  Graph the line y = -.5x + 2 with the window ZDecimal.
Without pressing TRACE , move the cursor to a point on the
line. Then move the cursor one unit to the right and down .5
unit to return to the line. If you start at a point on the line and
move 2 units to the right, how many units down will you have
to move the cursor to return to the line? Test your answer.
100.  Graph the three lines y = 2x + 1, y = x + 1, and
y = .5x + 1 together, and then identify each line without
using trace.
101. Repeat Exercise 99 for the line y = .7x – 2, using up instead
of down and .7 instead of .5.
Solutions to Check Your Understanding 1.2
1. By the steepness property, whenever x is increased by
1 unit, the value of y is increased by 4 units. Therefore,
each additional unit of production brings in $4 of revenue.
(The graph of y = 4x is called a revenue curve, and its
slope is called the marginal revenue of production.)
revenue
cost
2. When 2500 units are produced, the revenue equals the
cost. This value of x is called the break-even point. Since
profit = (revenue) – (cost), the company will make a
profit only if its level of production is greater than the
break-even point (Fig. 15).
2500
Figure 15
20 chapter 1 Linear Equations and Straight Lines
The Intersection Point of a Pair of Lines
1.3
y
L
S 5 (x, y)
x
M
Suppose that we are given a pair of intersecting straight lines L and M. Let us consider
the problem of determining the coordinates of the point of intersection S = (x, y). (See
Fig. 1.) We may as well assume that the equations of L and M are given in slope–intercept
or vertical form. First, let us assume that both lines are in slope–intercept form—that is,
that the equations are
L: y = mx + b,
M: y = nx + c.
Since the point S is on both lines, its coordinates satisfy both equations. In particular,
we have two expressions for its y-coordinate:
Figure 1
y = mx + b = nx + c.
The last equality gives an equation from which x can easily be determined. Then, the
value of y can be determined as mx + b (or nx + c). Let us see how this works in a particular example.
EXAMPLE 1
SOLUTION
y
Finding the Point of Intersection Find the point of intersection of the two lines
y = 2x – 3 and y = x + 1.
To find the x-coordinate of the point of intersection, equate the two expressions for y
and solve for x.
2x – 3 = x + 1
y5x11
x-3=1
(4, 5)
5
x=4
y 5 2x 2 3
Equate the two expressions for y.
Subtract x from both sides.
Add 3 to both sides.
To find the value of y, set x = 4 in either equation—say, the first. Then,
y = 2 # 4 – 3 = 5.
x
5
So the point of intersection is (4, 5). See Fig. 2.
Figure 2
EXAMPLE 2
SOLUTION
Now Try Exercise 1
Finding the Point of Intersection Find the point of intersection of the two lines
x + 2y = 6 and 5x + 2y = 18.
To use the method described above, the equations must be in slope–intercept form. Solving both equations for y, we get
y = – 12 x + 3
y = – 52 x + 9.
Equating the expressions for y gives
y
5
2x
– 12x + 3 = – 52 x + 9
– 12 x + 3 = 9
Combine x terms.
2x + 3 = 9
5
x 1 2y 5 6
Divide both sides by 2.
x=3
( 3, 32 )
x
5
Figure 3
Subtract 3 from both sides.
2x = 6
5x 1 2y 5 18
Add 52x to both sides.
Setting x = 3 in the first equation gives
y = – 12 (3) + 3 = 32.
So the intersection point is ( 3, 32 ) . See Fig. 3.
Now Try Exercise 3
1.3 The Intersection Point of a Pair of Lines
21
The preceding method works when both equations have the slope–intercept form
(y = mx + b). In case one equation has the form x = a, things are much simpler. The
value of x is then given directly without any work—namely, x = a. The value of y can
be found by substituting a for x in the other equation.
EXAMPLE 3
SOLUTION
Finding the Point of Intersection Find the point of intersection of the lines y = 2x – 1
and x = 2.
The x-coordinate of the intersection point is 2, and the y-coordinate is y = 2 # 2 – 1 = 3.
Now Try Exercise 5
Therefore, the intersection point is (2, 3).
The method just introduced may be used to solve systems of two equations in two
variables.
EXAMPLE 4
Solving a System of Equations Solve the following system of linear equations:
e
SOLUTION
2x + 3y = 7
4x – 2y = 9.
First, convert the equations to slope–intercept form:
Given equation
2x + 3y = 7
Subtract 2x from both sides.
3y = -2x + 7
y=
– 23x
+
7
3
Divide both sides by 3.
4x – 2y = 9
Given equation
-2y = -4x + 9
y = 2x –
Subtract 4x from both sides.
9
2
Divide both sides by -2.
Now, equate the two expressions for y and then solve for x and y.
2x 8
3x
–
9
2
9
2
8
3x
8
3x
= – 23 x +
=
=
=
x=
y=
y=
7
3
7
3
9
7
3 + 2
27
14
41
6 + 6 = 6
3 41
41
8 6 = 16
2x – 92 = 2 41
16
36
5
41
8 – 8 = 8
#
( ) – 92
Add 23x to both sides.
Add 92 to both sides.
Add fractions on right.
Multiply both sides by 38 .
Substitute value for x into second equation.
Perform arithmetic.
So the solution of the given system is x =
41
16 ,
y = 58 .
Now Try Exercise 9
Supply and Demand Curves
p
Supply curve
Figure 4
q
The price p that a commodity sells for is related to the quantity q available. Economists
study two kinds of graphs that express relationships between q and p. To describe these
graphs, let us plot quantity along the horizontal axis and price along the vertical axis.
The first graph relating q and p is called a supply curve (Fig. 4) and expresses the relationship between q and p from a manufacturer’s point of view. For every quantity q, the
supply curve specifies the price p for which the manufacturer is willing to produce the
quantity q. The greater the quantity to be supplied, the higher the price must be. So supply curves rise when viewed from left to right.
The second curve relating q and p is called a demand curve (Fig. 5) and expresses
the relationship between q and p from the consumer’s viewpoint. For each quantity q,
22 chapter 1 Linear Equations and Straight Lines
p
p
(a, b)
q
Demand curve
Figure 5
q
Figure 6
the demand curve gives the price p that must be charged in order for q units of the commodity to be sold. The greater the quantity that must be sold, the lower the price must be
that consumers are asked to pay. So demand curves fall when viewed from left to right.
Suppose that the supply and demand curves for a commodity are drawn on a single
coordinate system (Fig. 6). The intersection point (a, b) of the two curves has an economic significance: The quantity produced will stabilize at a units, and the price will be
b dollars per unit. This is the equilibrium point.
EXAMPLE 5
SOLUTION
Applying the Law of Supply and Demand Suppose that the supply curve for a certain
commodity is the straight line whose equation is p = .0002q + 2 ( p in dollars). Suppose
that the demand curve for the same commodity is the straight line whose equation is
p = -.0005q + 5.5. Determine both the quantity of the commodity that will be produced and the price at which it will sell in order for supply to equal demand.
We must solve the system of linear equations
.0002q + 2 = -.0005q + 5.5
e
p = .0002q + 2
p = -.0005q + 5.5.
.0007q + 2 = 5.5
Add .0005q to both sides.
.0007q = 3.5
q=
3.5
.0007
Equate the two expressions for p.
Subtract 2 from both sides.
= 5000
Divide both sides by .0007.
p = .0002(5000) + 2
Substitute the value for q into first equation.
p=1+2=3
Perform arithmetic.
Thus, 5000 units of the commodity will be produced, and it will sell for $3 per unit.
Now Try Exercise 19
INCORPORATING
TECHNOLOGY
Graphing utilities have commands that find the intersection point of a pair of lines.
Figure 7 shows the result of solving Example 4 with the intersect command of the calc
menu. Since the x-coordinate of the intersection point is assigned to ans, the
x-coordinate can be converted to a fraction by pressing MATH 1 ENTER from the home
screen. See Fig. 8.
Figure 7 [-3, 6] by [-3, 3]
Figure 8
1.3 The Intersection Point of a Pair of Lines
23
An instruction of the form
solve [first linear equation], [second linear equation]
graphs the corresponding lines and finds their intersection point. For instance, consider
Example 4. The instruction
solve 2x + 3y = 7, 4x − 2y = 9
produces a graph of the two lines and displays the result “x =
y = 58 ≈ 0.625000.”
Check Your Understanding 1.3
y
y 5 2 12 x 1 4
B
1. Use the method of this section to find the coordinates of the
point C.
C
2. Determine the coordinates of the points A and B by
inspection.
3x 1 y 5 9
3. Find the coordinates of the point D.
A
D
x Figure 9
EXERCISES 1.3
In Exercises 1–6, find the point of intersection of the given pair of
straight lines.
y = 4x – 5
y = -2x + 7
2x – 3y = 3
4. e
y=3
2. e
y = 3x – 15
y = -2x + 10
y = 13 x – 1
5. e
x = 12
3. e
14.
y
x – 4y = -2
x + 2y = 4
y 5 2 13 x 1 7
A
2x – 3y = 3
6. e
x
=6
B
y 5 2x 1 9
C
7. Does (6, 4) satisfy the following system of linear equations?
e
y 5 23x 1 19
x – 3y = -6
3x – 2y = 10
8. Does (12, 4) satisfy the following system of linear equations?
y = 13 x – 1
e
x = 12
E
11. e
5x – 2y = 1
2x + y = -4
15.
y
y 5 12 x 1 3
C
y 5 2x
x + 2y = 4
10. e 1
1
2x + 2y = 3
12. e
B
x + 2y = 6
x – 13 y = 4
y
x55
A
In Exercises 13–16, find the coordinates of the labeled points.
13.
x
D
In Exercises 9–12, solve the systems of linear equations.
2x + y = 7
9. e
x-y=3
x
D
y
16.
A
2x 1 y 5 14
x53
2x 1 3y 5 18 B
3x 1 2y 5 24
B
A
y52
C
x
≈ 2.56250 and
Solutions can be found following the section exercises.
Figure 9 shows a type of polygon that plays a prominent role in
Chapter 3; its four vertices are labeled A, B, C, and D.
1. e
41
16
x 1 2y 5 12
D
x
24 chapter 1 Linear Equations and Straight Lines
17. Supply Curve The supply curve for a certain commodity is
p = .0001q + .05.
(a) What price must be offered in order for 19,500 units of
the commodity to be supplied?
(b) What prices result in no units of the commodity being
supplied?
18. Demand Curve The demand curve for a certain commodity is
p = -.001q + 32.5.
(a) At what price can 31,500 units of the commodity be sold?
(b) What quantities are so large that all units of the commodity cannot possibly be sold no matter how low the price?
19. Supply and Demand Suppose that supply and demand for a
certain commodity are described by the supply and demand
curves of Exercises 17 and 18. Determine the equilibrium
quantity of the commodity that will be produced and the selling price.
20. Supply and Demand A discount book seller has determined
that the supply curve for a certain author’s newest paperback
1
book is p = 300
q + 13. The demand curve for this book is
p = -.03q + 19. What quantity of sales would result in supply exactly meeting demand, and for what price should the
book be sold?
21. Supply and Demand Suppose that the demand curve for corn
has the equation p = -.15q + 6.925 and the supply curve
for corn has the equation p = .2q + 3.6, where p is the price
per bushel in dollars and q is the quantity (demanded or produced) in billions of bushels.
(a) Find the quantities supplied and demanded when the
price of corn is $5.80 per bushel.
(b) Determine the equilibrium quantity of corn that will be
produced and the price at which it will sell.
22. Supply and Demand Suppose that the demand curve for soybeans has the equation p = -2.2q +19.36 and the supply
curve for soybeans has the equation p = 1.5q + 9, where p is
the price per bushel in dollars and q is the quantity (demanded
or produced) in billions of bushels.
(a) Find the quantities supplied and demanded when the
price of soybeans is $16.50 per bushel.
(b) Determine the equilibrium quantity of soybeans that will
be produced and the price at which it will sell.
23. Temperature Conversion The formula for converting Fahrenheit degrees to Celsius degrees is C = 59 (F – 32). For what
temperature are the Celsius and Fahrenheit values the same?
24. Temperature Conversion The precise formula for converting
Celsius degrees to Fahrenheit degrees is F = 95C + 32. An
easier-to-use formula that approximates the conversion is
F = 2C + 30.
(a) Compare the values given by the two formulas for a temperature of 5°C.
(b) Compare the values given by the two formulas for a temperature of 20°C.
(c) For what Celsius temperature do the two formulas give
the same Fahrenheit temperature?
25. Manufacturing A clothing store can purchase a certain style of
dress shirt from either of two manufacturers. The first manufacturer offers to produce shirts at a cost of $1200 plus $30
per shirt. The second manufacturer charges $500 plus $35 per
shirt. Write the two equations that show the total cost y of
manufacturing x shirts for each manufacturer. For what size
order will the two manufacturers charge the same amount of
money? What is that amount of money?
26. Time Apportionment A plant supervisor must apportion her
40-hour workweek between hours working on the assembly
line and hours supervising the work of others. She is paid $12
per hour for working and $15 per hour for supervising. If her
earnings for a certain week are $504, how much time does she
spend on each task?
27. Calling Card Options A calling card offers two methods of
paying for a phone call. Method A charges 1 cent per minute,
but has a 45-cent connection fee. Method B charges 3.5 cents
per minute, but has no connection fee. Write the equations
that show the total cost, y, of a call of x minutes for methods
A and B, and determine their intersection point. What does
the intersection point represent?
28. Towing Fees Sun Towing Company charges $50 plus $3 per
mile to tow a car, whereas Star Towing Company charges $60
plus $2.50 per mile. Write the equations that show the total
cost y of towing a car x miles for each company. For what
number of miles will the two companies charge the same
amount? What is that amount of money?
In Exercises 29 and 30, find the area of the shaded triangle. Each
triangle has its base on one of the axes. The area of a triangle is
one-half the length of its base times its height.
29.
y
3x 2 y 5 3
x1y55
x
30.
y
3x 1 4y 5 24
2x 2 4y 5 24
x
31. Weight Determination In a wrestling competition, the total
weight of the two contestants is 700 pounds. If twice the
weight of the first contestant is 275 pounds more than the
weight of the second contestant, what is the weight (in
pounds) of the first contestant?
32. Sales Determination An appliance store sells a 42″ TV for
$400 and a 55″ TV of the same brand for $730. During a oneweek period, the store sold 5 more 55″ TVs than 42″ TVs and
collected $26,250. What was the total number of TV sets sold?
TECHNOLOGY EXERCISES
In Exercises 33–36, graph the lines and estimate the point of intersection to two decimal pla…