Linear Equations and Inequalities: Hamilton Education Guides Manual 16 - Over 230 Solved Problems 9798887228310

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Linear Equations and Inequalities

Manual Title: Linear Equations and Inequalities Author: Dan Hamilton Editor: John Hamilton Cover design by: John Hamilton Copyright  1998 All rights reserved. Printed in the United States of America. No part of this manual may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior wri tten permission of the author. Request for per mission or further information should be addressed to Hamilton Education Guides via [email protected]. First published in 1998 Library of Congress Catalog Card Number 98-74114 Library of Congress Cataloging-in-Publication Data ISBN 979-8-88722-831-0

Hamilton Education Guides Book Series

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Hamilton Education Guides Manual Series

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eManual versions available in the Amazon Kindle Store

Hamilton Education Guides Manual Series

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eManual versions available in the Amazon Kindle Store

Contents Linear Equations and Inequalities Quick Reference to Problems 1 1.1

Introduction to Linear Equations.................................................................................. 2

1.2

Math Operations Involving Linear Equations ............................................................. 5

1.3

Solving Other Classes of Linear Equations .................................................................. 24

1.4

Formulas .......................................................................................................................... 41

1.5

Math Operations Involving Linear Inequalities .......................................................... 48

Appendix – Exercise Solutions................................................................................................... 67 Section 1.1 Section 1.2 Section 1.3 Section 1.4 Section 1.5

67 67 70 72 73

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Acknowledgments The primary motivating factor in writing the Hamilton Education Guides manual series is to provide students with specific subjects on mathematics. I am grateful to John Hamilton for his editorial comments, cover design, and suggestions on easier presentation of the topics. I would also like to acknowledge the original contributors of the Hamilton Education Guides book s for their editorial reviews. Finally, I would like to thank my family for their understanding and patience in allowing me to prepare this manual.

Hamilton Education Guides

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Introduction and Overview It is my belief that the key to learning mathematics is through positive motivation. Students can be greatly motivated if subjects are presented concisely and the problems are solved in a detailed step by step approach. Th is keeps students motivated and provides a great deal of encouragement in wanting to learn the next subject or to solve the next problem. During my teaching career, I found this method to be an effective way of teaching. I hope by presenting equations in this format, more students will become interested in the subject of mathematics. This manual is a chapter from my Mastering Algebra – Intermediate Level book with the primary focus on the subject of linear equations and inequalities . The scope of this manual is intended for educational levels ranging from the 9th grade to adult. The manual can also be used by students in home study programs, parents, teachers, special education programs, preparatory schools, and adult educational programs including coll eges and universities as a supplementary manual. A fundamental understanding of basic mathematical operations such as addition, subtraction, multiplication, and division is required. This manual addresses linear equations and inequalities and how they ar e simplified and mathematically operated. Students learn how to solve and verify one variable linear equations as well as add, subtract, multiply, and divide linear equations and inequalities. Detailed solutions to the exercises are provided in the Appendix. Students are encouraged to solve each problem in the same detail and step by step format as shown in the text. It is my hope that all Hamilton Education Guides books and manuals stand apart in their understandable treatment of the presented sub jects and for their clarity and special attention to detail. I hope readers of this manual will find it useful. With best wishes, Dan Hamilton

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Linear Equations and Inequalities

Quick Reference to Case Problems

Linear Equations and Inequalities Quick Reference to Case Problems 1.1

Introduction to Linear Equations............................................................................. 2

1.2

Math Operations Involving Linear Equations ........................................................ 5 Case I - Addition and Subtraction of Linear Equations, p. 5 4  3 u  0.48  1  2  ;  5 5

1 1  2    y  3  2  ;  3 5 2

x2

3 1 1 5 8

Case II - Multiplication and Division of Linear Equations, p. 12 1 3 1 x  2 3 5

. h2 ; 38

3 8

2 5

; 3   x

Case III - Mixed Operations Involving Linear Equations, p. 18 5x  20  3x

1.3

; 4 y  2  3 y  8 ; 5m  5  3m  2

Solving Other Classes of Linear Equations ............................................................. 24 Case I - Solving Linear Equations Containing Parentheses and Brackets, p. 24 2  3  x   5 x ;



2 x  5  3x  8  0 ;



  x  5  3   x  2  3

Case II - Solving Linear Equations Containing Integer Fractions, p. 29 x

1 2  x 3 3

1 4

; u u 

2 5

2 3

; 4y  2 1 y

Case III - Solving Linear Equations Containing Decimals, p. 35 3.4x  2.5  2.8x  0.5

1.4

.  x  0.2  0.5x  1  0 ; 55 .   x  0.2   x  5  0.45  0 ; 125

Formulas ..................................................................................................................... 41 V 

1.5

1 bh 3

9 5

; F  C  32 ;

A

1 h b1  b2  2

Math Operations Involving Linear Inequalities ..................................................... 48 Case I - Addition and Subtraction of Linear Inequalities, p. 48 3 2 2  w 1 4 3

; 5u  0.45  4u  1

1 3

1 2 1 ;    y   1  3

4

3

Case II - Multiplication and Division of Linear Inequalities, p. 54 

2  4y 3

2 3

; 1 w2

3 5

; 2.6 m  3

2 3

Case III - Mixed Operations Involving Linear Inequalities, p. 60 6t  10  9t  5 ;

Hamilton Education Guides

3x  25  8x ;

0.8n  10  12 . n

1

Linear Equations and Inequalities The objective of this manual is to improve the student’s ability to solve operations involving linear equations and linear inequalities. One variable linear equations and the process of determining the solution to an algebraic equation as well as how the solution set to a linear equation is verified is addressed in Section 1.1. Math operations involving addition, subtraction, multiplication, and division of linear equations are addressed in Section 1.2. In Section 1.3 linear equations containing parentheses and brackets, integer fractions , and decimals are introduced. Formulas, its definition, and the steps as to how they are solved for a specific variable is addressed in Section 1.4. Math operations involving linear inequalities are discussed in Section 1.5. Cases presented in each section are concluded by solving additional examples with practice problems to further enhance the student ability.

1.1

Introduction to Linear Equations

 ”. A numerical statement consists of two expressions which are separated by an equal sign “ The symbol “  ” implies that the left and the right hand side of the nume rical statement must equal to each other in order for the equal sign to hold true. For example, 3  5  8 , 6  4  2 ,and 1  5  6 are true statements where as 7  3  9 , 8  5  23 , and 1  0  2 are false. An algebraic equation consists of two algebraic expressions which are separated by an equal sign “  ”. For example, using y as variable, the statement y  5  6 ; 3 y  2  7 ; y 2  3 y  5 are called algebraic equations. A solution to an equation is a value that when substituted for the variable, 1 is substituted fo r y in the make the equation a true numerical equation. For example, if equation y  5  6 , we obtain 1  5  6 or 6  6 which is a true numerical statement. Therefore, we say that y  1 is the solution to the equation y  5  6 . The process of determining the solution to an algebraic equation is referred to as solving an equation. The set of all solutions to an algebraic equation is called its solution set. For example, the solution set of y  5  6 is 1 , which is expressed as  1 , and the solution set of 3y  5  7 is 2 3

2 , which is expressed as   . 3 



In the following sections we will learn how to solve an algebraic equation. However, we first need to learn the process as to how a solution to a linear equation is verified. To check a solution to an equation we need to use the following steps: Step 1:

Substitute the solution into the original equation in place of the variable.

Step 2:

Solve the equation.

Step 3:

If both sides of the equation become equal to each other then the solution satisfies the original equation. Otherwise, the solution does not satisfy the original equation.

The following examples show how the solution to an algebraic linear equation is verified: Example 1.1-1: Given the algebraic equation 5x  3  3x  5 , does x  3 , x  4 , and x  5 satisfy the original equation? 1. Substitute x  3 into the original equation and see if both sides of the equation become equal to each other.

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Linear Equations and Inequalities

1.1 Introduction to Linear Equations

?

?

5  3  3 3  3  5 ; 15  3 9  5 ; 12  14 Since the left hand side of the equation is not equal to the right hand side of the equation, therefore x  3 does not satisfy the original equation. This implies that the two sides are not equal to each other. 2. Substitute x  4 into the original equation and see if both sides of the equation become equal to each other. ?

5  4  3 3  4  5 ;

? 20  3 12  5 ; 17  17

Since the left hand side of the equation is equal to the right hand side of the equation, therefore x  4 does satisfy the original equation. This implies that the two sides are equal to each other. Therefore, x  4 is the solution to the equation 5 x  3  3 x  5 . 3. Substitute x  5 into the original equation and see if both sides of the equation become equal to each other. ?

5  5  3 3  5  5 ;

? 25  3 15  5 ; 22  20

Since the left hand side of the equation is not equal to the right hand side of the equation, therefore x  5 does not satisfy the original equation. This implies that the two sides are not equal to each other. Example 1.1-2: Given the algebraic equation y  3 y  2  4 , does y  1 , y  0 , and y  1 satisfy the original equation? 1. Substitute y  1 into the original equation and see if both sides of the equation become equal to each other. ? 1 3 1  2  4

?

?

; 1 3  3  4 ; 1 9  4 ; 1  5

Since the left hand side of the equation is not equal to the rig ht hand side of the equation, therefore y  1 does not satisfy the original equation. This implies that the two sides are not equal to each other. 2. Substitute y  0 into the original equation and see if both sides of the equation become equal to each other. ? 0  3 0  2  4

?

?

; 0  3  2  4 ; 0  6  4 ; 0  2

Since the left hand side of the equation is not equal to the right hand side of the equation, therefore y  0 does not satisfy the original equation. This implies that the two sides are not equal to each other. 3. Substitute y  1 into the original equation and see if both sides of the equation become equal to each other. ? 1  31  2  4

Hamilton Education Guides

?

?

; 1 3  1  4 ; 1 3  4 ; 1  1

3

Linear Equations and Inequalities

1.1 Introduction to Linear Equations

Since the left hand side of the equation is equal to the right hand side of the equation, therefore y  1 does satisfy the original equation. This implies that the two sides are equal to each other. Therefore, y  1 is the solution to the equation y  3 y  2  4 . Example 1.1-3: Determine if z  2 is the solution to each of the following equations. a. 3z  1  2 z Solution:

c. z  3  9 z  13

b. 6 z  2  4  2 z ?

d. 3z  13  10

?

a. Let z  2 in the equation 3z  1  2z , i.e., 3  2  1 2  2 ; 6  1 4 ; 5  4 . Therefore, z  2 is not the solution to 3z  1  2 z . ?

?

b. Let z  2 in the equation 6z  2  4  2z , i.e., 6  2  2  4  2  2 ; 12  2  4  4 ; 10  8 . Therefore, z  2 is not the solution to 6 z  2  4  2 z . ?

?

c. Let z  2 in the equation z  3  9 z  13 , i.e., 2  3  9  2  13 ; 5 18  13 ; 5  5 . Therefore, z  2 is the solution to z  3  9z  13 . ?

?

d. Let z  2 in the equation 3z  13  10 , i.e., 3  2  13 10 ; 6  13 10 ; 7  10 . Therefore, z  2 is not the solution to 3z  13  10 . Practice Problems - Introduction to Linear Equations Section 1.1 Practice Problems - Solve the following linear equations: 1. Determine whether 2 is the solution to each of the following equations: a. 3x  2  10 b. 2 x  3  x c. 6  x  2 x  1 d. 2 x  8  3x  2 2. Determine if y  2 is the solution to the following equations: a. y  3  2 y

b. 6 y  y  8 y  2

c. 6  3 y  0

d. 3 y  5  y

3. Given the algebraic equation 2 x  8   x  5  3 , does x  0 , x  1 , and x  6 satisfy the original equation? 4. Does a  2 satisfy any of the following equations? a. 3a  2  4a b. 3  7a  18 c. 5a  3  3a  1

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d. 8  a  3

4

Linear Equations and Inequalities

1.2

1.2 Math Operations Involving Linear Equations

Math Operations Involving Linear Equations

First degree equations of one variable, which are also referred to as linear equations, are solved by the proper use of addition, subtraction, multiplication, and division rules. It is important to learn how to apply these rules to linear equations in order to find the solution to the unknown variable. In this section, solving linear equations using either the addition and subtraction rules (Case I) or the multiplication and division rules (Case II) are disc ussed. Students are encouraged to learn how to solve first degree equations by properly organizing and applying the rules that are stated below in order to minimize mistakes. Case I

Addition and Subtraction of Linear Equations

To add or subtract the same positive or negative number to linear equations the following rules should be used: Addition and Subtraction Rules: The same positive or negative number can be added or subtracted to both sides of an equation without changing the solution: for all real nu mbers a , b , and c , 1. a  b if and only if a  c  b  c ab

2.

if and only if

ac  bc

The steps as to how linear equations are s olved, using the addition and subtraction rules, are as follows: Step 1 Isolate the variable to the left hand side of the equation by applying the addition and subtraction rules. Step 2

Find the solution by simplifying the equation. Check the answer by solution into the original equation.

substituting the

Examples with Steps The following examples show the steps as to how linear equations are solved using the addition and subtraction rules: Example 1.2-1 x 3 5

Solution: Step 1

x 3 5

Step 2

; x  3 3  5 3 ; x  0  8 ; x  8

Not Applicable

The solution set is  8 . ?

Check: 8  3  5 ; 5  5 Example 1.2-2 2 3 1 w2 3 5

Solution: Step 1

2 3 1 w2 3 5

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;

1 3  2  w  2  5  3 3

5

;

3 2 10  3 w 3 5

;

5 13 w 3 5

5

Linear Equations and Inequalities

;

5 5 13 5  w  3 3 5 3

w

Step 2

1.2 Math Operations Involving Linear Equations

13 5  5 3

; w

; 0 w 

13 5  5 3

13  3  5  5

; w

; w

53

13 5  5 3

39  25 15

; w

14 15

; w  0.93

The solution set is  0.93 . ?

2 3

Check: 1  0.93  2

? 2  5  3 ? 10  3 ? 13 3 1 3  2 5 3 2 ; ; ;  0.93   0.93   0.93  5 3 5 3 5 3 5 ?

; 167 .  093 .  2.6 ;

2.6  2.6

Example 1.2-3 4  3 u  0.48  1  2   5 5

Solution: Step 1

4  3 u  0.48  1  2   5 5

4 4  3  3 ; u  0.48  0.48  1  2   0.48 ; u  0  1  2   0.48  5

5

 5

5

4  3 ; u  1  2   0.48   5

Step 2

5

 1  5  3

4  3 u  1  2   0.48  5 5

; u

 ; u 

 ; u

8 14    0.48 5 5 



5



2  5  4   0.48 5

8  14    0.48  5 

 

; u

5  3 10  4     0.48  5 5 

 ; u

22  0.48 5

; u  4.4  0.48 ; u  4.88

The solution set is  4.88 . ? 1  5  3 2  5  4 ? 5  3 10  4 ? 8 14 4 ; 4.4  ; 4.4  ; 4.4     5 5 5 5 5 5 5 ? 8  14 ? 22 ; 4.4  ; 4.4  ; 4.4  4.4 5 5 ?

3 5

Check: 4.88  0.48 1  2

Example 1.2-4 

1 1  2   y  3  2   3 5 2



1 1  2   y  3  2   3 5 2

Solution: Step 1

1 1 1 1  2  2 ; y   y  y  3  2  ; y   0  3  2   3

5

2

 3

5

2

1  2 1 1 1  2 1 1 1 1  2 ; y   3  2  ; y    3  2   ; y  0  3  2   5

Hamilton Education Guides

 3

2

5

5

 3

2

5

 3

2

5

6

Linear Equations and Inequalities

1.2 Math Operations Involving Linear Equations

1 1  2 ; y  3  2    3

2

5  3  3  2

1 1  2 y  3  2    3 2 5

Step 2

 ; y 

; y

11 5  1   3 2 5

7  5  1 6

; y 

 2  2  1   1  

2

 11  2  5  3  1  3 2   5

35  6 30

; y

41 30

5

9  2 4  1 1    3 2  5

 ; y

 ; y 

; y

; y

65

3



; y 1

11 30

22  15  1  6  5

7 6

; y 

1 5

; y  1.37

The solution set is 1.37 . ? ?  3  3  2 2  2  1 1? 1  2  9  2 4  1 .   .   .   3  2  ; 0.2  137 Check:   137  ; 0.2  137 

 3

5

2



3

2

 3



2 

? ?  11  2  5  3  11 5   22  15  .  .    ; 0.2  137  ; 0.2  137   3 2   3 2 3 2  

 .  ; 0.2  137 ?

?

; 0.2  137 . 

? 7 ; 0.2  137 .  117 . ; 0.2  0.2 6

Example 1.2-5 2 3   x  3  4   5 7

Solution: 2 3   x  3  4   5 7

Step 1

2 2 2  3   x    3  4  5 5 5  7

;

2  3 ; 0  x    3  4  5 

7

 ; x     

28  3  7 

2  3 ; x    3  4  5 

Step 2

7

2  3 x    3  4  5  7  ; x     2 5

3 1

31  7

2 5

10 7

; x

; x 

2 5

3

4  7  3

1

7

; x     2 5

 3  7  31  1   1 7  

; x   

2  7  10  5 5 7

 

; x

2 5

3 1

21  31   7 

 ; x   

14  50 35

2 5

; x

64 35

; x  1.83

The solution set is  1.83 . Check:

? ? ? ? 4  7  3 ; 143 31 2 3 28  3 ; 143   183 .   3  4 ; 0.4  183 . 3 . 3 . 3 5 7 7 7 7 ?

; 143 .  3  4.43 ;

Hamilton Education Guides

1.43  1.43

7

Linear Equations and Inequalities

1.2 Math Operations Involving Linear Equations

Additional Examples - Addition and Subtraction of Linear Equations The following examples further illustrate how linear equations are solved using the addition and subtraction rules: Example 1.2-6 x 6  8 ; x 66  86 ; x 0  2 ; x  2

The solution set is  2 .

?

Check: 2  6  8 ; 8  8 Example 1.2-7 3  y  5 ; 3  3  y  5  3 ; 0  y  2 ; y  2 ?

The solution set is  2 .

?

Check: 3   2  5 ; 3  2  5 ; 5  5 Example 1.2-8 2  u5 ; 25  u55 ; 7  u0 ; 7  u ; u  7

The solution set is  7 .

?

Check: 2  7  5 ; 2  2 Example 1.2-9 5  x  10 ; 5  5  x  10  5 ; 0  x  15 ; x  15 ?

The solution set is  15 .

?

Check: 5   15  10 ; 5  15  10 ; 10  10 Example 1.2-10 24  10  h ; 24  10  10  10  h ; 14  0  h ; 14  h ; h  14 ?

The solution set is  14 .

?

Check: 24  10   14 ; 24  10  14 ; 24  24 Example 1.2-11 5  y  20 ; 5  20  y  20  20 ; 15  y  0 ; 15  y ; y  15

The solution set is  15 .

?

Check: 5  15  20 ; 5  5 Example 1.2-12 s  12  15 ; s  12  12  15  12 ; s  0  27 ; s  27

The solution set is  27 .

?

Check: 27  12 15 ; 15  15 Example 1.2-13 8.5  x  2.4 ; 8.5  8.5  x  2.4  8.5 ; 0  x  10.9 ; x  10.9 ?

The solution set is  10.9 .

?

Check: 8.5   10.9  2.4 ; 8.5  10.9  2.4 ; 2.4  2.4 Example 1.2-14 9  w  8 ; 9  8  w  8  8 ; 17  w  0 ; 17  w ; w  17

The solution set is  17 .

?

Check: 9  17  8 ; 9  9 Example 1.2-15 12 .  y  2.8 ; 12 .  12 .  y  2.8  12 . ; 0  y  16 . ; y  1.6 ?

?

.   16 .   2.8 ; 12 Check: 12 .  16 .  2.8 ;

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The solution set is  1.6 .

2.8  2.8

8

Linear Equations and Inequalities

1.2 Math Operations Involving Linear Equations

Example 1.2-16 x2

3 5

3 5

1 8

3 1 1 5 8

; x 2 2 1 2

9 8

; x

; x 

13 5

9  5  13  8

3 5

1 8

; x 0 1 2 45  104 40

; x

85

3 5

; x

; x

59 40

1  8  1  2  5  3 8

; x

5

59 40

; x  1

19 40

; x

8  1 10  3  8 5

; x  1.475

The solution set is  1.475 . 3 5

?

Check: 1475 .  2 1

2  5  3  1 8  1 ; 1475 1 10  3 8  1 13 ? 9 ; 1475 ; 1475 .  .   .   8 5 8 5 8 5 8 ?

?

; 1475 ; .  2.6 1125 . Example 1.2-17 2

3 1  y 3 4 4

; y

3 4

1125 .  1125 .

1 4

1 4

11  13 4

; y

; 2 3  y 3 3

11 13  4 4

; y

3? 4

Check: 2  0.5  3

1 4

3 4

1 4

3 4

1 4

; 2 3  y 0 ; 2 3  y ; y 

1 2 ; y   ; y  0.5 2 4 2

2  4  3  3  4  1 4

4

The solution set is  0.5 .

? 3  4  1 ; 8  3  1 2  4  3 ? 12  1 11 ? 13 ; ;  0.5   0.5  0.5  4 4 4 4 4 4 4

?

; 2.75  05 .  325 . ; Example 1.2-18

2.75  2.75

3 8

3 8

y  2.35  2

?

; y  2.35  2.35  2  2.35 ; y  0 

2  8  3  2.35 8

; y

16  3  2.35 8

; y

19  2.35 8

The solution set is 4.725 .

; y  2.375  2.35 ; y  4.725

? 2  8  3 ? 16  3 ? 19 3 ; 2.375  ; 2.375  ; 2.375  ; 2.375  2.375 8 8 8 8

?

Check: 4.725  2.35  2 Example 1.2-19 2.5  u  1

3 5

3 5

3 5

; 2.5  1  u  1  1

3 5

3 5

; 2.5  1  u  0 ; 2.5 

1  5  3  u 5

8 ; 2.5   u ; 2.5  1.6  u ; 41.  u ; u  4.1 5

?

3 5

?

Check: 2.5  41 .  1 ; 2.5  41 . 

1 5  3 ; 5

; 2.5 

53 u 5

The solution set is  4.1 . ?

2.5  41 . 

? ? 53 8 ; 2.5  41 .  ; 25 .  41 .  16 . 5 5

2.5  2.5

; Example 1.2-20

1 2 1 5  x  4 1 5 5 5

;

26 22  6 x 5 5

;

;

5  5  1  x  4  5  2  1 5  1 5

26 16 x 5 5

5

;

5

;

26 26 16 26  x  5 5 5 5

25  1 20  2 5  1 x  5 5 5

; 0 x 

16  26 5

;

; x 

26 22 6 x  5 5 5 10 5

; x

10 5

; x2

The solution set is  2 .

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9

Linear Equations and Inequalities

?

1 5

2 5

1.2 Math Operations Involving Linear Equations

1 5

Check: 5  2  4  1 ; ?

; 5.2  2  4.4  12 . ; Example 1.2-21 1 2 2 x 1  1 3 3 5 4 3

; x  ; x

; x

10  21 15

93 45

5

3

11 15

; x  2

1?2 3 3

5

4 3

3 45

2 5

; 2.07  133 .  0.66  14 . ; Example 1.2-22 ; 0.45  w 

4 3

31 2 5  2   3 3 5

11 4  15 3

; x0

4 3

2 3

; x  

4 3

; x 

; x

15  3

2  5  7  3 3 5

33  60 45

The solution set is  2.07 .

1 3  1 ? 2  1 5  2 ; 3

7 5

11 3  4 15

; x  2.07

?

2 6 3

5

; x

; x  

Check: 2.07  1   1 ; 2.07 

0.45  w  2

? 20  2 5  1 ? 22 6 25  1 26 ;  2   2  5 5 5 5 5 5

3.2  3.2

3

4 3

93 45

5

1 3  1  2  1 5  2

; x 

; x

5  5  1  2 ? 4  5  2  1 5  1 ;

3

5

2.07 

31 ? 2 5  2 4?2 7 ; 2.07      3 3 5 3 3 5

0.74  0.74

2  3  2  6 3

; 0.45  w 

62 6 3

8 3

; 0.45  w   6 ; 0.45  w  2.67  6

; 0.45  w  8.67 ; 0.45  0.45  w  8.67  0.45 ; 0  w  8.67  0.45 ; w  8.67  0.45 ; w  8.22 ?

?

2 3

Check: 0.45  8.22  2  6 ; 0.45  8.22  ?

; 0.45  8.22  2.67  6 ; Example 1.2-23 20  x  80  1

2 3

3

?

0.45  8.22 

?8 62  6 ; 0.45  8.22   6 3 3

8.67  8.67

4  5  x  16  5 

;

2  3  2  6 ;

The solution set is  8.22 .

1 3  2 3

;

22  5  x  42  5 

3 2 3

; 2 5x4 5

5 3

. ; 4.48  x  8.96  1.67 ; 4.48  x  10.63 ; 4.48  4.48  x  10.63  4.48 ; 2  2.24  x  4  2.24  167 . . The solution set is  615

. ;  x  615 . ; x  6.15 ; 0  x  615

Check:

?

20   615 .   80  1 ?

?

2 ; 3

; 2  2.4  615 . 4  2.24 

4  5   615 .   16  5 

1 3  2 ; 3

?

2 5  615 . 4 5 

3 2 3

? 5 ; 4.48  615 .  896 .  167 . ; 10.63  10.63 3

Example 1.2-24 3 3 1  h  2 5 4 8 5

;  

; 

11  h0 4

1 5  3  h  2  4  3 5

;

4

8  4  11 5  h

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; 

5 4

;

53 83  h 5 4

32  55 h 20

;

8 5

;   h

87 h 20

11 4

; h

8 5

;  

87 20

11 11 11  h  4 4 4

; h  4

7 20

; h  4.35

10

Linear Equations and Inequalities

1.2 Math Operations Involving Linear Equations

The solution set is 4.35 . 3? 5

Check: 1  4.35  2

? ? 1 5  3  2  4  3 ;  5  3  3 83 8? 11 ;  ;   4.35  4.35  4.35  4 5 4 5 4 5 4

?

; 16 .  435 .  2.75 ; Example 1.2-25 x 3

; x ; x

2 2 1  1 5 3 8

; x

1.6  1.6

3  5  2  2  1 8  1 5

3

8

; x

15  2 2 8  1   5 3 8

; x

17 2 9   5 3 8

17 16  27 17 17 11 17 11 17 17 11 17 2  8  9  3  ; x  ; x  ; x    ; x0   5 24 24 5 5 24 5 5 24 5 5 3 8

11 5  17  24 24  5

; x

55  408 120

; x

353 120

; x2

113 120

; x  2.942 The solution set is 2.942 .

2?2 5 3

Check: 2.94  3   1 ?

3  5  2 ? 2  1 8  1 ; 2.94  15  2 ? 2  8  1 ; 2.94  17 ? 2  9 1 ; 2.94  8 5 3 8 5 3 8 5 3 8

; 2.94  34 .  0.67  113 . ;

0.46  0.46

Practice Problems - Addition and Subtraction of Linear Equations Section 1.2 Case I Practice Problems - Solve the following linear equations by adding or subtracting the same positive or negative number to both sides of the equation: 5 x3

1.

x  13  12

2.

8  h  20

3.

4.

3  u  5

5.

2.8  x  3.7

6. x   2

7. 4.9  x  1 10.

2 3

y  2.38  3

1 3

8. u  2  2

3 5

3 8

2 3

3 8

9. 6  y  2

4 5

2 5

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11

Linear Equations and Inequalities

Case II

1.2 Math Operations Involving Linear Equations

Multiplication and Division of Linear Equations

To multiply or divide linear equations by the same positive or negative number the following rules should be used: Multiplication Rule: The same positive or negative number can be multiplied by both sides of an equation without changing the solution: for all real numbers a , b , and c , ab

if and only if a  c  b  c .

Division Rule: The same positive or negative number (except zero) can be divided by both sides an equation without changing the solution: for all real numbers a , b , and c , where c  0 ab

if and only if

a b  c c

.

The steps as to how linear equations are solved, using the multiplication and division rules, are as follows: Step 1 Isolate the variable to the left hand side of the equation by applying the addition and subtraction rules. Step 2

Find the solution by applying the multiplication or division rules. Check the answer by substituting the solution in the original equation. Examples with Steps

The following examples show the steps as to how linear equations are solved using the multiplication or division rules: Example 1.2-26 2x  

Solution: Step 1 Step 2

1 5

Not Applicable 2 x 

1 1 1   2 5 2

1 1 5 2

; x 

1 1 5 2

; x 

; x

1 1 52

1

; x   10

1 The solution set is   .  10 

1 1 2 ? 1 1 1 2 1 1 2 1 1 Check: 2    ;    ;   ;   ;    5 5  10 5 10 5 1 10 5 1 10 5 5 ?

?

?

Example 1.2-27 3  3w 8

Solution: 3  3w 8

;

; 3w  

3 8

Step 1

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3  3w  3w  3w 8

;

3  3w  0 8

;

3 3 3   3w  0  8 8 8

; 0  3w  

3 8

12

Linear Equations and Inequalities

3w  

Step 2 Check:

1.2 Math Operations Involving Linear Equations

3 1 8 3

1 3

3 8

1 The solution set is   .

1

; w8

; 3 w    

 8

?

3 3 3 1  3   ;  8 8 8 8

Example 1.2-28 1 3 1 x  2 3 5

Solution: Step 1

Not Applicable 1 3 1 x  2 3 5

Step 2

;

;

1 3  1 x   2  5  3 3

4 3 13 3 x    3 4 5 4

5

; x

13 3  5 4

;

; x

31 10  3 x 3 5 13 3  5 4

; x

?

1 3

13  3 5 4

4 13 x 3 5

; x

39 20

; x  1

19 20

The solution set is  1.95 .

; x  1.95 Check: 1  195 .   2

;

? 13 4  195 .  ? 13 7.8 ? 13 3 4 ;  195 ; ;   ; 2.6  2.6 .    5 3 5 3 5 3 5

Example 1.2-29 26  12y

Solution: Step 1

26  12y ; 26  12 y  12 y  12 y ; 26  12 y  0 ; 26  26  12 y  0  26

; 0  12 y  26 ; 12 y  26

12 y  26 ;

Step 2

13   y 26 1 12 26 13  26 .  ; y ; y ; y ; y  2 ; y  2166  6 12 12 12 6  12 6 . . The solution set is 2166

?

Check: 26  12  2166 ; 26  26 . Example 1.2-30 38 . h2

Solution: Step 1 Step 2

3 8

Not Applicable 38 . h2

;

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3 8

. h ; 38

.  h 2.375 38  .  38 38 .

2  8  3 8

; h  0.625

. h ; 38

16  3 8

; 3.8h 

19 8

; 3.8 h  2.375

The solution set is 0.625 .

13

Linear Equations and Inequalities

1.2 Math Operations Involving Linear Equations

?

Check: 38 .  0.625  2

? 19 3 ; 2.375  ; 2.375  2.375 8 8

Additional Examples - Multiplication and Division of Linear Equations The following examples further illustrate how linear equations are simplified using the above multiplication or division rules: Example 1.2-31 20 20  3 y 60 3 y  60 ;  ; y ; y  20 1 3 3

The solution set is  20 .

?

Check: 3 20  60 ; 60  60 Example 1.2-32 1 x8 5

1 x  5  8  5 5

;

The solution set is  40 .

; x  40

8 ?  ? 8? 40 1 Check:  40  8 ; 8 ; 8 ; 8  8 5 1 5

Example 1.2-33 5  3h 7

; h

5  3h  3h  3h 7

;

1  5 3 7

;

5  3h  0 7

;

5 5 5   3h  0  7 7 7

; 0  3h  

5 7

; 3h  

5 7

;

1 1 5  3 h    3 3 7

5 The solution set is   .

5

; h   21

 21 

Note that another way of solving for h is by not isolating the variable to the left hand si de of the equation. However, in the very last step, we should write the variable to the left hand side of the equation and the solution to the right hand side of the equation as shown below. 5  3h 7

1 5 1    h 3 3 7 3

;

; 

1 5 h 3 7

; 

5 h 21

5

; h   21

5 5 5  5? 5 5? 3 5 5 ? 3 5 5 ? 15 5 ? 15 Check:  3   ;   ;  ;  ;  ;  7 7  7 21 7 1 21 7 1 21 7 21 7 21 7

Example 1.2-34 u  3 5

;

u  5  3  5 5

; u  15

The solution set is  15 .

3  ? 3? 15 15 ? Check:  3 ;   3 ;   3 ; 3  3 5 1 5

Example 1.2-35 2w  28

;

2 w 28  2 2

14 14  28 ; w ; w ; w  14 1 2

The solution set is 14 .

?

Check: 2 14  28 ; 28  28

Hamilton Education Guides

14

Linear Equations and Inequalities

1.2 Math Operations Involving Linear Equations

Example 1.2-36  y  15 ;

 y 15  1 1

; y  

15 1

; y  15

2 5

2 5

The solution set is  15 .

?

Check:  15 15 ; 15  15 Example 1.2-37 3  

;

2 x 5

2 5

2 5

2 5

2 5

; 3  x   x  x ; 3  x  0 ; 3  3  x  0  3 ; 0  x  3 ;

2 5 5 x   3 5 2 2

3 5 1 2

; x 

; x

3 5 1 2

; x

15 2

; x7

1 2

; x  7.5

2 x3 5

The solution set is 7.5 .

Second Approach: Keep the variable x to the right hand side of the equation. 3  

2 x 5

2 5

5 2

; 3     x   ?

3 5 1 2

;   x ;

35 x 1 2

? 2  7.5 2 7.5 ; 3  ; 5 1 5 1

?

2 5

5 2

Check: 3   7.5 ; 3  

1 15 15 x ; x ; x7 2 2 2 3 ? 3 ? 15  ; 3  ; 3  3 3  1 5

;

; x  7.5

Example 1.2-38 1 x  3 8

1 x  8  3  8 8

;

The solution set is  24 .

; x  24

3 ?  ? 3? 24 1 Check:  24  3 ;   3 ;   3 ; 3  3 8 1 8

Example 1.2-39 h 7 5

Check:

h  5  7 ; h  35 5 7  ? 35 ? 7? 35 7 ;  7 ; 7 5 1 5

The solution set is  35 .

; 5 

; 77

Example 1.2-40 20 x  30

;

 x 20 30   20 20

3 3 1  30 ; x ; x   ; x  1 ; x  1.5 The solution set is  1.5 . 2 2  20 2

?

.   30 ; 30  30 Check: 20   15 Example 1.2-41 5 

; 3 

w 3

; 5  

w  3 5 3

w 3

; 5 

w w w   3 3 3

; 5 

w 0 3

; 5  5 

w  05 3

; 0

w 5 3

;

w 5 3

The solution set is 15 .

; w  15

Second Approach: Keep the variable w to the right hand side of the equation. 5 

w 3

;  5  3 

w  3 3

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; 15  w ; w  15

15

Linear Equations and Inequalities

1.2 Math Operations Involving Linear Equations

5 ? 15 ? 5  15 Check: 5  ; 5  ; 5  ; 5  5 3 1 3 ?

Example 1.2-42 12  6y ; 12  6 y  6 y  6 y ; 12  6 y  0 ; 12  12  6 y  0  12 ; 0  6 y  12 ; 6 y  12 2 2  12 ; y ; y  ; y2 1 6

1 1 6 y   12  6 6

;

The solution set is  2 .

Second Approach: Keep the variable y to the right hand side of the equation. 2 2  12  y ;  y ; 2 y ; y  2 ; 1 6

1 1 12   6 y  6 6

12  6y ;

?

Check: 12  6  2 ; 12  12 Example 1.2-43 3 y  18 ;

1 1 3 y   18   3 3

;

18 1 y  1 3

;

6 6  18 ; y   ; y   ; y  6 1 3

18  1 y 1  3

The solution set is  6 . ?

Check: 3   6 18 ; 18  18 Example 1.2-44 2 5 x 3 8

Check:

;

2 3 5 3 x    3 2 8 2

; x

53 82

; x

15 16

The solution set is  0.94 .

; x  0.94

? 5 2   0.94 ? 2 188 . ?   0.94  ;  0.625 ;   0.625 ; 0.625  0.625 3 8 3 3

Example 1.2-45 3.6 x  0.22

3 .6 x 0.22  3 .6 3.6

;

22 11 11 22  10  220 100 ; x   36 ; x   ; x ; x ; x  0.06 100  36 180  3600 180 10 The solution set is  0.06 .

?

.   0.06  2.2 ; 0.22  0.22 Check: 36 Example 1.2-46 

1 y  23 3

1 3

The solution set is  69 .

;  y  3  23  3 ; y  69

23 ?  ? 1 1  69 ? 23 ? 69 1 69 ? Check:   69  23 ;    23 ;   23 ;   23 ;   23 ; 23  23 3 3 1 3 1 1 3

Example 1.2-47 2

3 1  1 x 5 3

;

13 4  x0 5 3

;

2  5  3   1 3  1 x 5

;

3

13 13 4 13   x  0 5 5 3 5

Hamilton Education Guides

;

10  3 31  x 5 3 4 3

; 0 x  

13 5

;

;

13 4  x 5 3

4 13 x 3 5

;

;

13 4 4 4  x x x 5 3 3 3

4 3 13 3 x    3 4 5 4

; x

13  3 5 4

16

Linear Equations and Inequalities

; x

39 20

1.2 Math Operations Involving Linear Equations

The solution set is  1.95 .

; x  1.95

Second Approach: Keep the variable x to the right hand side of the equation. 2

3 1  1 x 5 3

; 

;

39 x 20

2  5  3   1 3  1 x 5

3

10  3 31  x 5 3

;

;

13 4  x 5 3

13 3 4 3    x 5 4 3 4

;

; 

13  3 x 5 4

.  x ; x  1.95 ; 195

3? 5

1 3

Check: 2  1  195 .  ;

.  13 ? 7.8 13 ? 4 13 ? 4   195 ;  ; 2.6  2.6   195 .  ;  5 3 5 3 5 3

Example 1.2-48 2y  

3 8

1 2

3 1 8 2

; 2 y     ?

Check: 2  01875 .  

; y

3 1 82

; y

3 16

. ; y  01875

The solution set is  0.1875 .

? 3 3 ; 0.375  ; 0.375  0.375 8 8

Example 1.2-49 x 

2 3

 x 0.67  1 1

;  x  0.67 ; ?

Check: 0.67 

; x  

0.67 1

The solution set is  0.67 .

; x  0.67

2 ; 0.67  0.67 3

Example 1.2-50 3 1 2 x  1 5 5

;

2  5  3 x   1 5  1 5

5

;

10  3 5 1 x 5 5

13 6 x 5 5

;

 13 5 6 5 x     5 13 5 13

; x

6 13

The solution set is  0.462 .

; x  0.462 3 5

;

?

1 5

Check: 2  0.462  1 ;

? 6 13   0.462 ? 6 6 6 13   0.462  ;  ;    5 5 5 5 5 5

Practice Problems - Multiplication and Division of Linear Equations Section 1.2 Case II Practice Problems - Solve the following linear equatio ns by applying the multiplication or division rules: 1. 3 y   4.

2 3

x  2 8

7.  w  1

4 5

1 2

2.  x  1

2 3

3.

3  2 h 8 1 8

1 2

5.  x  35

6. 2 u  1

1 2

9. 2.8 x  1.4

8.  y  12

3 5

10. 2 x  4.3

Hamilton Education Guides

17

Linear Equations and Inequalities

Case III

1.2 Math Operations Involving Linear Equations

Mixed Operations Involving Linear Equations

In Cases I and II we learned how to solve linear equations by either applying: 1. The addition and subtraction rules or, 2. The multiplication or division rules. In this section, solution to linear equations which may involve using all four rules is discussed. Addition and Subtraction Rules: The same positive or negative number can be added or subtracted to both sides of an equation without changing the solution: for all real numbers a , b , and c , 1. a  b if and only if a  c  b  c 2. a  b if and only if a  c  b  c . Multiplication Rule: The same positive or negative numbe r can be multiplied by both sides of an equation without changing the solution: for all real numbers a , b , and c , ab

if and only if a  c  b  c .

Division Rule: The same positive or negative number (except zero) can be divided by both sides an equation without changing the solution: for all real numbers a , b , and c , where c  0 ab

if and only if

a b  c c

.

The steps as to how linear equations are solved using the addition and subtraction, multiplication, and division rules are as follows: Step 1

Isolate the variable to the left hand side of the equation by applying the additio n and subtraction rules.

Step 2

Find the solution by applying the multiplication or division rules. Check the answer by substituting the solution into the original equation. Examples with Steps

The following examples show the steps as to how linear equati ons are solved using the addition, subtraction, multiplication, and division rules: Example 1.2-51 5x  20  3x

Solution: Step 1

5 x  20  3x

; 5x  3x  20  3x  3x ; 2 x  20  0 ; 2 x  20

10 10  2 x 20 2 x  20 ;  ; x ; x  10 1 2 2

Step 2 ?

The solution set is 10 .

?

Check: 5 10  20  3 10 ; 50  20  30 ; 50  50 Example 1.2-52 Solution: Step 1

4 y  2  3y  8 4 y  2  3y  8 ; 4 y  3 y  2  3 y  3 y  8 ; y  2  0  8 ; y  2  8

Hamilton Education Guides

18

Linear Equations and Inequalities

1.2 Math Operations Involving Linear Equations

y  2  2  8  2 ; y  0  10 ; y  10

Step 2

The solution set is 10 .

Not Applicable ?

?

Check: 4 10  2  3 10  8 ; 40  2  30  8 ; 38  38 Example 1.2-53 u  3  2 5

Solution: Step 1

u  3  2 5

Step 2

u  5 5

;

u  3  3  2  3 5

u  0  5 5

;

;

u  5 5

u ; 5   5  5 ; u  25 5

The solution set is  25 .

5 ? ? ? ?  25 25 5 Check:  3  2 ;   3  2 ;   3  2 ; 5  3  2 ; 2  2 5 1 5

Example 1.2-54 5m  5  3m  2

Solution: Step 1

5m  5  3m  2

; 5m  3m  5  3m  3m  2 ; 2 m  5  0  2 ; 2 m  5  2

; 2m  5  5  2  5 ; 2m  0  3 ; 2 m  3 Step 2

2 m  3 ? 3 2 5 5 ;   2 2

;

2 m 3  2 2

3 2

Check: 5   5  3   2 ; 

3

; m2

3 The solution set is   . 2

15 5 ? 9 2  15  1  5  2 ?  9  1  2  2 15  10 ? 9  4 ;    ;   2 1 2 1 2 1 2 1 2 2

Example 1.2-55 15 x  5  3x  2

Solution: Step 1

15 x  5  3x  2

; 15x  3x  5  3x  3x  2 ; 18 x  5  0  2 ; 18 x  5  2

; 18 x  5  5  2  5 ; 18 x  0  7 ; 18 x  7 Step 2

18 x  7 ?

;

 x 18 7   18 18

; x

7 18

; x  0.389

The solution set is 0.389 .

?

Check: 15  0.389  5  3  0.389  2 ; 584 .  5  116 .  2 ; 0.84  0.84

Hamilton Education Guides

19

Linear Equations and Inequalities

1.2 Math Operations Involving Linear Equations

Additional Examples - Mixed Operations Involving Linear Equations The following examples further illustrate how to solve linear equations using the addition, subtraction, multiplication, and division rules: Example 1.2-56 5 5  3 x 15 3x  5  10 ; 3x  5  5  10  5 ; 3x  0  15 ; 3x  15 ;  ; x ; x5 1 3 3

The solution set is  5 .

?

?

Check: 3  5  5 10 ; 15  5 10 ; 10  10 Example 1.2-57 4 y  2  3y  5 ; 4 y  3y  2  3y  3y  5 ; y  2  0  5 ; y  2  5 ; y  2  2  5  2 ; y  0  7 ; y  7 The solution set is  7 . ?

?

Check: 4  7  2 3  7  5 ; 28  2  21  5 ; 26  26 Example 1.2-58 2 x 5 3 3

;

2 x  5 5  3 5 3

;

2 x  0  2 3

;

2 x  2 3

;

3 2 3  x   2 2 3 2

; x

3 1

; x  3

The solution set is  3 . Check:

?

?

?

?

2 2 2   3  5  3 ;   3  5  3 ;   1  5  3 ; 2  5  3 ; 3  3 3 3 1

Example 1.2-59 4

;

u  5u 2

u 2

u 2

; 4  4   5  4  u ; 0   1 u ;

1 u  2  u  1 2 1

;

u  2u 1 2

;

3u 1 2

;

3 u 1 2

;

u  1 u 2

;

2 3 2  u  1 3 2 3

u  u  1 u  u 2

; u

2 3

;

u u   1 0 2 1

; u  0.67

The solution set is  0.67 . Check: 4 

? 0.67 ?  5  0.67 ; 4  0.33  4.33 ; 4.33  4.33 2

Example 1.2-60 5 x  3  10

; 5x  3  3  10  3 ; 5x  0  7 ; 5 x  7 ; ?

5 x 7  5 5

; x

7 5

; x  1.4

The solution set is  1.4 .

?

.   3  10 ; 7  3 10 ; 10  10 Check:  5  14 Example 1.2-61 8t  5  2t  3 ; 8t  2t  5  2t  2t  3 ; 10t  5  0  3 ; 10t  5  3 ; 10t  5  5  3  5

; 10t  0  8 ; 10t  8 ;

Hamilton Education Guides

 t 10 8   10 10

; t

8 10

; t  0.8

The solution set is  0.8 .

20

Linear Equations and Inequalities

1.2 Math Operations Involving Linear Equations

?

?

.  14 . Check:  8  0.8  5 2  0.8  3 ; 6.4  5 16 .  3 ; 14 Example 1.2-62 5  3 y  5 y ; 5  5  3 y  5 y  5 ; 0  3 y  5 y  5 ; 3 y  5 y  5 ; 3 y  5 y  5 y  5 y  5 ; 2 y  0  5

; 2 y  5 ;

2 y 5  2 2

; y

5 2

; y2

?

1 2

The solution set is 2.5 .

; y  2.5

?

Check: 5  3  2.5  5  2.5 ; 5  7.5 12.5 ; 12.5  12.5 Example 1.2-63 6 x  10  8 x  4 ; 6 x  10  10  8 x  4  10 ; 6 x  0  8 x  6 ; 6x  8 x  6 ; 6 x  8 x  8 x  8 x  6 ; 2 x  0  6 ; 2 x  6 ;

3 3 6 ; x ; x ; x3 1 2

2 x 6  2 2

?

The solution set is  3 .

?

Check: 6  3  10 8  3  4 ; 18  10  24  4 ; 28  28 Example 1.2-64 w  10  4 5

;

w  10  10  4  10 5

;

w  0  4  10 5

;

w  14 5

; 5 

w  5  14 5

; w  70

The solution set is  70 . 14 ? ? ? ?  14 70 70 Check:  10  4 ;  10  4 ; 14  10  4 ; 4  4  10  4 ; 5 1 5

Example 1.2-65 0.25 x  3.5  1.2  0.5

;

  x 2.8 0.25   0.25 0.25

;

.  35 .  0.7  35 . ; 0.25x  0  2.8 ; 0.25x  2.8 ; 0.25x  3.5  0.7 ; 0.25x  35

2.8 x 0.25

28 10 280 28  10  28 100 10 ; x ; x ; x = x = x  11.2 25 1  25 25    25 10 100 The solution set is  11.2 .

?

?

.   35 .  12 .  0.5 ; 28 Check: 0.25   112 .  35 .  0.7 ; 0.7  0.7 Example 1.2-66 5m  6  3m  1 ; 5m  3m  6  3m  3m  1 ; 8 m  6  0  1 ; 8 m  6  1 ; 8 m  6  6  1  6

; 8 m  0  7 ; 8 m   7 ;

8 m 7  8 8

7 The solution set is   .

7

; m8

 8

? ? 21 35 35 6 ? 21 1 35 1  6  8 ? 211  1 8 7 7  6 1 ;     ;  8 8 8 8 8 1 8 1 8 1 8 1 35  48 ? 21  8 13 13 ; ;   8 8 8 8

Check: 5    6  3    1 ; 

Example 1.2-67 x

1 2  3x  2 5

1 2

; x  3 x   3x  3 x 

Hamilton Education Guides

2 5

1 2

; 2 x   0 

2 5

1 2

; 2 x   

2 5

1 2

1 2

2 5

; 2 x     

1 2

21

Linear Equations and Inequalities

1.2 Math Operations Involving Linear Equations

1 1 1 1 1 2 x 10  ; ; ; ; ; x  10 ; x ; x   20  2 10  2 2 2 5 2 1 1 The solution set is   .  20  ? ? ? ? 1 1 1 2 1  2  1  20 3 2 2  20 3 2 22 3  5  2  20 Check:    3      ;   ;   ;    20  5 20 2 20  2 20 5 40 20 5 40 20  5 11 11 11 11   ? 55  22 ? 15  40 22  ;   ;   ;  20 20   40 100 40 100 20 20

2  2  1 5 2x  0 

4  5 2 x  10

1 2 x  10

Example 1.2-68 6a  3  4a  4

; 2a  7 ;

; 6a  4a  3  4a  4a  4 ; 2a  3  0  4 ; 2a  3  4 ; 2a  3  3  4  3 ; 2a  0  7

2 a 7  2 2

; a

7 2

; a3

?

1 2

The solution set is 3.5 .

; a  3.5

?

Check: 6  35 .  3  4  35 .  4 ; 21  3  14  4 ; 18  18 Example 1.2-69 0.4 m  5  0.6 m ; 0.4m  0.6m  5  0.6m  0.6m ; 0.2m  5  0 ; 0.2m  5  5  0  5 ; 0.2m  0  5 ; 0.2m  5 ;

0 .2 m 5  0 .2 0.2

;

5 m 0.2

5 25 25 5  10  50 1 ; m 2 ; m ; m ; m ; m  25 1 1 2 2 10

The solution set is  25 .

?

?

Check: 0.4  25  5  0.6  25 ; 10  5 15 ; 15  15 Example 1.2-70 5 5  4 x 20 8 x  20  4 x ; 8 x  4 x  20  4 x  4 x ; 4 x  20  0 ; 4 x  20 ;  ; x ; x5 1 4 4

The solution set is  5 .

?

?

Check: 8  5  20  4  5 ; 40  20  20 ; 40  40 Example 1.2-71 5 z  3  2 z  8 ; 5z  2 z  3  2 z  2 z  8 ; 3z  3  0  8 ; 3z  3  8 ; 3z  3  3  8  3 ; 3z  0  5 ; 3z  5 ;

3 z 5  3 3

; z

5 3

; z  1

2 3

The solution set is  1.667 .

. ; z  1667

?

?

.  11335 . .   3 2  1667 .   8 ; 8335 Check: 5  1667 .  3  3335 .  8 ; 11335 Example 1.2-72

7y 1  1

2 3

2 3

; 7 y 11  1 1 ; 7y  0 

Hamilton Education Guides

1 3  2  1 3

5 1 3 1

; 7y  

; 7y 

5 1  1 3 3 1

; 7y 

53 3

22

Linear Equations and Inequalities

;

2 7y  3

1.2 Math Operations Involving Linear Equations

2 2 2 2 1 7 y 3  ; ; y  73 ; y  ; y  21 3 7 7 7 1

2 The solution set is   .  21 

5 ? 2  ? 5 5 5 2 14 1 ? 1  3  2 14  1  1  21 ? 3  2 14  21 ? 5 35 ? 5 35 Check: 7   1 1 ; ; ;     ;  ;  ;   3 3 3 21 3 21 1 3 21  1 3 21 3 21 3 21 3

Example 1.2-73 2 z  1  12

; 2 z  1  1  12  1 ; 2 z  0  11 ; 2 z  11 ;

2 z 11   2 2

; z

11 2

; z  5

1 2

; z  5.5

The solution set is  5.5 . ?

?

.   1  12 ; 11  1  12 ; 12  12 Check:  2    55 Example 1.2-74 6 x  3  5x

; 6 x  5x  3  5 x  5x ; x  3  0 ; x  3  3  0  3 ; x  0  3 ; x  3 The solution set is  3 . ?

?

Check: 6  3  3  5  3 ; 18  3  15 ; 15  15 Example 1.2-75 2 y  8  5 y  13 ; 2 y  5 y  8  5 y  5 y  13 ; 7 y  8  0  13 ; 7 y  8  13 ; 7 y  8  8  13  8

; 7 y  0  21 ; 7 y  21 ;

7 y 21  7 7

3 3  21 ; y ; y   ; y  3 1 7

The solution set is  3 . ?

?

Check:  2  3  8 5  3  13 ; 6  8  15  13 ; 2  2 Practice Problems - Mixed Operations Involving Linear Equations Section 1.2 Case III Practice Problems - Solve the following linear equations by applying the addition, subtraction, multiplication, and division rules: 1. 3x  20  5x  8

2. 6 y  2  3  10 y

4. 5x  3  15

5.

7. 25  3y  2 y

8. 10 y  2  8 y

1 2

10. m   4m 

y  4  3 4

3.

x 35 2

6. 5  9.

w  10 2

2 x  5  12 3

2 3

Hamilton Education Guides

23

Linear Equations and Inequalities

1.3

1.3 Solving Other Classes of Linear Equations

Solving Other Classes of Linear Equations

In many instances linear equations contain parentheses and brackets, fractions, or decimals. To simplify and solve these classes of equations students should be familiar with the rules that are applicable to each class. For example, solving linear equations that contain either parentheses and brackets or integer fractions require familiarity with the commutative, associative, and distributive rules (see Section 1.1 in Mastering Algebra – Intermediate Level ) a s well as the fractions rules (see Section 1.2 in Mastering Algebra – Intermediate Level ). In this section students learn how to solve linear equations containing parentheses and brackets (Case I), fractions (Case II), and decimals (Case III). Case I

Solving Linear Equations Containing Parentheses and Brackets

To solve linear equations containing parentheses and brackets students need to be familiar with the concepts of signed numbers and the proper use of parentheses and brackets. Linear equations containing parentheses and brackets are solved using the following steps: Step 1

Simplify the linear equation by properly multiplying the negative sign inside the parentheses or brackets.

Step 2

Isolate the variable to the left hand side of the equation by apply ing the addition and subtraction rules (see section 1.2, Case I).

Step 3

Find the solution by applying the multiplication or division rules. Check the answer by substituting the solution into the original equation (see section 1.2 Case II). Examples with Steps

The following examples show the steps as to how linear equations containing parentheses and brackets are solved using the addition, subtraction, multiplication, and division rules: Example 1.3-1 2  3  x   5 x

Solution: Step 1

2  3  x   5x ; 2  3  x  5 x ; 1  x  5x

Step 2

1  x  5x ; 1    x  5x   5x  5x ; 1  6 x  0 ;  1  1  6x  0  1 ; 0  6 x  1

; 6 x  1 Step 3

6 x  1

;

6 x 1  6 6

?

; x

1 6

; x  0.166 ?

.   5  0166 . Check: 2  3  0166 ; 2  3  0166 .  083 . ; 0.83  0.83

Example 1.3-2

 x  2  3x  1  4

Solution: Step 1

 x  2  3x  1  4 ;  x  2  3x  1  4 ;   x  3x    1  2  4 ; 2 x  1  4

Hamilton Education Guides

24

Linear Equations and Inequalities

1.3 Solving Other Classes of Linear Equations

Step 2

2 x  1  4

; 2 x  1  1  4  1 ; 2 x  0  5 ; 2 x  5

Step 3

2 x  5

2 x 5  2 2

;

; x

5 2

; x  2

?

1 2

; x  2.5 ?

?

Check:  2.5  2  3  2.5  1 4 ;  4.5  7.5  1  4 ; 45 .  85 .  4 ; 4  4 Example 1.3-3

2 x  5  3x  8  0

Solution: Step 1

2 x  5  3x  8  0 ; 2 x  5  3x  8  0 ; 2 x  3x   8  5  0 ;  x  3  0

Step 2

x  3  0

;  x  3  3  0  3 ;  x  0  3 ;  x   3

Step 3

 x  3

 x 3  1 1

;

; x3

?

?

?

?

Check: 2  3  5  3  3  8  0 ; 6  5  9  8  0 ; 6  5  1 0 ; 1  1 0 ; 0  0 Example 1.3-4









  x  5  3   x  2  3

Solution: Step 1

  x  5  3   x  2  3 ;  x  5  3  x  2  3 ;  x  2  x  1 ;  x  2  x  1

Step 2

 x  2  x  1 ;  x  x  2   x  x  1

; 2 x  2  0  1 ; 2 x  2  1

; 2 x  2  2  1  2 ; 2 x  0   1 ; 2 x  1 Step 3

2 x  1

;

2 x 1  2 2

; x

1 2

; x  0.5 ?

?

.  15 . .  3 ;  15 .   15 . ; 15 Check: 0.5  5  3 0.5  2  3 ;  4.5  3  15 ?

Example 1.3-5









  x  1   x  5  2x

Solution: Step 1





  x  1   x  5  2x ;  x  1  x  5  2 x ;   x  x  5  1  2x ;  0  4  2x

; 4  2x Step 2

; 4  2x ; 4  2 x  2 x  2 x ; 4  2 x  0 ; 4  4  2 x  0  4 ; 0  2 x  4 ; 2 x  4

Hamilton Education Guides

25

Linear Equations and Inequalities

1.3 Solving Other Classes of Linear Equations

2 x  4

Step 3

;

2 x 4  2 2

2 2 4 ; x   ; x   ; x  2 1 2 ?

Check:  2  1   2  5  2  2 ; 3   7  4 ;  3  7  4 ; 4  4 ?

?

Additional Examples - Solving Linear Equations Containing Parentheses and Brackets The following examples further illustrate how to solve linear equations containing parentheses and brackets using the addition, subtraction, multiplication, and division rules: Example 1.3-6 2 x  4  5 ; 2 x  8  5 ; 2 x  8  8  5  8 ; 2 x  0  13 ; 2 x  13 ;

1 2 x 13 13  ; x ; x  6 ; x  6.5 2 2 2 2

?

?

Check: 26.5  4  5 ; 2  2.5  5 ; 5  5 Example 1.3-7 3  4 x  1  5 x  2 ; 3  4 x  4  5 x  10 ; 4x  3  4  5x  10 ; 4 x  7  5 x  10 ; 4 x  7  7  5 x  10  7 ; 4 x  0  5 x  17 ; 4 x  5 x  17 ; 4 x  5 x  5 x  5 x  17 ;  x  0  17 ;  x  17 ; ?

?

 x 17  ; x  17 1 1

?

Check: 3  417  1  517  2 ; 3  4 18  5 15 ; 3  72  75 ; 75  75 Example 1.3-8 2   x  1  2 x  5 ; 2  x  1  2 x  5 ; 2  1  2 x  x   5 ; 3  x  5 ; 3  3  x  5  3 ; 0  x  2 ; x  2 ?

?

?

Check: 2  2  1  2  2  5 ; 2  1  4  5 ; 1  4  5 ; 5  5 Example 1.3-9 3 y   y  5  10  0 ; 3y  y  5  10  0 ; 3 y  y    5  10  0 ; 2 y  5  0 ; 2 y  5  5  0  5 2 y 5  2 2

; 2 y  0  5 ; 2 y  5 ;

; y

5 2

; y  2

?

1 2

; y  2.5 ?

?

?

Check: 3  2.5   2.5  5  10  0 ; 7.5   2.5  10  0 ; 7.5  2.5  10  0 ; 10  10  0 ; 0  0 Example 1.3-10





2 36x  2  5  3  0 ; 218 x  6  5  3  0 ; 218 x  1  3  0 ; 36 x  2  3  0 ; 36 x  5  0  x 36 5   36 36

; 36 x  5  5  0  5 ; 36 x  0  5 ; 36 x  5 ;

; x

5 36

; x  0.139

.  2  5  3  0 ; 230.834  2  5  3  0 ; 23 1166 .   5  3  0 Check: 236  0139 ?

?

?

?

?

?

.  5  3  0 ; 2 15 ; 2 35 .  3  0 ; 3  3  0 ; 0  0 Example 1.3-11









5  4 3x  5  2x  0 ; 5  4 3x  5  2 x   0 ; 5  4 3x  2x  5  0 ; 5  4 x  5  0 ; 5  4 x  20  0

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26

Linear Equations and Inequalities

1.3 Solving Other Classes of Linear Equations

4 x 15  4 4

; 4 x   20  5  0 ; 4 x  15  0 ; 4 x  15  15  0  15 ; 4 x  0  15 ; 4 x  15 ; ; x

15 4

; x  3

3 4

; x  3.75 ?

.  5   7.5  0 ; 5  4 6.25  7.5  0 Check: 5  43  3.75  5  2  3.75  0 ; 5  4 1125 ?

?

?

?

; 5  4 125 . 0 ; 5  5 0 ; 0  0 Example 1.3-12 3  2 x  2  5  3x  4 ; 3  2 x  4  5  3x  4 ; 2 x  3  4  5  3x  4 ; 2 x  12  3x  4

; 2 x  12  12  3x  4  12 ; 2 x  0  3x  16 ; 2 x  3x  16 ; 2 x  3x  3x  3x  16 ; 5x  0  16 ; 5x  16 ;

5 x 16  5 5

; x

16 5

; x3

1 5

?

; x  3.2 ?

?

?

.   5  9.6  4 ; 3  2.4  5  56 Check: 3  23.2  2  5  3  3.2  4 ; 3  2 12 . ; 0.6  5  56 . ; 5.6  5.6 Example 1.3-13









10  2x  4  x  3x  1 ; 10   2 x  4  x   3x  1 ; 10  2x  x  4  3x  1 ; 10   3x  4  3x  1 ; 10  3x  4  3x  1 ; 10  4  3x  3x  1 ; 6  3x  3x  1 ; 6  6  3x  3x  1  6 ; 0  3x  3x  5 ; 3x  3x  5 ; 3x  3x  3x  3x  5 ; 6 x  0  5 ; 6 x  5 ;

6 x 5 5  ; x   ; x  0.83 6 6 6 ?

?

.  4.83  2.5  1 ; 10  6.49  35 Check: 10  2  0.83  4  0.83  3  0.83  1 ; 10   166 . ; 3.5  3.5 Example 1.3-14 ?

5x  2 x  1  3 x  5 ; 5 x  2 x  1  3x  15 ; 7 x  1  3x  15 ; 7 x  1  1  3x  15  1

; 7 x  0  3x  16 ; 7 x  3x  16 ; 7 x  3x  3x  3x  16 ; 10 x  0  16 ; 10 x  16 ; ; x

16 10

; x 1

6 10

  x 16 10   10 10

; x  1.6 ?

?

?

.  2 16 .  1  316 .  5 ; 8  32 .  1  3 3.4 ; 8  2.2 10.2 ; 10.2  10.2 Check: 5 16 Example 1.3-15









8  2 3x  2  x  5 x  1  2 ; 8  23x  2  x   5x  5  2 ; 8  2 3x  x  2  5x  5  2

; 8  23x  x  2  5x  7 ; 8  2 2 x  2  5x  7 ; 8  4 x  4  5x  7 ; 8  4  4 x  5x  7 ; 12  4 x  5x  7 ; 12  12  4 x  5x  7  12 ; 0  4 x  5x  5 ; 4 x  5x  5 ; 4 x  5x  5x  5x  5 ; x  0  5 ; x  5

Hamilton Education Guides

27

Linear Equations and Inequalities

1.3 Solving Other Classes of Linear Equations

?

Check: 8  23  5  2  5  5 5  1  2 ; 8  215   3  5 6  2 ; 8  2 15  3  30  2 ?

?

?

?

; 8  2 12  30  2 ; 8  24  32 ; 32  32 Practice Problems - Solving Linear Equations Containing Parentheses and Brackets Section 1.3 Case I Practice Problems - Solve the following linear equations by applying the addition, subtraction, multiplication, and division rules: 1. x  2 x  3  3

2. 2  3 x  1  3   x  5

3. 2  3 x  1  5x  0

4. 4  x  1  3x  2 x  1

5. 25   x  2   x  3  0

6.  x  5  3 x  1  2  2

7. 3   x  1  2  3x  5

8. 5  x  3  4x  8

9. 3  2 x  5  4 x  3  3x

10. 6 x  2  2 x  1  3 x  2

Hamilton Education Guides

28

Linear Equations and Inequalities

Case II

1.3 Solving Other Classes of Linear Equations

Solving Linear Equations Containing Integer Fractions

A class of linear equations contains integer fractions. To solve these type of problems students need to be familiar with the fraction rules (review fraction concepts discussed in Section 1.2 in Matering Algebra – Intermediate Level). Note that the method used in solving linear equations with integer fractions is similar with what is addressed in section 1.2. However, these type of problems require more attention due to computations involvin g with integer fractions. Linear equations containing fractions are solved using the following steps: Step 1

Isolate the variable to the left hand side of the equation by applying the addition and subtraction rules (see section 1.2, Case I).

Step 2

Find the solution by applying the multiplication or division rules. Check the answer by substituting the solution into the original equation (see section 1.2, Case II). Examples with Steps

The following examples show the steps as to how linear equations contain ing integer fractions are solved using the addition, subtraction, multiplication, and division rules: Example 1.3-16 x

1 2  x 3 3

x

2 1 2 2 1 2  x ; x  x   x  x ; 1  2  x  1  0 ;  1  2  x  1  0   1 3 3 3 3 3 3 3 3 3 3

Solution: Step 1

 1  3  2 1  1 1 1 1 1 1 1  x   0 ;  3  2  x  1  0 ; x   0 ; x    0  1 3 3  3  3 3 3 3 3 3 3  

; 

;

Step 2

1 1 1 1 x0 ; x  3 3 3 3

1 1 1 1 x  ; 3  x  3  ; x  1  3 3 3 3 1?2 3 3

Check: 1   1 ;

1 1 ? 2 1  3  1  1 ? 2 3  1 ? 2 2 2   ;  ;  ;  1 3 3 1 3 3 3 3 3 3

Example 1.3-17 1 2 u u  4 5

Solution: Step 1

 1  4  1  1  2 1 2  u  ;  4  1 u  2 u  u  ; 1  1  u  2 ;  1  1  u  2 ;  1 4 5  4   1 4 4 5  4 5 5 5  

;

Hamilton Education Guides

3 2 u 4 5 29

Linear Equations and Inequalities

1.3 Solving Other Classes of Linear Equations

4 3 2 4 24 3 2 8 u ;  u  ; u ; u ; u  0.53 3 4 5 3 53 4 5 15

Step 2

?

1 4

.   053 .  Check: 053

? 053 . ?2 2  ; 053 ; 0.53  .  013 .  0.4 ; 5 4 5

0.4  0.4

Example 1.3-18 2 4y  2 1 y 3

Solution: 2 1 3  2 y 4 y  2  3  2 y 4 y  2  5 y 4 y  2  1 y ; 4y  2  ; ; 3 3 3 3

Step 1

5

5

5

5   4 5 ; 4 y  y  2  y  y ; 4   y  2  0 ;    y  2  0     3 3 3 1

3

3

 4  3  5 1  7 7  y  2  0 ;  12  5  y  2  0 ; y  2  0 ; y  2  2  0  2 1 3  3  3 3  

; 

;

7 7 y02 ; y  2 3 3 7 3 7 3 3 2 6 y  2 ;  y  2 ; y  ; y  ; y  0.857   3 7 3 7 7 7

Step 2

?

2 3

?

.  2 Check: 4  0.857  2 1  0.857 ; 343 ?

.  ; 143

? 4.285 5  0.857 .  ; 143 ; 3 3

? 3 2 ?5 1  3  2  0857 . .   0.857 ; 143 .   0.857 ; 143

3

3

3

143 .  143 .

Example 1.3-19 1 1 1 m  m 4 3 2

Solution: 1 1 1 1 1 1 1 1 1 1 1 m   m ; m  m   m  m ; m  m   0 ;  1  1  m  1  4 2 4 2 3 2 2 4 3 2 4 2 3 3

Step 1

 1  2  1  4  1 3 1   m  ;  2  4  m  1 ; 6 m  1 ; m  ;  42 3  8  4 3 8 3 3   3

4

3 1 4 3 4 1 4 1 4 m ;  m  ; m ; m  ; m  0.44 4 3 3 4 3 3 3 3 9

Step 2 Check:

? ?1 1 1 0.44 ? 1 0.44  0.44    0.44 ;   ; 011 .  033 .  022 . ; 4 3 2 4 3 2

Hamilton Education Guides

011 .  011 .

30

Linear Equations and Inequalities

1.3 Solving Other Classes of Linear Equations

Example 1.3-20 x  5  4x  2

3 5

x  5  4x  2

3 3 2  5  3 ; x  4 x  5  4 x  4 x  2 ; 3x  5  0  5 5 5

Solution: Step 1

; 3 x  5  

; 3 x 

13 5 13 13 13 1  5  5 ; 3x  5  5    5 ; 3x  0    ; 3x  5 1 5 5 5 1

13  25 38 ; 3x   5 5 38

38

 38 3 x 3x 3x    5 ;  5 ; 3 5 3 3 3

Step 2

; x

38  1 8 38 ; x ; x  2 ; x  2.53 53 15 15

1 ?

?

3 Check: 2.53  5 4  2.53  2 ; 753 . 1012 .  2.6 ; 7.53  7.52 5

Additional Examples - Solving Linear Equations Containing Integer Fractions The following examples further illustrate how to solve linear equations containing integer fractions using the addition, subtraction, multiplication, and division rules: Example 1.3-21  2  2  1 3  1 2 1 1 1 1 2 1 1 x  5  x  ; x  x  5  x  x  ;  2  1  x  5  0  1 ;  x5 3 2 5 3 2 2 2 5  3 2 3 2 5 5   1  4  3 x 5   6 5

;   ;

;

1 1 1 1 1 1 5 11  5  5 1 x  1  25 1 x 5  ; x 55  5 ; x 0   ; x  ; 6 5 6 5 6 5 1 6 5 6 5 1

1 24 1 24 24 6 24  6 144 4 x ; x  6    6 ; x    ; x   ; x ; x  28 ; x  28.8 6 5 6 5 5 1 5 5 5 1

Check:

?1 ? 1 28.8 1 ? 28.8 1 2 1 2  28.8 57.6  28.8  5   28.8  ;  5  ;   5   3 2 5 3 2 5 3 2 5

?

; 192 .  5  14.4  0.2 ; 14.2  14.2 Example 1.3-22  1  5  1  3  2 2u 4 1 1  u  4 ;  5  3 u  4 ; u4 ;  u  u  4 ;  1  1  u  4 ;  ; 2u  1  15  4 3 5  3 5  15  15 15 1 3 5   30 2 u 60 30  60  ; 2u  60 ; ; u ; u ; u  30 2 2 1 2

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Linear Equations and Inequalities

1.3 Solving Other Classes of Linear Equations

4 ? 30  5  30  3 ?  ?  60 1 1 30 30 ? 150  90 ? 4? 4 ; 4 ;  4 ; 4 ; 4 ; Check:  30   30  4 ;  3 5 3 5 3 5 15 15 1

44

Example 1.3-23

 1  1  2  5  1 1 y 9  0 y  9  2 y ; y  2 y  9  2 y  2 y ;  1  2 y  9  0 ;  1  2  y  9  0 ;  5 1 5   5 1 5 5  

9 9 9 9 1  10  ;   y  9  0 ;  y  9  0 ;  y  9  9  0  9 ;  y  0  9 ;  y  9 

5 

5 9

5

9 5

;    y  9   Check:

5

5

5

5 5 ; y  ; y5 9 1

? ? ? 5 1  5  9  2  5 ;  9  10 ; 1  9 10 ; 5 5

10  10

Example 1.3-24  1  5  1  3  1 2 1 1 1 1 2m 1 m m  m  ;  1  1  m  1 ;    m  ;  5  3 m  1 ; ; 3 5 2  15  15 2 3 5 2  3 5 15 2 2 2  

; 2m 2  1 15 ; 4m  15 ; Check:

3 4 m 15 15  ; m  ; m  3 ; m  3.75 4 4 4 4

? ?1 1 1 375 . 375 . ?1  375 .   375 .  ;   ; 125 .  075 .  05 . ; 3 5 2 3 5 2

0.5  0.5

Example 1.3-25 x

2 1 2 2 1 2 2 1  6  x ; x  x   6  x  x ; x  x   6  0 ; 1  2  x  1  6 ;  1  2  x  1  6   1 3 3 3 3 3 3 3 3 3 3 3 3

 1  3  1  2  1 1 1 1 1 1 1 1 6 1  x   6 ;  3  2  x  1  6 ; x   6 ; x    6  ; x  0   1 3 3  3  3 3 3 3 3 3 3 1 3 3  

; 

;

6  3  11 1 x  18  1 x  19 3  x  3  19 1 x ; ; ; ; 3 3 3 3 3 3 3 1 3 1? 3

2 3

1? 3

Check: 19   6  19 ; 19   6 

x  19

? 2 19 1? 38 ; 19   6  ; 19  033 .  6  12.67 ; 3 3 3

18.67  18.67

Example 1.3-26 3 1 2 3 2 1 3 1 1 2 1 2  3  1 5 3 x  6  5 3 x  1 3 x  x0  ; x    ; x ; ; ; 10 3 5 5 3 10 10 3 3 5 3 10 10 15 10 15 5 3

;

  x 10 3x 1 45 10   ; 3x 15  1 10 ; 45 x  10 ; ; x ; x  0.22  10 15 45 45 45

Check:

? 3 1 ? 2 3  0.22 1 ? 2 0.66 1 ? 2  0.22   ;   ;   ; 0.066  0333 .  0.4 ; 10 3 5 10 3 5 10 3 5

Hamilton Education Guides

0.4  0.4

32

Linear Equations and Inequalities

1.3 Solving Other Classes of Linear Equations

Example 1.3-27 1 1 1 1 1 2 1 w  8  2w  ; w  2w  8  2w  2w  ; w  w  8  0  ;  1  2  w  8   1 5 3 5 3 5 1 3  5 1 3

 1 1  2  5  1 9 1 9 1  w  8   ;  1  10  w  8   1 ;  w  8   ;  w  8  8    8 5 1 3  5  5 3 5 3 3  

; 

1 8 11  8  3  9 w  1  24  9 w   25  9 w   5   25   5 9 ;  w ; ; ; 5 3 5 9 3 9 3 1 5 3 5 3 1

9 5

;  w0   ; w

25  5 125 17 ; w ; w  4 ; w  4.629 3 9 27 27

Check:

? ? ? 1 1 4.629 1  4.629  8  2  4.629  ;  8  9.258  ; 093 ; .  8  9258 .  0333 . 5 3 5 3

8.93  8.93

Example 1.3-28 x

 1  4  1  1  2 1 2 1 1 2 1 x   0  x ; x  x   x  x ; 1  1  x  2  0 ;  1  1  x  2  0 ;  1 4 5  4  1 4 4 5 4 4 5 4 5 5  

2  4  1 x   0 4  5

;  

;

3 2 2 2 3 4 2 4 3 2 3 2 3 2 x   0 ; x    0 ; x 0   ; x   ; x    4 5 5 5 4 4 3 5 3 4 5 5 4 5

24 8 ; x   ; x  0.533 53 15 ? 0533 2?1 . .  0133 . .    0533 . .  0.4  Check: 0533 ; 0533 ; 0133 5 4 4

; x

Example 1.3-29  2  2  1  3  1 2 2 1 1 1 1 2 2 1 1 2 x   x  ; x  x   x  x  ;  2  1  x  1  0  2 ;  x   3 2 4 3 3 2 4 2 2 3  3 2 3 4 2 3 4 3   1 2  4  3 x   6  4 3

;  

;

1 1 2 1 1 1 2 1 1 2  4  1 3 1 x  8  3 1 x  11 x  ; x    ; x ; ; 6 4 3 6 4 4 3 4 6 12 6 12 6 3 4

11 11 ; x  ; x  5.5 2  12 2 .  3  2  2 2 1?1 2 2  55 . 1 ? 1  55 . 2 11 1 ? 55 . 2 11  4  1  3 ? 55 .    55 .  ;    ;    ;  Check:  55 3 4 2 3 3 4 2 3 3 4 2 3 3 4 23 1 6

; 6  x  6 

;

44  3 ? 16.5  4 41 ? 20.5  ;  ; 12 6 12 6

342 .  342 .

Example 1.3-30  1  3  2  2  1 1 2 1 7 1  t  5  t ;  3  4  t  5  1 t ; t  5  t t  t  5  t ;  1  2  t  5  1 t ;  23 4  2 3  6  2 3 4 6 4 4 4  

Hamilton Education Guides

33

Linear Equations and Inequalities

1.3 Solving Other Classes of Linear Equations

 7  4  1  6  7 1 1 1 17  7 1  t  5 ;  28  6  t  5 ; 34 t  5 ; t 5 ; t  t  5  t  t ;    t  5  0 ;  64  6 4  24  6 4 4 4 12  24   17

12

;

  17  12 12 12  5 60 9  t  5 ; t  ; t ; t 3 ; t  3.53   12  17 17 17 17 17

Check:

? 1 2 1 1  353 . 2  353 . ? 1  353 . 353 . 7.06 ? 353 .  353 .   353 .  5   353 . ;  5   5  ; 2 3 4 2 3 4 2 3 4

?

.  412 . ; 177 .  235 .  5  088 . ; 412 Practice Problems - Solving Linear Equations Containing Integer Fractions Section 1.3 Case II Practice Problems - Solve the following linear equations by applying the addition, subtraction, multiplication, and division rules: 1.

1 1 y  y5 2 5

2. x  3  2 3

3 5

1 3

5. s  1  2 s

2 1 1 x 3 4

8. t  5  t

4. u  u  6 7. x 

x 2

1 2

2 3

2 2 y 1 3 3

3.

y

6.

w 2 1 w  4 3 3 2 3

3 5

9. 4 z  2 z  1

1 4

2 5

10. 6  t  t  1 t

Hamilton Education Guides

34

Linear Equations and Inequalities

Case III

1.3 Solving Other Classes of Linear Equations

Solving Linear Equations Containing Decimals

Another class of linear equations contains decimals. To solve these type of problems students need to be familiar with conversion of decimal numbers to integer fraction form (review decimal fraction concepts discussed in Chapter 4 of the Mastering Fractions book). Note that the method used in solving linear equations with decimal numbers is similar with what is addressed in section 1.2. However, these type of problems require more attention due to computations involving with decimals. Linear equations c ontaining decimals are solved using the following steps: Step 1

Isolate the variable to the left hand side of the equation by applying the subtraction rules (see section 1.2, Case I).

addition and

Step 2

Find the solution by applying the multiplication or division rules. Check the answer by substituting the solution into the original equation (see section 1.2, Case II). Examples with Steps

The following examples show the steps as to how linear equations containing decimals are solved using the addition, subtraction, multiplication, and division rules: Example 1.3-31

3.4x  2.5  2.8x  0.5

Solution: Step 1

3.4x  2.5  2.8x  0.5 ; 3.4x  2.8x  2.5  2.8x  2.8x  0.5 ; 6.2x  2.5  0  0.5 ; 6.2x  2.5  0.5 ;

6.2x   2.5  2.5  0.5  2.5

; 6.2x  0  3 ; 6.2x  3

3

Step 2

6 .2 3 3 3  10 30 x 6.2x  3 ; ; x ; x 1 ; x ; x ; x  0.484 62 6.2 6 .2 6.2 1 62 62 10 ?

Check: 3.4  0.484  2.5  2.8  0.484  0.5 ; 165 .  25 .  135 .  05 . ; 0.85  0.85 ?

Example 1.3-32 1.25 x  0.2  0.5x  1  0

Solution: Step 1

1.25 x  0.2  0.5x  1  0

; 0.75x  0.75  0 ;

. x  0.25  0.5x  1  0 ; 1.25x  0.5x  0.25  1  0 ; 125

0.75x   0.75  0.75  0  0.75

; 0.75x  0  0.75 ; 0.75x  0.75 75

Step 2

 0 .75 0.75 0.75 75  100 x 0.75x  0.75 ; ; x ; x   100 ; x   75  0 .75 0.75 0.75 100  75 100

; x

Hamilton Education Guides

7500 ; x  1 7500 35

Linear Equations and Inequalities

Check:

1.3 Solving Other Classes of Linear Equations





?

?

?

.  12 .    0.5  1  0 ; 15 .   15 . 0 ; 1.25 1  0.2   0.5  1  1  0 ; 125

?

15 .  15 . 0 ; 0  0

Example 1.3-33 8.4 x  0.5  0.2 x   1.25x

Solution: Step 1

8.4 x   0.5  0.2 x   1.25x

. x ; 8.4x  0.2x  0.5  1.25x ; 8.4x  0.5  0.2x  125

. x ; 8.6x  1.25x  0.5  1.25x  1.25x ; 7.35x  0.5  0 ; 8.6x  0.5  125 ;

7.35x   0.5  0.5  0  0.5

; 7.35x  0  0.5 ; 7.35x  0.5 5

Step 2

 7 .35 0.5 0.5 5  100 500 x 7.35x  0.5 ; ; x ; x  10 ; x  ; x 735  7 .35 7.35 7.35 10  735 7350 100

; x  0.068 Check: 8.4  0.068  0.5   0.2  0.068 125 .  0.068 ; ?

?

0.57   0.5  0.014  0.09

?

; 057 .  0.48  0.09

; 0.09  0.09 Example 1.3-34









55 .   x  0.2   x  5  0.45  0

Solution: Step 1

55 .   x  0.2   x  5  0.45  0 ;

;

5.5  x  0.2   x  4.55  0

; 5.5  x  0.2  x  4.55  0

;  x  x   5.5  4.55  0.2  0 ; ;

5.5  x  0.2   x  5  0.45  0

2 x  10.25  10.25  0  10.25

2 x  5.5  4.55  0.2  0

; 2 x  10.25  0

; 2 x  0  10.25 ; 2 x  10.25 1025

Step 2

1025 2 10.25 10.25 1025  1 x 2 x  10.25 ; ; x ; x  100 ; x  ; x 2 200 2 2 2 100  2 1

; x  5.125 Check:





?

5.5  5125 .  0.2  5125 .  5  0.45  0

;

?

5.5  4.925   0125 .  0.45  0

?

; 5.5  4.925  0.575  0

?

; 0575 .  0575 . 0 ; 0  0

Hamilton Education Guides

36

Linear Equations and Inequalities

1.3 Solving Other Classes of Linear Equations

Example 1.3-35 0.5x   x  2.5  0.45x  1

Solution: Step 1

0.5x   x  2.5   0.45x  1

;

0.5x  0.55x  15 .

; 0.5x  x  2.5  0.45x  1 ;

; 0.5x  0.55x  0.55x  0.55x   15 . ;

0.5x   x  0.45x   1  2.5

0.05x  0  15 .

;

0.05x  15 .

15

 0 .05 15 . 15 . 15  100 1500 x 0.05x  15 . ; ; x ; x  10 ; x  ; x  5 0 .05 0.05 0.05 10  5 50

Step 2

100

; x  30 Check:



?

;

0.5  30 30  2.5  0.45  30  1

?

0.5  30  27.5  135 .  1

?

; 15  275 .  125 . ; 15  15

Additional Examples - Solving Linear Equations Containing Decimals The following examples further illustrate how to solve linear equations containing decimals using the addition, subtraction, multiplication, and division rules: Example 1.3-36 0.2 x  2.4  0.52 x  35 . ; 0.2 x  0.52 x   2.4  0.52 x  0.52 x   3.5 ; 0.32 x  2.4  0  35 . ; 0.32 x  2.4  35 . 11

x 11 . 0.32 x 11 .   . ; 0.32 x  11. ; ; 0.32 x  2.4  2.4  3.5  2.4 ; 0.32 x  0  11 ; ; x   10 32 0.32 0.32 0.32 1

100

; x

11  100 1100 ; x ; x  3.44 320 10  32

Check:

?

0.2   3.44  2.4  0.52   3.44  35 .

?

.  171 . ; 0.69  2.4  179 .  35 . ; 171

Example 1.3-37 0.65 x  0.2  0.25x

.  0.25x ; 0.65x  0.25x   013 . 0 ; 0.65x  013 .  0.25x  0.25x ; 0.9 x  013 13

013 . 013 . . ; 0.9 x  013 .  013 .  0  013 . ; 0.9 x  0  013 . ; x ; 0.9 x  013 ; x ; x   100 9 0.9 0.9 10

; x

13  10 130 ; x ; x  0.144 100  9 900

Check:

?

0.65 0144 .  0.2  0.25   0144 . 

?

; 0.65  0.056  0.036 ; 0.036  0.036

Example 1.3-38 4.5x  0.2 x  01 .   0.3  0

Hamilton Education Guides

; 4.5x  0.2 x  0.02  0.3  0 ; 4.3x  0.32  0 ; 4.3x  0.32  0.32  0  0.32

37

Linear Equations and Inequalities

1.3 Solving Other Classes of Linear Equations

32

0.32 32  10 320 ; 4.3x  0  0.32 ; 4.3x  0.32 ; x  ; x   100 ; x   ; x ; x  0.074 43 4.3 100  43 4300 10

Check:

?

4.5   0.074  0.2 0.074  01 .   0.3  0

?

?

; 033 .  0.01  0.02  03 .  0 ; 033 .  033 . 0 ; 0  0

Example 1.3-39 4 x  0.45  3.9 x  0.005

.  39 . x  0.005 ; 4x  3.9x   18 ; 4 x  18 .   3.9 x  3.9 x   0.005

.  0.005 ; .  0  0.005 ; 7.9 x  18 ; 7.9 x  18

7.9 x   18 .  18 .   0.005  18 .

. . ; 7.9 x  0  1805 ; 7.9 x  1805

1805

18050 1805 . 1805  10 ; x ; x  1000 ; x  ; x ; x  0.23 79 79000 7.9 1000  79 10

Check:

?

40.23  0.45  39 .  0.23  0.005

?

; 092 .  18 .  089 .  0.005 ; 0.88  0.88

Example 1.3-40 01 . 0.4 x  1  0.2 x  0.28

; 0.04 x  01.  0.2 x  0.28 ; 0.04 x  0.2 x   01.  0.2 x  0.2 x   0.28

. x  01.  0  0.28 ; 016 . x  01.  0.28 ; ; 016

016 . x   01 .  01 .   0.28  01 .

. x  0  0.38 ; 016

38

0.38 3800 0.38 38  100 . x  0.38 ; x  ; 016 ; x ; x   100 ; x   ; x ; x  2.375 16 1600 016 . 016 . 100  16 100

Check:





?

01 . 0.4   2.375  1  0.2   2.375  0.28

.  0195 . ; 0195 Example 1.3-41







;

?

01 .  0.95  1  0.475  0.28



0.5  x  0.2  0.45x  2.5x ; 0.5  x  0.45x  0.2  2.5x ;

;

?

01 .   195 .   0195 .

; 0.275x  01.  2.5x

0.5 0.55x  0.2  2.5x

; 0.275x  2.5x   01.  2.5x  2.5x ; 2.225x  01.  0 ; 2.225x  01.  01.  0  01. ; 2.225x  0  01. 1

1  1000 1000 01 . 01 . ; 2.225 x  01. ; x  ; x ; x   10 ; x   ; x ; x  0.045 2225 10  2225 22250 2.225 2.225 1000

Check:





?

0.5  0.045  0.2  0.45   0.045  2.5   0.045

.  011 . ; 011 Example 1.3-42

0.2  x  1  0.2 x  2  2.4x

Hamilton Education Guides

;

?

0.5 0.245  0.02  011 .



;

?

0.5   0.25  011 .



; 0.2  x   1  0.2 x  0.4  2.4 x ; 0.2  x  1  0.4  0.2x  2.4x

38

Linear Equations and Inequalities

1.3 Solving Other Classes of Linear Equations

. x  2.4 x ; 0.2  0.6   x  0.2 x   2.4 x ; 0.8  12

;

0.2  x   0.6  0.2 x   2.4 x

;

0.8   1.2 x  2.4 x   2.4 x  2.4 x

; 0.8  3.6 x  0 ; 0.8  0.8  3.6x  0  0.8 ; 0  3.6 x  0.8 8

0.8 0.8 80 8  10 ; 3.6 x  0.8 ; x  ; x ; x  10 ; x  ; x ; x  0.22 36 3.6 360 3.6 36  10 10

Check: 0.2  0.22  1  0.20.22  2  2.4  0.22 ; ?

?

0.02  1  0.2  2.22  0.53

;

?

0.02  1  0.444  0.53

?

; 0.02  0556 .  053 . ; 0.54  0.53 Example 1.3-43 0.01x  0.25 x  01 .   1 ; 0.01x  0.25 x  0.025  1 ; 0.26 x  0.025  1 ;

0.26 x  0.025  0.025  1  0.025

975

0.975 97500 975 100 ; 0.26 x  0  0.975 ; 0.26 x  0.975 ; x  ; x  1000 ; x  ; x ; x  3.75 26 26000 0.26 1000  26 100 ?

Check: 0.01 3.75  0.253.75  01.  1 ; 0.0375  0.25  385 . 1 ; 1  1 .  1 ; 0.0375  09625 Example 1.3-44 . x ; 0.3x  0.5x   0.03  2.35  1.2 x 0.3 x  01 .   0.5x  2.35  1.2 x ; 0.3x  0.03  0.5 x  2.35  12 ?

?

. x ; 0.2 x  1.2 x   0.03  2.35  1.2 x  1.2 x  ; x  0.03  2.35  0 ; x  0.03  2.35 ; 0.2 x  0.03  2.35  12 ;

x  0.03  0.03  2.35  0.03

Check:

; x  0  2.32 ; x  2.32 ?

0.32.32  01 .   0.5  2.32  2.35  12 .  2.32

?

?

; 0.3  2.42  116 .  0.434 .  2.35  2.784 ; 0.726  116

; 0.434  0.434 Example 1.3-45

0.2x  23.   0.2x 1 18. x  0 ; 0.2x  2.3  0.2x  0.2  18. x  0 ; 0.2x  0.2x  0.2  23.  18. x  0 . x  2.5  0 ; ;  0.4x  2.5  18 . x  0 ;  0.4 x  18 . x   2.5  0 ; 14

1.4 x   2.5  2.5  0  2.5

25

2.5 250 25  10 . x  0  2.5 ; 14 . x  2.5 ; x  ; 14 ; x   10 ; x   ; x ; x  1.78 14 140 14 . 10  14 10

.   2.3  0.2 178 .  1   18 .  178 .   0 ;  0.35  2.3  0.2  2.78  3.21  0 0.2  178 ?

Check:

?

?

;  2.65  0.56  3.21  0 ; 321 .  321 . 0 ; 0  0 ?

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Linear Equations and Inequalities

1.3 Solving Other Classes of Linear Equations

Practice Problems - Solving Linear Equations Containing Decimals Section 1.3 Case III Practice Problems - Solve the following linear equations by applying the addition, subtraction, multiplication, and division rules: 1. 0.35  0.2 x  0.5  0.1x

2.

5.2 x  01 .  x  0.25  0.2 x

3.

0.4 x  2  0.2 x  1  0.25

4. 1.2 x  0.56  0.6 x  1.25x

5.

 x  05.    x  01.   3x  x

6.

5 0.02 x  0.002  0.5x  1.25

8.

135 .  0.5 x  0.2  0

. 08 . x  0.2  5  2.2x 9. 05

7. 10.

0.5x  2  2.5x   2.8





0.25x  13 .  1.2 x  1.7  2.8

Hamilton Education Guides

40

Linear Equations and Inequalities

1.4

1.4 Formulas

Formulas

Formulas are rules that are stated using symbols, called variables, and are expressed as equations. For example, to find the area of a circle with a radius equal to 3 cm we multiply the constant  “pronounced pi” by the square of the radius. Thus, the area is   32  9 cm 2 . Note that in this case, the stated rule can then be expressed as a formula A    r 2 which is an equation involving two variables, A and r , and a constant,  . In this section we learn how to solve for a specific variable in a given formula by using the following steps: Step 1:

Isolate the variable either to the left or right hand side of the equation by applying the addition and subtraction rule (see Section 1.2, Case I).

Step 2:

Solve for the variable by applying the multiplication or division rules (see Section 1.2, Case II).

Note - If the variable is isolated to the right hand side of the equ ation, to be consistent with the steps used in the previous sections, at the very last step move the variable to the left hand side of the equation (see examples 1.4-2, 1.4-6, and 1.4-7). Examples with Steps The following examples show the steps for solving a specific variable in a given formula: Example 1.4-1 1 3

Solve V  bh for b . Solution: Step 1 Step 2 Example 1.4-2

Not Applicable V 

1 bh 3

3V bh  h h

1 3

; 3  V  3  bh ; 3V  bh ;

;

3V b h

3V

; b h

Solve a  b  c  2d for c and d .

Solution: I.

Step 1

a  b   c  2d ; a  b  cd  2 d ; a  b  2 d  cd ; a  b  b  b  2d  cd

; a  b  0  2d  cd ; a  b  2 d  cd ; a  b  2d  2d  2d   cd ; a  b  2d  0  cd ; a  b  2 d  cd Step 2 II. Step 1

a  b  2 d  cd

;

a  b  2 d cd  d d

;

a  b  2d c d

; c

a  b  2d d

a  b   c  2d ; a  b  b  b  c  2d ; a  b  0  c  2d ; a  b  c  2d

Hamilton Education Guides

41

Linear Equations and Inequalities

Step 2

1.4 Formulas

c  2 d a  b  d d  a  b ab  ; ; c2 c  2 c  2  c  2

a  b   c  2 d ;

Example 1.4-3 Solve A  2r 2  2rh for  and h . Solution: I. Step 1

Not Applicable

Step 2

A  2r  2rh 2

; 



 2r  2rh 2 r A



2

  2 rh 

 2 r 2  2 rh  2

A

;

2r  2rh 2



A 2r  2rh 2

A  2r 2  2rh

II. Step 1



; A   2r  2rh ; 2

; A  2r 2  2r 2  2r 2  2rh ; A  2r 2  0  2rh

; A  2rr 2  2rh A  2r 2  2rh

Step 2

;

A  2r 2 A  2r 2 A  2r 2 2  rh   h ; h  ; 2r 2r 2r 2  r

Example 1.4-4 Solve A  p  prt for p and Solution: I. Step 1

t

.

Not Applicable

A  p  prt ; A  p1  rt  ;

Step 2

p1  rt A A   A  p ; p  ; 1  rt 1  rt  1  rt  1  rt

A  p  prt ; A  p  p  p  prt ; A  p  0  prt ; A  p  prt

II. Step 1

A  p  prt ;

Step 2

A p A  p prt A p    t ; t ; pr pr pr pr 

Example 1.4-5 Solve

1 2 1 x y  z 2 3 5

for x , y , and z .

Solution: I.

Step 1

1 2 1 x y z 2 7 5

;

Step 2

1 1 2 x z y 2 5 7

1 2  1 ; 2  x  2 z  y ; x  5 z  7 y  2 5 7 

Hamilton Education Guides

1 2 2 1 2 x y y z y 2 7 7 5 7

1 1 2 x0 z y 2 5 7

; 2

;

1 1 2 x z y 2 5 7

4

42

Linear Equations and Inequalities

1.4 Formulas

II. Step 1

1 2 1 x y z 2 7 5

;

1 1 2 1 1 x x y z x 2 2 7 5 2

Step 2

2 1 1 y z x 7 5 2

;

7 7 71 1  7 2 71 1  z x  y   z  x ; y   z  x  ; y  10 4 2 7 25 2  25 2 

III. Step 1

2 7

1 5

1 2

; 0 y  z  x ;

2 1 1 y z x 7 5 2

Not Applicable 1 2 1 x y z 2 7 5

Step 2

  ; 5   x  y  5  z ; 1 2

2 7 

5 10 x yz 2 7

1 5

5

10

; z 2 x 7 y

Additional Examples - Formulas The following examples further illustrate how to solve for a specific variable in a given formula: Solve y  mx  b for x and b .

Example 1.4-6

y  mx  b ; y  b  mx  b  b ; y  b  mx  0 ; y  b  mx ;

Solution: I.

yb y  b mx  x  ; m m m 

yb

; x m

II. y  mx  b ; y  mx  mx  mx  b ; y  mx  0  b ; y  mx  b ; b  y  mx Solve s  2t a  b for

Example 1.4-7

Solution: I. s  2t a  b ;

t

and b .

2 t a  b  s s s  t ; t ; 2a  b 2a  b 2 a  b  2a  b

II. s  2t a  b ; s  2at  2bt ; s  2at  2at  2at  2bt ; s  2at  0  2bt ; s  2at  2bt ; Example 1.4-8

s  2at s  2at s  2at 2 bt b ; b  ; 2t 2t 2t 2 t

Solve 2s  3t  4 s  2t   5 for s and

t

.

Solution: I. 2s  3t  4 s  2t   5 ; 2s  3t  4s  8t  5 ; 2s  3t  3t  4s  8t  3t   5 ; 2s  0  4s  11t  5 ; 2s  4s  11t  5 ; 2s  4s  4s  4s  11t  5 ; 2s  0  11t  5 ; 2s  11t  5 ;

2 s 11t  5  2 2

; s

11t  5 2

II. 2s  3t  4 s  2t   5 ; 2s  3t  4s  8t  5 ; 2s  2s  3t  4s  2s  8t  5 ; 0  3t  2s  8t  5 ; 3t  2s  8t  5 ; 3t  8t  2s  8t  8t   5 ; 11t  2s  0  5

Hamilton Education Guides

43

Linear Equations and Inequalities

1.4 Formulas

 t 2s  5 11   11 11

; 11t  2s  5 ;

2s  5

; t   11

Solve I  prt for p , r , and

Example 1.4-9

.

I p r t I I  p ; p  ; rt rt r t rt

I  prt ;

Solution: I.

t

I p r t I I r ; r   ; p t pt p t pt

II. I  prt ;

I p r t I I t ; t   ; p r pr p r pr

III. I  prt ;

Solve A  r 2 for  and r .

Example 1.4-10

A  r 2

Solution: I.

r 2  r2 r 2 A

;

A

II. A  r 2 ;



;

A r

2



; 

A r2

A A  r 2 ;  r2 ; r2  ;   



r2  

A

; r



A



Solve P  2l  2 w for l and w .

Example 1.4-11

P  2l  2 w

Solution: I.

P  2w l 2

;

; P  2 w  2l  2 w  2 w ; P  2 w  2 l  o ; P  2 w  2 l ; P  2w 2

; l

II. P  2l  2 w ; P  2l  2l  2l  2w ; P  2l  0  2 w ; P  2l  2 w ; P  2l w 2

;

P  2l 2

1 2

A

Solution:

1 bh 2

bh 2

; A

; 2  A  2 

bh 2

; 2 A  bh ;

 2 A bh  b b

;

2A h b

2A

; h b

5 9

Solve C   F  32 for F .

Example 1.4-13 C

; Example 1.4-14

; w

P  2l 2 w  2 2

Solve A  bh for h .

Example 1.4-12

Solution:

P  2w 2 l  2 2

5  F  32 9

;

9 C  32  F  0 5

9 9 5  C    F  32 5 5 9

;

9 C  32  F 5

;

9 C  F  32 5

;

9 C  32  F  32  32 5

9

; F  5 C  32

Solve A  h b1  b2  for h and b2 .

Hamilton Education Guides

1 2

44

Linear Equations and Inequalities

A

Solution: I.

;

1.4 Formulas

1 h b1  b2  2

; 2  A  2  h b1  b2  ; 2 A  h b1  b2  ; 1 2

h b1  b 2  2A  b1  b2  b1  b 2 

2A 2A h ; h b b1  b2 1  b2

II. A  h b1  b2  ; 2  A  2  h b1  b2  ; 2 A  h b1  b2  ; 2 A  b1h  b2 h 1 2

1 2

; 2 A  b1h   b1h  b1h  b2 h ; 2 A  b1h  0  b2 h ; 2 A  b1h  b2 h ; ;

2 A  b1 h 2 A  b1h  b2 ; b2  h h

Example 1.4-15

Solve

x4 y5  8 10

x4 y 5  8 10

Solution: I.

;

;

; 8 

for x and y .

x4 4 y 5  8  8 10 5

4 x  0   y  5  4 5 x4 y 5  8 10

II.

2 A  b1h b2 h  h h

4 4 4  4 20 ; x  y   4 ; x  y  4  4 ; x  5 y 5 5 5

5 x4 y 5   10  8 10 4

  ; 10

5  x  4  5  y  0 4

4 5

4 5

; x  4   y  5 ; x  4  4   y  5  4

;

;

5  x  4  5  y 4

5  x  4  y  5 4

;

5  x  4  5  y  5  5 4

5 5 5  5 20 ; x   5  y ; x  5  5  y ; y  4 x 4 4 4

Note that in some instances formulas are solved for a specific variable while numerical values for the remaining variables are given. The following are few examples of this case: Solve A  h b1  b2  for h , if A  50 , b1  3 , and b2  5 . 1 2

Example 1.4-16 A

Solution:

;

1 h b1  b2  2

; 2  A  2  h b1  b2  ; 2 A  h b1  b2  ; 1 2

h b1  b 2  2A   b1  b2   b1  b 2 

2A 2A h ; h b1  b2 b1  b2

Substituting the given numerical values in the above equation we obtain a specific value for h . 25 1 2  50 25 2A  100 h ; h ; h ; h ; h  12 ; h  12.5 2 3 5 2 b1  b2 8 2

Hamilton Education Guides

45

Linear Equations and Inequalities

9 5

Solve F  C  32 for C , if F  100 , F  25 , and F  10 .

Example 1.4-17 Solution:

1.4 Formulas

F

;

9 C  32 5

9 5

9 5

9 5

; F  32  C  32  32 ; F  32  C  0 ; F  32  C

5 5 9   F  32   C 9 9 5

;

5  F  32  C 9

5 9

; C   F  32

5 9

5 9

I. If F  100 , then C  100  32 ; C   68 ; C 

340 9

; C  37

7 9

; C  37.78

Note that 100 degrees Fahrenheit corresponds with 27.78 degrees Centigrade. 5 9

5 9

II. If F  25 , then C  25  32 ; C    7 ; C   5 9

5 9

35 9

; C  3

III. If F  10 , then C   10  32 ; C    42 ; C  

210 9

8 9

; C  3.89

; C  23

3 9

; C  23.33 1 3

Solve V  bh for h , if V  150 and b  2 .

Example 1.4-18 Solution:

V 

1 bh 3

1 3

; 3  V  3  bh ; 3V  bh ;

 3V bh  b b

;

3V h b

; h

3V b

Substituting the given numerical values in the above equation we obtain a specific value for h . h

Example 1.4-19 Solution:

3V b

; h

3  150 2

; h

450 2

; h  225

Solve S  2r 2  2rh 1. For S , if r  2 and h  5 . 2. For h , if S  40 and r 2. I. If r  2 and h  5 , then S  2r 2  2rh ; S  2  2 2  2  2  5 ; S  2  4  2 10 ; S  8  20 ; S  28 II. S  2r 2  2rh ; S  2r 2  2r 2  2r 2  2rh ; S  2r 2  0  2rh ; S  2r 2  2rh ;

S  2r 2 S  2r 2 S  2r 2 2  rh  h ; h  ; 2r 2  r 2r 2r

If S  40 and r  2 , then h  ; h

Hamilton Education Guides

4 10  2  40  8 40  2  2 2 ; h ; h 4  2  2 4 

10  2



46

Linear Equations and Inequalities

1.4 Formulas

5 9

Solve C   F  32 for F , if C  37.78 , C  0 , and C  10 .

Example 1.4-20 C

Solution: ;

5  F  32 9

;

9 C  32  F  0 5

9 9 5  C    F  32 5 5 9

;

9 C  32  F 5

;

9 C  F  32 5

;

9 C  32  F  32  32 5

9 5

; F  C  32

9 5

I. If C  37.78 , then F   37.78  32 ; F 

340  32 5

; F  68  32 ; F  100

0 9 II. If C  0 , then F   0  32 ; F   32 ; F  0  32 ; F  32 5

5

Note that zero degree Centigrade corresponds with 32 degrees Fahrenheit. 9 5

III. If C  10 , then F   10  32 ; F  

90  32 5

; F  18  32 ; F  14

Practice Problems - Formulas Section 1.4 Practice Problems - Solve each formula for the indicated variable. 1. V  r 2 h 3. C  2r 5.

yb

7. m 

for r and h

yb x

9. E  mc 2

4. d  rt

for r

1 2 x b 3 3

2. 2 x  2 y  3 x  y   5

for x and b

for y and b for c and m

Hamilton Education Guides

6.

y

for

abc 3

1 3

8. V  r 2 h 10.

t

for x and y

and r for c

for  , r , and h

y  2 x  3 y   3  5 y  x

for x and y

47

Linear Equations and Inequalities

1.5

1.5 Math Operations Involving Linear Inequalities

Math Operations Involving Linear Inequalities

Just as the symbol “  ” represents equality, the symbols “  ” and “  ” represent less than or greater than , respectively. In general, an equation states that two algebraic expressions are equal. An inequality, in the other hand, states that an algebraic expression is either greater than or less than another one. Note that an inequality with numbers on both sides is a numerical statement of inequality. Numerical statements of inequality are either true or false. For example, 6  5  8 , 4  3  10 , and 15  3  2 are true statements where as 5  3  10 , 8  2  2 , and 6  3  30 are false statements. Algebraic inequalities are inequalities that contain one or more variables. For example, x  3  5 , 2x  5  x  8 , and 3 y  5 y  2 are referred to as algebraic inequalities. An algebraic inequality becomes either a true or false numerical statement, each time a number is substituted for the variable. For example, the algebraic inequality y  3  5 is a false numerical statement if y  3 , because by substituting y  3 in y  3  5 we obtain 3  3  5 which implies 0  5 . On the other hand, y  3  5 is a true numerical statement if y  3 , because y  3  5 becomes 3  3  5 which implies 6  5 . Note that the set of all solutions to an inequality that make an algebraic inequality to become a true numerical statement is referred to as its solution set. For example, the solution set for y  3  5 is the set of all real numbers that are greater than 2 . This is









expressed as y y  2 . The notation y y  2 is read as “the set of all y such that y is greater than 2 .” In this section students learn how to solve algebraic inequalities. The rules for solving inequalities, with only one exception, are the same ones we have learned for solving equations. Solving inequalities using addition and subtraction rules (Case I), and multiplication or division rules (Case II) are addressed below: Case I

Addition and Subtraction of Linear Inequalities

To add or subtract the same positive or negat ive number to linear inequalities the following rules should be used: Addition and Subtraction Rules: The same positive or negative number, or variable, can be added or subtracted to both sides of an inequality without changing the solution: for all real numbers a , b , and c , 1. a  b if and only if a  c  b  c 2.

ab

if and only if

a c  bc.

Linear inequities are added or subtracted using the following steps: Step 1

Isolate the variable to the left hand side of the inequality by applying the addition and subtraction rules.

Step 2

Find the solution by simplifying the inequality. Check the answer by substituting different values, that are greater than or less than the solution found for the inequality, into the original inequality. Examples with Steps

The following examples show the steps as to how linear inequalities are solved using the addition and subtraction rules: Hamilton Education Guides

48

Linear Equations and Inequalities

1.5 Math Operations Involving Linear Inequalities

Example 1.5-1 x 58

Solution: Step 1

x 5 8

Step 2

x 55  85

;

;

x  0  13

x  13

;



Check 1: Let x  2 which is less than 13 . Then, is “yes” because 7  8 .

?

2  5  8

Check 2: Let x  15 which is greater than 13 . Then, is “no” because



The solution set is x x  13 .

Not Applicable

;

?

7  8 ?

?

15  5  8

;

The answer is

?

10  8 ?

The answer is

10  8 .

Example 1.5-2 3 2 2  w 1 4 3

Solution: 3 2  2  4  3 1 3  2 8  3  w  3  2 11  w  5 2  w 1 ; w ; ; 4 3 4 3 4 3 4 3

Step 1

;

5 11 11 11 5 11 5 11  w  ; 0w  ; w   3 4 4 4 3 4 3 4

w

Step 2

5 11 5  4  3 11 w  20  33 w   13 w   1.08  ; w ; ; ; 3 4 12 12 3 4





. . The solution set is w w  108 ?

?

2 11 5 . . Then, is 2 3  108 . 1 ;  108 .  ; Check 1: Let w  108 4 3 answer is “yes” because 167 .  167 . .

4

3

?

2.75  108 .  167 . ?

The

? ? . . Then, is 2 3  2  1 2 ; 11  2  5 Check 2: Let w  2 which is greater than 108

4

;

?

2.75  2  167 . ?

The answer is “yes” because

3

4

3

4.75  167 . .

Example 1.5-3 5u  0.45  4u  1

1 3

Solution: Step 1

5u  0.45  4u  1

1 1 3  1 u  0.45  0  4 ; 5u  4u  0.45  4u  4u  ; 3 3 3

. ; u  0.45  0.45  133 .  0.45 ; u  0  178 . ; u  1.78 ; u  0.45  133

Step 2

Not Applicable

Hamilton Education Guides





. . The solution set is u u  178

49

Linear Equations and Inequalities

1.5 Math Operations Involving Linear Inequalities

?

?

4 1 . . Then, is 5 178 .   0.45  4 178 .   1 ; 8.9  0.45  712 .  Check 1: Let u  178 3

;

?

8.45  712 .  133 . ?

3

The answer is “yes” because 8.45  8.45 . ?

?

. . Then, is 5  0  0.45  4  0  1 1 ; 0  0.45  0  4 Check 2: Let u  0 which is less than 178 3

3

?

4 ; 0.45  ? The answer is “yes” because 0.45  133 . . 3 ?

. . Then, is 5  5  0.45  4  5  1 1 Check 3: Let u  5 which is greater than 178

3 ? ? 4 ; 25  0.45  20  ; 24.55  20  133 . ; 24.55  2133 . ? The answer is “no” because we 3 ?

can not choose u  5 since

5  178 . .

Example 1.5-4 

1 2 1   y   1  3 4 3



1 2 1   y   1  3 4 3

Solution: Step 1

1 4

1 4

1

1  3  2

3

3

;   y   y y 

 ;  1  y  0   1  3  2 3 4 3  

1 1 6 1 2 1 1 5 1 1 5 2 1 ;  y ;    y   ; 0 y   ; y   1 4 3 3 4 3 4 4 3 4 1 4

;  y  y

Step 2

2 1 2  4  11 y  8  1 y  9 y  2 1 y  2.25  ; y ; ; ; ; 1 4 4 4 4 1 4





The solution set is y y  2.25 . Check: Let y  10 which is greater than 2.25 . Then, is ;



1?  1 5   10      3 3 4

is “yes” because

1? 4

;    10 

0.25   8

or



1? 2 1   10    1  3 4 3

? 1 5 1? 6 ;    10  ; 0.25   10  2 ? The answer 3 4 3

0.25  8 .

Example 1.5-5 2 2  y  2 1 3 7

Solution: Step 1

2 2 2 2 2 2 2 2 2 2  y  2 1 ;   y  2 1  ; 0  y  2 1  ; y  2 1  3 7 3 3 7 3 7 3 7 3

Step 2

 2 1  7  2  2 2 2   ; y   2  7  2   2 y  2  1  ; y   2  1 2   2 ; y    7  3  1 7 3 7 3 7  3 1

Hamilton Education Guides

50

Linear Equations and Inequalities

1.5 Math Operations Involving Linear Inequalities

; y        2 1

; y

9 7

 2  7  9  1  2 5 2   ; y   14  9   2 ; y   1 7  7  3 7 3   3

2 3

; y

5  3  2  7

; y

73

15  14 1 ; y  ; y  0.05 21 21





The solution set is y y  0.05 . ?

?

? 2 2 9  0.05  2  1 ; 0.67  0.05  2  ; 0.72  2  128 . ? The 3 7 7 answer is “yes” because 0.72  0.72 .

Check 1: Let y  0.05 . Then, is

? ? 2 2 9 Check 2: Let y  5 which is greater than 0.05 . Then is  5  2  1 ; 0.67  5  2 

3

;

?

5.67  2  128 . ?

7

7

The answer is “yes” because 5.67  0.72 .

Additional Examples - Addition and Subtraction of Linear Inequalities The following examples further illustrate how linear inequalities are solved using the addition and subtraction rules: Example 1.5-6 x  8  12

;

x  8  8  12  8

;

x04

;



Check: Let x  3 which is less than 4 . Then, is Example 1.5-7 5  y   8

;

5  5  y   8  5

;

0 y  3

?

3  8  12 ?

The answer is “yes” because



11  12 .



The solution set is y y   3 .

y  3

;

Check 1: Let y  1 which is greater than 3 . Then, is answer is “yes” because



The solution set is x x  4 .

x4

?

5  1   8

;

?

6   8

;

6 ? 8  ? The 1 1

68. ?

Check 2: Let y  5 which is greater than 3 . Then, is 5  5   8 ? The answer is “yes” because 0   3 . Example 1.5-8 5   w9 ; 5 w   w w9 ; 5 w  09 ; 5 w  9 ; 55 w  95 ; 0 w  4 ; w  4





The solution set is w w  4 . Check: Let w  8 which is greater than 4 . Then, is 5  1. Example 1.5-9 6  x   20

;

6  6  x   20  6

;

0  x   26

;

?

5 89?

x   26



?

6  20   20

;

?

14   20

;

14 ? 20  ? 1 1

14  20 .

Check 2: Let x  5 which is greater than 26 . Then, is

Hamilton Education Guides



The solution set is x x   26 .

Check 1: Let x  20 which is greater than 26 . Then, is The answer is “yes” because

The answer is “yes” because

?

6  5   20 ?

The answer is “yes”

51

Linear Equations and Inequalities

because

1.5 Math Operations Involving Linear Inequalities

11   20 .

Check 3: Let x  50 which is less than 26 . Then, is

?

6  50   20

;

?

44   20

;

44 ? 20  1 1

?

; 44  20 ? The answer is “no” because we can not choose x  50 since Example 1.5-10 m  12  15

;

m  12  12  15  12

;

m  0  27

m  27

;





The solution set is m m  27 .

Check 1: Let m  5 which is less than 27 . Then, is is “yes” because

50   26 .

?

5  12  15

;

?

7  15

;

7 ? 15  ? The answer 1 1

7   15 . ?

Check 2: Let m  20 which is less than 27 . Then, is 20  12  15 ? The answer is “yes” because 8  15 . Example 1.5-11 . ; x   11.6 9.2  x   2.4 ; 9.2  9.2  x   2.4  9.2 ; 0  x   116





. . The solution set is x x  116

. . Then, is Check 1: Let x  15 which is less than 116

?

?

;

5.8   2.4

. . Then, is 9.2  5   2.4 ; Check 2: Let x  5 which is greater than 116 is “no” because we can not choose x  5 since 5   116 . . Example 1.5-12 2.3  w   4.8 ; 2.3  2.3  w   4.8  2.3 ; 0  w   2.5 ; w   2.5

14.2   2.4 ?

The answer is “yes” because

9.2  15   2.4

;

58 . ? 2.4  ? 1 1

5.8  2.4 . ?

?

The answer





. . The solution set is w w   25

Check 1: Let w  2.5 . Then, is

?

2.3  2.5   4.8 ?

The answer is “yes” because 4.8  4.8 .

Check 2: Let w  10 which is greater than 2.5 . Then, is because 7.7   4.8 . Example 1.5-13

?

2.3  10   4.8 ?

The answer is “yes”

2 3 1 3  2  2  7  3 h  3  2  14  3 h  5  17 h  5  5  17  5 h  0  17  5 h 1  2 ; h  ; ; ; ; 3 7 7 3 3 7 3 7 3 3 7 3 3 7

; h

17  3  5  7 73

; h

51  35 16 ; h ; h  0.76 21 21





The solution set is h h  0.76 .

2? 3 2 ? 3 1 3  2 ? 2  7  3  Check: Let h  0 which is less than 0.76 . Then, is 0  1  2 ; 1  2 ; 3 7 3 7 3  2 ? 14  3 5 ? 17  ; ;  ? The answer is “yes” because 167 .  2.43 . 3 7 3 7

3

7

Example 1.5-14 y  3.85  1

3 1 8  3 y  3.85  8  3 y  3.85  11 .  .  138 . ; y  385 ; ; ; y  385 8 8 8 8

Hamilton Education Guides

52

Linear Equations and Inequalities

1.5 Math Operations Involving Linear Inequalities



?

?

3 8

.  1 ; 138 .  Check 1: Let y  5.23 . Then, is 5.23  385

? 83 ? 11 1 8  3 ; 138 .  .  ? The ; 138

8

8

8

answer is “yes” because 138 .  138 . .

 3 . . Then, is 2  385 .  1 ; 185 .  Check 2: Let y  2 which is greater than 178 ?

1 8  3

?

8

?

.  ; 185



The solution set is y y  5.23 .

.  385 .  138 .  385 . ; y  0  5.23 ; y  5.23 ; y  385

8

83 .  ? The answer is “yes” because 185 ; 185 .  138 . . 8 8 ? 11

Example 1.5-15 1 3 1 3  5  1  w  2  5  3  1 5  1 15  1  w  10  3  5  1 16  w  13  6 3  w  2 1 ; ; ; 5 5 5 5 5 5 5 5 5 5 5 5

;

16 13  6 16 19 16 19 16 19 16 16 19  16 3 w  w  ; w ; ; 0 w   ; w  ; w 5 5 5 5 5 5 5 5 5 5 5 5





. . The solution set is w w  06

; w  0.6

? 3 ? 13 6 1 1 16 Check 1: Let w  0.6 . Then, is 3  0.6  2  1 ;  0.6   ;

5

.  38 . . answer is “yes” because 38

5

5

5

5

5

?

3.2  0.6  2.6  12 . ?

? 3 1 1 16 ? 13 6   ; Check 2: Let w  0 which is less than 0.6 . Then, is 3  0  2  1 ;

5

The answer is “yes” because

5

5

5

5

5

The

?

32 .  2.6  12 . ?

3.2  38 . .

? 3 ? 13 6 1 1 16 Check 3: Let w  5 which is greater than 0.6 . Then, is 3  5  2  1 ;  5  

5

;

?

3.2  5  2.6  12 .

since

;

?

8.2  38 . ?

5

5

5

5

5

The answer is “no” because we can not choose w  5

5  0.6 .

Practice Problems - Addition and Subtraction of Linear Inequalities Section 1.5 Case I Practice Problems - Solve the following linear inequalities by adding or subtracting the same positive or negative number to both sides of the inequality. 1.

2. 3   u  8

x  10  12

5. 0.65  t  2

.  w   2.8 4. 32

1 3

7. 0.8  w  1  0.9 2 3

2 5

10. x  1  2 

2 7

3.

2 3

8. 1   h  2

8  x 5

3 5

6. s   1 3 8

3 5

9. y  1.25   2

3 4

2 7

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Case II

1.5 Math Operations Involving Linear Inequalities

Multiplication and Division of Linear Inequalities

To multiply or divide linear inequalities by the same positive or negative number the following rules should be used: Multiplication Rule: a. The same positive number can be multiplied by both sides o f an inequality without changing the solution: for all real numbers a , b , and c , with c  0 (a positive number), a  b if and only if a  c  b  c .

b. The same negative number can be multiplied by both sides on an inequality, however, the direction of the inequality must be changed in order to keep the same inequality: for all real numbers a , b , and c , with c  0 (a negative number), a  b if and only if a  c  b  c .

Division Rule: a. The same positive number (except zero) can be divided by both sides of an inequality without changing the solution: for all real numbers a , b , and c , with c  0 (a positive number), a  b if and only if

a b  c c

.

b. The same negative number can be divided by both sides of an inequality, however, the direction of the inequality must be changed in order to keep the same inequality: for all real numbers a , b , and c , with c  0 (a negative number), a  b if and only if

a b  c c

.

Linear inequalities are multiplied or divided using the following steps: Step 1

Isolate the variable to the left hand side of the inequality by applying the addition and subtraction rules.

Step 2

Find the solution by applying the multipl ication or division rules. Check the answer by substituting different values, that are greater than or less than the solution found for the inequality, into the original inequality. Examples with Steps

The following examples show the steps as to how linear inequalities are solved using the multiplication or division rules: Example 1.5-16 3x  

Solution: Step 1 Step 2

1 5

Not Applicable 3x  

Hamilton Education Guides

1 5

;

1 1 1  3 x    3 5 3

; x

1 1 53

; x

1 15

; x   0.06

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Linear Equations and Inequalities

1.5 Math Operations Involving Linear Inequalities

The solution set is  x x   0.06 . ?

?

1 5

1 Check 1: Let x  10 which is less than 0.06 . Then, is 3  10   ; 30   ? The answer 5

30 0.2  1 1

is “yes” because 30   0.2 or

; 30  0.2 . ?

?

1 Check 2: Let x  0 which is greater than 0.06 . Then, is 3  0   ; 0   0.2 ? The answer

is “no” because we can not choose

x0

5 since 0   0.06 .

Example 1.5-17 

2  4y 3



2  4y 3

Solution: Step 1

; 0  4y  Step 2

4 y 

2 3

2 3

2 3

2 3

2 3

2 3

;   4y  4y  4y ;   4y  0 ;   4y  0 ;    4y  0  2 3

; 4 y 

; 4 y 

2 3

2 3

1 2 1   4 3 4

; y

2 1 3  4

; y

2 12

; y

1 6

. ; y   0166

The solution set is  y y   0166 . . 2 3

?

?

. Check 1: Let y  0166 . Then, is   4  0166 ; 0.66  4  0166 ? The answer is “yes” . .

because 0.66  0.66 . ?

?

2 2 . Check 2: Let y  2 which is less than 0166 . Then, is   4  2 ;    8 ? The answer

is “yes” because 0.66   8 or

0.66 8  1 1

3

3

; 0.66  8 . ?

?

?

2 2 . Check 3: Let y  2 which is greater than 0166 . Then, is   4  2 ;   8 ; 0.66  8 ? 3

3

The answer is “no” because we can not choose y  2 since 2   0166 . . Example 1.5-18 2 3 1 w2 3 5

Solution: Step 1 Step 2

Not Applicable 2 3 1 w2 3 5

;

;

1 3  2 w   2  5  3

3 5 13 3  w  5 3 5 5

3

; w

5

13  3 55

; w

;

3 2 10  3 w 3 5

39 25

;

5 13 w 3 5

; w   1.56 The solution set is w w   156 . .

? ? 13 2 3 5 . . Then, is 1  10   2 ;  10   Check: Let w  10 which is less than 156

3

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5

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Linear Equations and Inequalities

; 

1.5 Math Operations Involving Linear Inequalities

16.6 2.6 50 ? 13  ; 16.6  2.6 .   ? The answer is “yes” because 16.6   2.6 or 1 1 3 5

Example 1.5-19 2   5t

Solution: Step 1

2   5t ; 2  5t   5t  5t ; 2  5t  0 ; 2  2  5t  0  2 ; 0  5t  2 ; 5t  2

Step 2

5t  2 ;

5 t 2  5 5

The solution set is t t  0.4 .

; t  0.4

?

Check: Let t  5 which is greater than 0.4 . Then, is 2   5  5 ? The answer is “yes” because 2   25 or

2 25  1 1

; 2  25 .

Example 1.5-20 2.6 m  3

Solution: Step 1

2 3

Not Applicable 2.6 m  3

Step 2

2 3

; 2.6m 

3  3  2 3

; 2.6m 

92 3

; 2.6m 

11 3

; 2.6m  3.66 ;

2 .6 3.66 m 2.6 2.6

366 366  10 3660 100 ; m  26 ; m  ; m ; m  1.4 100  26 2600 10

The solution set is m m  14 . . ?

?

2 11 Check 1: Let m  1 which is less than 1.4 . Then, is 2.6 1  3 ; 2.6  ? The answer is 3

3

“yes” because 2.6  3.7 . ?

?

?

2 11 . ? The Check 2: Let m  2 which is greater than 1.4 . Then, is 2.6  2  3 ; 5.2  ; 5.2  37 3

3

answer is “no” because we can not choose m  2 since 2  1.4 . Additional Examples - Multiplication and Division of Linear Inequalities The following examples further illustrate how linear inequalities are simplified using the above multiplication or division rules: Example 1.5-21 5 y  75 ;

5 y 75  5 5

; y   15

The solution set is  y y   15 . ?

Check: Let y  40 which is less than 15 . Then, is 5   40  75 ? The answer is “yes” because 200  75 .

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1.5 Math Operations Involving Linear Inequalities

Example 1.5-22 2 x 4 3

;

3 2 3  x   4 2 3 2

4 3 1 2

; x 

; x

43 1 2

12 2

; x

; x  6

The solution set is  x x   6 .

? 2 12 ? Check 1: Let x  6 . Then, is  6   4 ;    4 ? The answer is “yes” because 4   4 .

3

3

? 2 20 ? Check 2: Let x  10 which is less than 6 . Then, is  10   4 ;    4 ? The answer is

6.67 4  1 1

“yes” because 6.67   4 or

3

3

; 6.67  4 .

Example 1.5-23 4   2h 7

; h

;

4 14

4  2h   2h  2h 7

;

4  2h  0 7

;

4 4 4   2h  0  7 7 7

4 7

; 0  2h  

; 2h  

4 7

1 4 1  2 h    2 7 2

;

The solution set is h h   0.28 .

; h   0.28

Note that another way of solving these type of inequalities is as shown below: 4   2h 7

;

4 1 1    2 h  7 2 2

; 

4 h 14

; 0.28  h or h  0.28

? 4 ? Check 1: Let h  0.28 . Then, is   2  0.28 ; 0.56   2  0.28 ? The answer is “yes” because

7

0.56  0.56 .

4? Check 2: Let h  7 which is greater than 0.28 . Then, is   2  7 ? The answer is “yes” 7

because 0.57   14 . Example 1.5-24 w   9 ;

 w 9  1 1

The solution set is w w  9 .

; w9 ?

Check: Let w  2 which is less than 9 . Then, is 2   9 ? The answer is “yes” because 2 9  1 1

; 29.

Example 1.5-25 5  

; x

2 x 5

25 2

2 5

2 5

2 5

2 5

2 5

2 5

; 5  x   x  x ; 5  x  0 ; 5  5  x  0  5 ; 0  x  5 ;

2 x5 5

;

5 2 5  x  5 2 5 2

The solution set is x x  12.5 .

; x  12.5

Note that another way of solving these type of inequalities is as shown below: 5  

2 x 5

5 2

2 5

5 2

; 5       x ;

25 x 2

; 12.5  x or x  12.5 ?

?

2 2 Check: Let x  5 which is less than 12.5 . Then, is 5    5 ; 5    5 ? The answer is “yes” 5

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1.5 Math Operations Involving Linear Inequalities

because 5   2 or

5 2  ; 52. 1 1

Example 1.5-26 1 k3 8

The solution set is k k  24 .

1 8

; 8  k  8  3 ; k  24 3

?  ? 24 1 3? Check 1: Let k  24 . Then, is  24  3 ;  3 ;  3 ? The answer is “yes” because 3  3 .

8

8

1

Check 2: Let k  8 which is less than 24 . Then, is

1 ? 1 ?  8  3 ;  8  3 ? The answer is “yes” 8 8

because 1  3 . Example 1.5-27 5 

w 6

; 5 

w w w   6 6 6

; 5 

w 0 6

; 5  5 

w  05 6

; 0

w 5 6

;

w 5 6

; 6 

w  5 6 6

; w  30

Or the inequality can be solved in the following way: 5 

w 6

;  5  6 

w  6 6

; 30 

w  6 6

The solution set is w w  30 .

; 30  w or w  30 ?

?

10 10 Check: Let w  10 which is less than 30 . Then, is 5  ; 5   ? The answer is “yes”

because 5   166 . or

5 166 .  1 1

6

6

; 5  166 . .

Example 1.5-28 1 5 x 5 7

1 5

5 7

; 5  x    5 ; x  

25 7

The solution set is  x x   3.57 .

; x   3.57

? 5 1 357 . ? 5 Check 1: Let x  3.57 . Then, is  357 .  ;    ? The answer is “yes” because

5

7

5

7

0.714  0.714 ? 5 ? 5 ? 5 1 1 . . Then, is  5   ;  5   ; 1   ? The Check 2: Let x  5 which is less than 357

Check 3:

5 7 5 7 7 1 0.7  answer is “yes” because 1   0.71 or ; 1  0.7 1 1 ? ? 5 1 ? 5 . . Then, is  0   ; 0   ; 0   0.71 ? Let x  0 which is greater than 357 5 7 7 answer is “no” because we can not choose x  0 since 0   357 . .

The

Example 1.5-29 

1 y5 4

4 1

1 4

;    y  5 

4 1

; y

20 1

; y   20

The solution set is  y y   20 .

5 ?  ? 20 5 ? 1 Check 1: Let y  20 . Then, is   20  5 ;  5 ;  5 ? The answer is “yes” because 5  5 . 4 1 4 2 ? 8 ? 2? 1 Check 2: Let y  8 which is greater than 20 . Then, is   8  5 ;  5 ;  5 ? The answer 4 1 4 is “yes” because 2  5 .

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1.5 Math Operations Involving Linear Inequalities

10 ? ?  ? 40 1 10 ? Check 3: Let y  40 which is less than 20 . Then, is   40  5 ;  5 ; 10  5 ? 5 ; 4 1 4 The answer is “no” because we can not choose y  40 since 40   20 .

Example 1.5-30

2  3  2   1 5  1 x

2

2 1  1 x 3 5

3

5

;

8 8 6 8   x  0 3 3 5 3

; 0 x  

;

6 5

8 3

;

;

62 5 1  x 3 5

6 8 x 5 3

;

;

8 6  x 3 5

5 6 8 5  x  6 5 3 6

;

8 6 6 6  x x x 3 5 5 5

; x

40 18

;

8 6  x0 3 5

; x   2.22

Or the inequality can be solved in the following way: 2

2 1  1 x 3 5

;

2  3  2   1 5  1 x 3

5

;

62 5 1  x 3 5

;

8 6  x 3 5

5 6

6 5

5 8 6 3

;      x ; 

40 x 18

The solution set is  x x   2.22 .

; 2.22  x or x   2.22 2 3

?

1 5

Check 1: Let x  2.22 . Then, is 2   1  2.22 ;

8 ? 6 8 ? 1332 . ? The answer is   2.22 ;  3 5 3 5

“yes” because 2.66  2.66 . 8? 2? 1 Check 2: Let x  0 which is greater than 2.22 . Then, is 2   1  0 ;  0 ? The answer is 3

5

3

“yes” because 2.66  0 . 2 ? 1 8 ? 90 8? 6 Check 3: Let x  15 which is less than 2.22 . Then, is 2   1  15 ;  15 ;  3

5

3

5

3

5

?

; 2.66  18 ? The answer is “no” because we can not choose x  15 since 15   2.22 . Practice Problems - Multiplication and Division of Linear Inequalities Section 1.5 Case II Practice Problems - Solve the following linear inequalities by applying the multiplication or division rules: 1. 4 y   4.

2 3

2 3

2.  x  1

w  3 7

5. 3 4

7. 2 x  2  1 1 3

2 3

w  5 2

8. 328 . x  2.4

3.

2   2h 5 1 4

6. 2 u   1

1 5

1 4

9.  y   2

3 4

10. 5   2 x

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Linear Equations and Inequalities

Case III

1.5 Math Operations Involving Linear Inequalities

Mixed Operations Involving Linear Inequalities

In Cases I and II we learned how to solve linear inequalities by either applying: 1. The addition and subtraction rules, or 2. The multiplication or division rules. In this section, solution to linear inequalities which may involve using all four rules is discussed. Addition and Subtraction Rules: The same positive or negative number, or variable, can be added or subtracted to both sides of an inequality without changing the solution: for all r eal numbers a , b , and c , 1. a  b if and only if a  c  b  c 2. a  b if and only if a  c  b  c . Multiplication Rule: a. The same positive number can be multiplied by both sides of an inequality without changing the solution: for all real numbers a , b , and c , with c  0 (a positive number), a  b if and only if a  c  b  c .

b. The same negative number can be multiplied by both sides on an inequality, however, the direction of the inequality must be changed in order to keep the same inequality: for all real numbers a , b , and c , with c  0 (a negative number), a  b if and only if a  c  b  c .

Division Rule: a. The same positive number (except zero) can be divided by both sides of an in equality without changing the solution: for all real numbers a , b , and c , with c  0 (a positive number), a  b if and only if

a b  c c

.

b. The same negative number can be divided by both sides of an inequality, however, the direction of the inequality must be changed in order to keep the same inequality: for all real numbers a , b , and c , with c  0 (a negative number), a  b if and only if

a b  c c

.

Linear inequalities are solved using the following steps: Step 1

Isolate the variable to the left hand side of the inequality by applying the addition and subtraction rules.

Step 2

Find the solution by applying the multiplication or division rules. Check the answer by substituting different values, that are greater than or less than the solution found for the inequality, into the original inequality. Examples with Steps

The following examples show the steps as to how linear inequalities are solved using the addition, subtraction, multiplication, and division rules:

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1.5 Math Operations Involving Linear Inequalities

Example 1.5-31 6y 1  2

3 7

Step 1

6y 1  2

3 7

Step 2

3 6 y  2 1 7

Solution:

; 6y 

17  7 7

3 7

3 7

3 7

; 6 y 11  2 1 ; 6 y  0  2 1 ; 6 y  2 1 ; 6y 

2  7  3  1

; 6y 

7

10 7

; 6y 

. ; ; 6 y  143

14  3 1 7

6 y 1.43  6 6

; 6y 

17 1  7 1

; 6y 

17 1  1 7 7 1

; y  0.24

The solution set is  y y  0.24 . ?

?

?

3 17 17 Check 1: Let y  4 which is greater than 0.24 . Then, is 6  4  1  2 ; 24  1  ; 25  ? 7

7

7

The answer is “yes” because 25  2.43 . ?

?

?

3 17 Check 2: Let y  0 which is less than 0.24 . Then, is 6  0  1  2 ; 0  1  ; 1  2.43 ? The 7

7

answer is “no” because we can not choose y  0 since 0  0.24 . Example 1.5-32 Solution: Step 1

6t  10  9t  5 6t  10  9t  5 ; 6t  9t  10  9t  9t  5 ; 3t  10  0  5 ; 3t  10  5

; 3t  10  10  5  10 ; 3t  0   5 ; 3t   5 Step 2

3t   5 ;

3 t 5  3 3

; t  1.66

The solution set is t t  166 . .

?

?

.  10  9 166 .  5 ; 9.9  10  14.9  5 ? The answer is “yes” Check 1: Let t  1.66 . Then, is 6 166

because 19.9  19.9 . ?

?

. . Then, is 6 1  10  9 1  5 ; 6  10  9  5 ? The Check 2: Let t  1 which is less than 166 answer is “yes” because 16  14 . ?

?

. . Then, is 6  3  10  9  3  5 ; 18  10  27  5 Check 3: Let t  3 which is greater than 166 ?

; 28  32 ? The answer is “no” because we can not choose t  3 since 3  166 . . Example 1.5-33 Solution: Step 1 Step 2

3x  25  8x 3x  25  8x ; 3x  8x  25  8x  8x ; 5x  25  0 ; 5x  25 5x  25 ;

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5 x 25  5 5

; x  5

The solution set is  x x   5 .

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?

?

Check 1: Let x  5 . Then, is 3  5  25  8  5 ; 15  25  40 ? The answer is “yes” because 15   15 . ?

?

Check 2: Let x  2 which is greater than 5 . Then, is 3  2  25  8  2 ; 6  25  16 ? The answer is “yes” because 6  9 . Example 1.5-34 0.8n  10  12 . n

Solution: Step 1

0.8n  12 . n  10  12 . n  12 . n ; 0.4n  10  0 ; 0.4n  10  0 ; 0.4n  10  10  0  10

; 0.4n  0   10 ; 0.4n   10 Step 2

0.4n   10 ;

0.4 n 10  0.4 0.4

10 n 0.4

;

10 10  10 100 ; n  14 ; n  ; n ; n  25 1 4 4 10

The solution set is n n  25 .

?

?

.  25 ; 20  10  30 ? The answer is “yes” Check 1: Let n  25 . Then, is 0.8  25  10  12 because 30  30 . ?

?

.  20 ; 16  10  24 ? The Check 2: Let n  20 which is less than 25 . Then, is 0.8  20  10  12 answer is “yes” because 26  24 . ?

?

.  30 ; 24  10  36 Check 3: Let n  30 which is greater than 25 . Then, is 0.8  30  10  12 ?

; 34  36 ? The answer is “no” because we can not choose n  30 since 30  25 . Example 1.5-35 z

2 1  4z  3 5

z

2 1  4z  3 5

Solution: Step 1

2 3

2 3

2 3

1 5

; 3z      Step 2

3 z  

;

1 2  5 3

1 5

; 3 z   0 

1 5

2 3

; z  4z   4z  4z  2 3

; 3z 

1 7 1  3 z   3 15 3

; 3 z  0   

1  3  2  5

; z

53 7 1 15  3

2 3

1 5

; 3 z   

; 3 z 

; z

7 45

1 5

3  10 15

2 3

; 3 z   

1 5

2 3

; 3 z 

7 15

. ; z   0155

The solution set is z z   0155 . . ?

?

2 1 . . Then, is 0155 .  0.66   0.62  0.2 ? The Check 1: Let z  0155 .   4  0155 .   ; 0155 3

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5

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1.5 Math Operations Involving Linear Inequalities

answer is “yes” because 0.82   0.82 . ?

?

2 1 . . Then, is 5   4  5  ; 5  0.66  20  0.2 ? Check 2: Let z  5 which is greater than 0155 3

5

The answer is “yes” because 4.34  19.8 . ?

2 1 . . Then, is 5   4  5  Check 3: Let z  5 which is less than 0155 ?

?

; 5  0.66   20  0.2 ; 5.66   20.2 or ;

3 5.66 20.2  1 1 ?

5

?

; 5.66  20.2 ? The answer

is “no” because we can not choose z  5 since 5   0155 . . Additional Examples - Mixed Operations Involving Linear Inequalities The following examples further illustrate how to solve linear inequalities using the addition, subtraction, multiplication, and division rules: Example 1.5-36 2x  8  10 ; 2 x  8  8  10  8 ; 2 x  0  18 ; 2 x  18 ;

2 x 18  2 2

; x9

The solution set is  x x  9 . ?

?

Check 1: Let x  20 which is greater than 9 . Then, is 2  20  8  10 ; 40  8  10 ?. The answer is “yes” because 32  10 . ?

?

?

Check 2: Let x  0 which is less than 9 . Then, is 2  0  8  10 ; 0  8  10 ; 8  10 ? The answer is “no” because we can not choose x  0 since 0  9 . Example 1.5-37 6w  5  2w  8 ; 6w  5  5  2w  8  5 ; 6w  0  2w  13 ; 6w  2w  13 ; 6w  2w  2w  2w  13

; 4w  0  13 ; 4w  13 ;

4 w 13  4 4

The solution set is w w  3.25 .

; w  3.25 ?

?

.  5  2  325 .  8 ; 19.5  5  6.5  8 ? The answer is “yes” Check 1: Let w  3.25 . Then, is 6  325

because 14.5  14.5 . ?

?

Check 2: Let w  2 which is less than 3.25 . Then, is 6  2  5  2  2  8 ; 12  5  4  8 ?. The answer is “yes” because 7  12 . Example 1.5-38 2 x 38 5

;

2 x  3 3  8  3 5

;

2 x05 5

;

2 x5 5

;

5 2 5  x  5 2 5 2

; x

25 2

; x  12.5

The solution set is x x  12.5 . ? ? ? 2 2 Check: Let x  5 which is less than 12.5 . Then, is  5  3  8 ;  5  3  8 ; 2  3  8 ? The answer

5

5

is “yes” because 5  8 .

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1.5 Math Operations Involving Linear Inequalities

Example 1.5-39 4

;

u  5  2u 3

1 u  3  2u  1 3 1

u 3

u 3

; 4  4   5  4  2u ; 0   1  2 u ; ;

u  6u 1 3

;

7u 1 3

;

7 u 1 3

;

u  1  2u 3

u  2u  1  2 u  2 u 3

;

3 7 3  u  1 7 3 7

; u

3 7

;

u 2u   1 0 3 1

; u  0.43

The solution set is u u  0.43 . ?

?

0.43 .  5  0.86 ? The answer is “yes” Check 1: Let u  0.43 . Then, is 4   5  2  0.43 ; 4  014 3

because 414 .  414 . . ?

?

6 Check 2: Let u  6 which is greater than 0.43 . Then, is 4   5  2  6 ; 4  2  5  12 ?. The 3

answer is “yes” because 6   7 . ?

?

?

0 0 Check 3: Let u  0 which is less than 0.43 . Then, is 4   5  2  0 ; 4   5  2  0 ; 4  5 ? 3

3 0  0.43 .

The answer is “no” because we can not choose u  0 since Example 1.5-40 25   18  h ; 25  h   18  h  h ; 25  h   18  0 ; 25  h   18 ; 25  25  h   18  25 ; 0  h  7 ; h  7 ;

h 7  1 1

; h 7

The solution set is h h   7 .

? ? 25 ? 19 Check 1: Let h  1 which is greater than 7 . Then, is 25   18  1 ; 25   19 ; ? 

1

1

The answer is “yes” because 25  19 . ? ? 25 ? 16 Check 2: Let h  2 which is greater than 7 . Then, is 25   18  2 ; 25   16 ; ? 

1

1

The answer is “yes” because 25  16 . Example 1.5-41 5  y  35 ; 5  y  y  y  35 ; 5  y  0  35 ; 5  y  35 ; 5  5  y  35  5 ; 0  y  30 ;  y  30 ;

 y 30  1 1

; y   30

The solution set is  y y   30 . ?

Check 1: Let y  40 which is less than 30 . Then, is 5   40  35 ? The answer is “yes” because 5   5 . ?

?

Check 2: Let y  0 which is greater than 30 . Then, is 5  0  35 ; 5  35 ? The answer is “no” because we can not choose y  0 since 0   30 . Example 1.5-42 8  w  6 ; 8  w  w  w  6 ; 8  w  0  6 ; 8  w  6 ; 8  8  w  6  8 ; 0  w  6  8 ; w  14 ;

 w 14  1 1

; w   14

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The solution set is w w   14 .

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Linear Equations and Inequalities

1.5 Math Operations Involving Linear Inequalities

? ? 8 ? 14 Check 1: Let w  20 which is less than 14 . Then, is 8   20  6 ; 8   14 ; ? The 

1

1

answer is “yes” because 8  14 . ?

?

Check 2: Let w  10 which is greater than 14 . Then, is 8  10  6 ; 8  16 ? The answer is “no” because we can not choose w  10 since 10   14 . Example 1.5-43 x 75 4

x 7 7  5 7 4

;

x  0  12 4

;

x  12 4

;

x 4

; 4   12  4 ; x  48 The solution set is  x x  48 .

? ? 50 Check 1: Let x  50 which is greater than 48 . Then, is  7  5 ; 12.5  7  5 ? The answer is

4

“yes” because 5.5  5 . ? ? ? 12 Check 2: Let x  12 which is less than 48 . Then, is  7  5 ; 3  7  5 ; 4  5 . The answer is

4

“no” because we can not choose x  12 since 12  48 . Example 1.5-44 2

3 1  k2 5 4

;

2  5  3  k  2  4  1 5

4

;

10  3 8 1 k 5 4

;

13 9 k 5 4

; 2.6  k  2.3 ; 2.6  k  k  k  2.3

; 2.6  k  0  2.3 ; 2.6  k  2.3 ; 2.6  2.6  k  2.3  2.6 ; 0  k   0.3 ;  k   0.3 ;

 k 0.3  1 1

; k  0.3

The solution set is k k  0.3 .

2  4  1 3? 1 2  5  3 ? Check: Let k  01. which is less than 0.3 . Then, is 2  01 . 2 ;  01 .  5

10  3 8 1  01 .  5 4 ?

;

;

?

13 9  01 .  5 4

4

5

4

?

.  2.3 ? The answer is “yes” because 2.6  2.4 . ; 2.6  01

Example 1.5-45 3 4

2.8  u  2

; 2.8  u  u  u 

2  4  3 4

; 2.8  u  0 

. ; u  555 . ; ; 2.8  2.8  u  2.75  2.8 ; 0  u  555

83 4

; 2.8  u 

 u 5.55  1 1

11 4

; 2.8  u  2.75

; u   5.55

The solution set is u u   5.55 .

?

?

3 Check 1: Let u  5.55 . Then, is 2.8   555 .  2 ; 2.8   555 .  4

;

?

11 2.8   555 .  4

2  4  3 ; 4

?

2.8   555 . 

83 4

?

.  2.75 ? The answer is “yes” because 2.8  2.8 . ; 2.8   555 ?

?

3 Check 2: Let u  10 which is less than 5.55 . Then, is 2.8   10  2 ; 2.8   10  4

2  4  3 4

11 83 ; 2.8   10  ; 2.8   10  2.75 ? The answer is “yes” because 4 4 2.8 7.25  ; 2.8  7.25 . 2.8   7.25 or 1 1 ?

; 2.8   10 

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?

?

65

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1.5 Math Operations Involving Linear Inequalities

Practice Problems - Mixed Operations Involving Linear Inequalities Section 1.5 Case III Practice Problems - Solve the following linear inequalities by applying the addition, subtraction, multiplication, and division rules: 1. 2x  9  9x  20

2. 15x  3  20x

3. 4 x  5  10

4. 12t  4  4t  8

5. 4w  5  8w  17

6. 10 y  4  4 y  12

7.

y 2 3 5 1 3 3 5

4 5

8. 3  t  2

1 3

9. 3.4  w  2

3 5

10. 0.48x  2.5  15 . x  0.35

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Appendix – Exercise Solutions: Section 1.1 Solutions - Introduction to Linear Equations 1.

Determine whether 2 is the solution to each of the following equations: ? ? a. 3  2  2 10 ; 6  2 10 ; 4  10 . Therefore, 2 is not the solution to 3x  2  10 . ? ? b. 2  2  3 2 ; 4  3  2 ; 1  2 . Therefore, 2 is not the solution to 2 x  3  x . ? ? c. 6  2  2  2  1 ; 4  4  1 ; 4  5 . Therefore, 2 is not the solution to 6  x  2 x  1 . ? ? d. 2  2  8  3  2  2 ; 4  8  6  2 ; 4  4 . Therefore, 2 is the solution to 2 x  8  3 x  2 .

2.

Determine if y  2 is the solution to the following equations: ? a. 2  3  2  2 ; 1  4 . Therefore, y  2 is not the solution to y  3  2 y . ? ? b. 6  2  2 8  2  2 ; 12  2  16  2 ; 14  14 . Therefore, y  2 is the solution to 6 y  y  8 y  2 .

3.

c.

? ? 6  3  2  0 ; 6  6  0 ; 0  0 . Therefore, y  2 is the solution to 6  3 y  0 .

d.

? ? 3  2  5   2 ; 6  5  2 ; 6  7 . Therefore, y  2 is not the solution to 3y  5  y .

Given the algebraic equation 2 x  8   x  5  3 , does x  0 , x  1 , and x  6 satisfy the original equation? a.

? ? Let x  0 in 2 x  8   x  5  3 . Then, 2  0  8  0  5  3 ; 0  8  5  3 ; 8  2 . Therefore, x  0 does not satisfy 2 x  8   x  5  3 .

b.

? ? Let x  1 in 2 x  8   x  5  3 . Then,  2  1  8  1  5  3 ; 2  8  6  3 ; 10  3 . Therefore, x  1 does not satisfy 2 x  8   x  5  3 .

c.

? ? Let x  6 in 2 x  8   x  5  3 . Then, 2  6  8  6  5  3 ; 12  8 1  3 ; 4  4 . Therefore, x  6 satisfies 2 x  8   x  5  3 .

4.

Does a  2 satisfy any of the following equations? ? ? a. 3  2  2  4  2 ; 6  2  8 ; 8  8 . Therefore, a  2 is the solution to 3a  2  4a . ? ? b. 3  7  2 18 ; 3  14 18 ; 17  18 . Therefore, a  2 is not the solution to 3  7a  18 . ? ? c. 5  2  3  3  2  1 ; 10  3  6  1 ; 7  7 . Therefore, a  2 is the solution to 5a  3  3a  1 . ? d. 8  2  3 ; 8  5 . Therefore, a  2 is not the solution to 8  a  3 .

Section 1.2 Case I Solutions - Addition and Subtraction of Linear Equations 1.

x  13  12 ; x  13  13  12  13 ; x  0  25 ; x  25

2.

8  h  20 ; 8  8  h  20  8 ; 0  h  12 ; h  12

3.

5  x 3 ; 5 3 x 3 3 ; 8  x  0 ; 8  x ; x  8

4.

3  u  5 ;  3  5  u  5  5 ; 2  u  0 ; 2  u ; u  2

5.

2.8  x  3.7 ; 2.8  2.8  x  3.7  2.8 ; 0  x  6.5 ; x  6.5

6.

x

 2  8  3  3 16  3 3 19  3 22 3 3 3 3 3 3 19 3 2 ; x  2  ; x0  ; x  ; x ; x ; x ; x  2.75 8 8 8 8 8 8 8 8 8 8 8 8 8 8

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Linear Equations and Inequalities

Solutions

1  3  2  4.9 2 5 2 3 2  4.9 ; x   4.9 ; x  1.67  4.9 ; x  3.23 ; 4.9  4.9  x  1  4.9 ; 0  x  ; x 3 3 3 3 3

7.

4.9  x  1

8.

2  5  3  2  3  1 1 1 3 1 3 1 10  3 6  1 1 3 u  2  2 ; u  2  2  2  2 ; u  0  2  2 ; u    ; u 5 3 3 5 3 3 5 3 5 3 5 3 ; u

9.

6

2 4 4 4  6  3  2  2  5  4 18  2 10  4 20 14 2 4   y0 ;  y2 ; 6 2  y2 2 ;   y0 ;  y 3 5 3 5 3 5 5 5 3 5 3 5

20  5  14  3  y

;

10.

 13  3   7  5 13 7 39  35 74  ; u ; u ; u ; u  4.93 53 5 3 15 15

35

y  2.38  3

;

100  42 142 y ;  y ; 9.46  y ; y  9.46 15 15

3  5  2  2.38 15  2 17 2 2  2.38 ; y    2.38 ; y  2.38  2.38  3  2.38 ; y  0   ; y 5 5 5 5 5

; y  34 .  2.38 ; y  1.02

Section 1.2 Case II Solutions - Multiplication and Division of Linear Equations 2 1 2 2 1 2 1 ; 3 y     ; y   ; y 3 3 3 3 3 3 9

1.

3y  

2.



3.

3 1 1 3 0187 . h 3 1 3 .  h ; .  h ; h  0.187  2h ;   2 h  ;  h ;  h ; 0187  ; 0187 8 8 2 2 8  2 16 1 1

x 1  3  2 x 5 x 5 10 x 3 2  x 3.33 1 2 x 1 ;    ;   ;   ;   2   2 ;  x  ;  x  3.33 ; ; x  3.33 2 3 2 3 2 3 2 3 3 1 1 2 3

1. Note that in cases where the variable, in this case h , is in the right hand side of the equation we can solve for h without isolating the variable to the left hand side by applying the multiplication or division rules. However, in the very last ste p we should move the variable h to the left hand side of the equation and the solution to the right hand side of the equation. 2. Another method of solving for h is by applying the addition or subtraction rules as shown below: 3 3 3 3 3 3 1 3 1 3 3  2h ;  2h  2h  2h ;  2 h  0 ;   2h  0  ; 0  2h   ; 2 h   ; 2 h      8 8 8 8 8 8 2 8 2 8 8 ; h

3 1 3 ; h ; h  0.187 82 16

4.

x x  2 ;  8  2  8 ; x  16 8 8

5.

 x  35 ;

6.

7. 8.

9.

 x 35  ; x  35 1 1

4 1  2  1 16  1  3  8  2  1 17 3 3 4 1 1  2  8  1 8 17 3 8 u 2 u  1 ; u u ;  u  ; ; ; u ; u 8 2   8 2  17 8 2 8 2 8 2 17 2 17 1  17 12 ; u ; u  0.71 17 1  5  4 9  w 18 . 4 5 4 . ; w  1 ; w   ; w  ;  w  ;  w  18 ; w  1.8 5 5 5 1 1 5  y 24 1 1 y  12 ;  y  2  12  2 ;  y  24 ;  ; y  24 2 2 1 1 14 14  140  2 .8 x 14 . 14  10 1 14 10 2.8 x  1.4 ;  ; x ; x ; x ; x ; x   ; x  0.5 28   2 .8 2.8 10  28 280 2 28 2 28 10 

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Linear Equations and Inequalities

Solutions

 2  5  3 x  4.3 10  3  13 4.3 13 5 4.3 5 3 4.3  5 x  4.3 ;  x    10. 2 x  4.3 ;  ;  ;  x   ; x 5    5 5 5 1 5 13 1 13 1  13 ; x

215 . ; x  1.654 13

Section 1.2 Case III Solutions - Mixed Operations Involving Linear Equations 1.

3x  20  5x  8 ; 3x  20  20  5 x  8  20 ; 3x  0  5 x  12 ; 3x  5 x  12 ; 3x  5 x  5 x  5 x  12 ; 2 x  0  12

; 2 x  12 ; 2.

2 x 12 12  ; x ; x  6 2 2 2

6 y  2  3  10 y ; 6 y  10 y  2  3  10 y  10 y ; 16 y  2  3  0 ; 16 y  2  3 ; 16 y  2  2  3  2

; 16 y  0  5 ; 16 y  5 ;

y 16 5 5  ; y ; y  0.313  16 16 16

3.

x x x x x  35 ;  3353 ; 02 ;  2 ; 2   2  2 ; x  4 2 2 2 2 2

4.

5x  3  15 ; 5x  3  3  15  3 ; 5x  0  12 ; 5x  12 ;

5.

y y y y y  4  3 ;  4  4  3  4 ;  0  7 ;  7 ;  4  7  4 ; y  28 4 4 4 4 4

6.

5

7.

25  3 y  2 y ; 25  3 y  2 y  2 y  2 y ; 25  5 y  0 ; 25  25  5 y  0  25 ; 0  5 y  25 ; 5 y  25 ;

w w w w w  10 ; 5  5   10  5 ; 0   5 ;  5 ; 2   5  2 ; w  10 2 2 2 2 2

; y 8.

5 y 25  5 5

25 ; y5 5

10 y  2  8 y ; 10 y  2  2  8 y  2 ; 10 y  0  8 y  2 ; 10 y  8 y  2 ; 10 y  8 y  8 y  8 y  2 ; 2 y  0  2 ; 2 y  2

; 9.

5 x 12 12  ; x ; x  2.4 5 5 5

2 y 2 2  ; y   ; y  1 2 2 2 2 2 2 3 2 3 3 7 3 7 2 21 x  5  12 ; x  5  5  12  5 ; x  0  12  5 ; x  7 ;  x   7 ; x   ; x  ; x ; x  10.5 3 3 3 3 2 3 2 2 1 2 1 2

10. m 

1 2 1 2 1 2 1 2 1 1 2 1  4m  ; m  4m   4m  4m  ; 3m   0  ; 3m    ; 3m      2 3 2 3 2 3 2 3 2 2 3 2

; 3m  0  

 2  2   1  3 2 1 4  3 7 1 7 1 7 1 7  ; 3m  ; 3m  ; 3m  ;   3m     ; m  ; m 3 2 3 2 6 6 3 6 3 63 18

; m  0.388

Section 1.3 Case I Solutions - Solving Linear Equations Containing Parentheses and Brackets x 6  ; x  6 1 1

1.

x   2x  3  3 ; x  2 x  3  3 ;  x  3  3 ;  x  3  3  3  3 ;  x  0  6 ;  x  6 ;

2.

2  3 x  1  3   x  5 ; 2  3x  3  3  x  5 ;  2  3  3x   3  5  x ; 1  3x  8  x ; 1  1  3x  8  1  x

; 0  3x  7  x ; 3x  7  x ; 3x  x  7  x  x ; 4 x  7  0 ; 4 x  7 ; 3.

4 x 7  ; x  1.75 4 4

2  3 x  1  5x  0 ; 2  3x  3  5x  0 ;  2  3   3x  5x   0 ; 5  2 x  0 ; 5  5  2 x  0  5 ; 0  2x  5

Hamilton Education Guides

69

Linear Equations and Inequalities

; 2 x  5 ; 4.

Solutions

2 x 5 5  ; x   ; x  2.5 2 2 2

4  x  1  3x  2 x  1 ; 4 x  4  3x  2 x  2 ; 4x  3x   4  2 x  2 ; x  4  2 x  2 ; x  2 x  4  2 x  2 x  2

;  x  4  0  2 ;  x  4  2 ;  x  4  4  2  4 ;  x  0  2 ;  x  2 ; 5.





2 5   x  2   x  3  0 ; 25  x  2  x  3  0 ; 2 7  x   x  3  0 ; 14  2 x  x  3  0 ; 14  3   2 x  x   0

; 17  3x  0 ; 17  17  3x  0  17 ; 0  3x  17 ; 3x  17 ; 6.

 x  5  3 x  1  2  2 ;  x  5  3x  3  2  2



3x 17 17  ; x ; x  5 .67 3 3 3

; x  5  3x  1  2 ; x  5  3x  1  2 ;  x  3x    5  1  2

; 2 x  4  2 ;  2 x  4  4  2  4 ;  2 x  0  6 ; 2 x  6 ; 7.

x 2  ; x  2 1 1

2 x 6 6  ; x   ; x  3 2 2 2



3    x  1  2  3x  5 ; 3    x  1  2  3x  5 ; 3    x  3  3x  5 ; 3  x  3  3x  5 ; 3  3  x  3x  5

; 0  x  3 x  5 ; x  3x  5 ; x  3x  3 x  3x  5 ;  2 x  0  5 ;  2 x   5 ; 8.









 5  x   3  4 x   8 ; 5  x  3  4x   8 ;  5  3    x  4 x   8 ; 8  5x   8 ; 8  5 x  8

; 8  8  5 x  8  8 ; 0  5 x  16 ; 5 x  16 ; 9.

2 x 5 5  ; x  ; x  2.5 2 2 2

5 x 16  ; x  3.2 5 5

3  2 x  5  4x  3  3x ; 3  2 x  5  4 x  3  3x ; 3  5  2 x  4x   3  3x ; 8  2 x  3  3x ; 8  8  2 x  3  8  3x

; 0  2 x  5  3x ; 2 x  5  3x ; 2 x  3x  5  3x  3x ; x  5  0 ; x  5 10. 6 x  2  2 x  1  3 x  2 ; 6 x  12  2 x  2  3x  6 ; 6 x  2 x    12  2  3x  6 ; 4 x  14  3x  6 ; 4 x  3x  14  3x  3x  6 ; x  14  0  6 ; x  14  6 ; x  14  14  6  14 ; x  0  20 ; x  20

Section 1.3 Case II Solutions - Solving Linear Equations Containing Integer Fractions

 y  5  2  y   5 5 y  2 y y y 3y 5 3y 1 1 1 1 1 1 y y5 ; y y y y5 ;  05 ; 5 ; 5 ;  ; 25 2 5 2 5 5 5 2 5 10 10 10 1

1.

; 3y  1  5  10 ; 3y  50 ;

3 y 50 50  ; y ; y  1667 . 3 3 3

A second way to solve this problem is as follows:  1  5   2  1  y y 1 1 1 1 1 1  1 1  5  2  y  5 ;  y  y5 ; y  y  y  y 5 ;   0  5 ;    y  5 ;  y5  2 5  10  2 5 2 5 5 5 2 5 25  

;

3 y  1  5  10  0 3 y  50 3y 5 3 y  50 0 3y 3 y5 ; 5  55 ;  0 ; 0 ;  ; 3 y  50  1  10  0 ; 10  1 10 10 10 1 10 10 1

; 3y  50  0 ; 3y  50  50  0  50 ; 3y  0  50 ; 3y  50 ;

2.

x  3

3.

y

3 y 50 50  ; y ; y  1667 . 3 3 3

 x  2  1  x   3 2 x  x x x x x x 3x 2 3 2 x 3  3 0 ; 3 ;  3 ; x  3 ;  x  3  ; x  2 ; x   3  ; ; 1 2 2 2 2 2 1 2 2 2 2 3 2 3

2 2  y  1 ; 1  3 3 

2 1 3  2  1 ;   y   1 3 3

Hamilton Education Guides

5 5 3  2 5 5 2 3  2  1  3  1  2   y  ;  ;  y  y ; y    3 3 3 3 3 3 3 3  

70

Linear Equations and Inequalities

;

Solutions

3 5 5 3  y   ; y 1 5 3 3 5

4.

 u  3  1  u  6 3u  u  u u 2u 6 2 u 18 1 9 u u  6 ;  6 ; 6 ;  ; 2u  1  6  3 ; 2u  18 ;  ; ; u ; u9 1 3 3 3 3 1 2 2 1 3 1

5.

s 1

1  3  2  2  5  3 s 3  2 10  3 5 13 5 5 13 5 13 5 2 3 2 s ; s  s ; s  s ; s   s ; s0 s ; s 3 5 3 5 3 5 3 5 3 3 5 3 5 3

; s

5 5s  13s 13 5 13 13 13 5 13 5 s 13s 5 5  s  13s  1 5  ; s ; s s s s ; s s0 ;   ;  1 5 3 5 3 5 5 5 3 5 3 1 5 3 5 3

;

 s 8s 5 24 25 25   ; 8 s  3  5  5 ; 24 s  25 ;  ; s ; s  1.04  5 3 24 24 24

w 1  3  2 w 3 2 w 5w 4 w 4 w 2 w  5w  w4 ; 1 w  4 ;  w4 ;  4 ; 4 ;  ; 4 w  1  3  4 3 3 3 3 3 3 3 3 3 3 1

6.

; 4 w  12 ;

7.

x

 12 4 w 12 3  ; w ; w   ; w  3 4 4 4 1

2 1  4  1 2 5 5 2 5 5 2 1 2 4 1 5 2 x ; x  1 x ; x   x ; x  x ; x x  x x ; x x 0 3 4 3 4 3 4 3 4 4 3 4 4 4 3

; x

5 2 2 2 5 2 5 2  x   0 ; x  x 0 ; x x  ; 1  4 3 3 3 4 3 4 3 

5 2 1 x  ;   4 3 1

2 5 2  1  4  1  5   x   x  ;  1 4 3 4 3  

8 2 x 2 1 2 2  4  5 ;   x  ;  x  ;   4   4 ; x   ; x  2 ; x  2.67  4  4 3 4 3 3 3 3

8.

t  5

;

9.

2 2 2 2 2  t ; t  t  5  t  t ; t  t  5  0 ; 1   3 3 3 3 3

2 1 t  5 ;    1 3

 1  3  1  2  3 2 2  t  5 ; t 5  t  5 ;   1  3 3 3  

5 3 3 5 3 t  5 ;  t  5  ; t  ; t  3 3 5 3 5 1

2  5  3 z  1  4  1 12  2 10  3 4  1 14 13 5  14 13 5 2 3 1  4  3  2 z 4 z 2 z 1 ; z z z z  ;   z  ; ; 3 5 4 3 5 4 3 5 4 3 5 4  3 5 4  14  5  13  3   5   31 5 15 5 15 5  15 75 70  39  5 31  z  ;  z ;  z  ;  ; z ; z ; z  0.605 z  ;   15  3 5 4  15  4 31 4  31 4 31 124 4 15  

10. 6 

6

1  5  2 t 1 1 7 1 52 1 2 1 7 1   1 7 t  t 1 t ; 6  t  t  t ; 6  t  t  t ; 6  t  1   t ; 6  t     t ; 6 t  t   1 5  5 2 5 2 5 2 5 2 5 2 2  1  5  1  7  1 1 2 1 2 2 2 1 2 1 5  7  t ; 6  t   t   t ; 6  t   t ; 6  t  t   t  t ; 6  t  t  0   2 1  5 2 5 2 5 5 5 2 5 2 5  

; 66

;

 1  5   2  2  1 2 1 2 1 2 5  4  1 2  t  6 ;  t  t  0  6 ; 0  t  t  6 ; t  t  6 ;    t  6 ;   t  6  2 5  2  5 2 5 2 5 2 5 10   

  9 9 10 10 6  10 60 6 10 t  6 ;  t  6  ; t  ; t ; t ; t  6.67  10 9 10 9 1 9 1 9 9

A second way to solve this problem is as follows: 6

1  5  2 t 1 1 7 1 52 1 2 1 7  t  t 1 t ; 6  t  t  t ; 6  t  t  t ; 6  t  1   t ; 6  0.5t  1  14 . t ; 6 t  t  2 5  5 2 5 2 5 2 5 2

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Solutions

8.

8 4 8 1 2 x 0.8 4 10  ; 2 x  0.8  0 ; 2 x  0.8 ; ; x ; x ; x ; x  0.4 2 10  2 2 2 10 1 .  0.5 x  01 .  0 ; 0.5x  135 .  0 ; 0.5 x  125 .  125 .  0  125 . 135 .  0.5 x  0.2  0 ; 135 .  01 .   0 ; 0.5 x  125

9.

125 25    10  125 25 0 .5 x 125 . 125 . 100 . ; 0.5 x  125 . ;  ; 0.5 x  0  125 ; x ; x ; x ; x ; x  2.5 5       100 5 0.5 0.5 0.5 10 10 10 0.5  0.8 x  0.2  5  2.2 x ; 0.5 0.8 x  0.2  5  2.2 x ; 0.5 0.8 x  4.8  2.2 x ; 0.4 x  2.4  2.2 x





; 0.4 x  2.2 x  2.4  2.2 x  2.2 x ; 2.6 x  2.4  0 ; 2.6 x  2.4  2.4  0  2.4 ; 2.6 x  0  2.4 ; 2.6 x  2.4 24 12   10  24 2 .6 x 2.4 12 10  ; ; x ; x ; x ; x  0.923 26    26  2 .6 2.6 10 13 13 10 .  12 . x  17 .  2.8 ; 0.25x  12 . x  3  2.8 10. 0.25x  13 .  12 . x  17 .   2.8 ; 0.25 x  13 . x    17 .  13 .   2.8 ; 145

2 10  .   x 0.2 2  100 145 20 10 . x  3  3  2.8  3 ; 145 . x  0  0.2 ; 145 . x  0.2 ;  ; 145 ; x ; x ; x ; x  0.138 145    145 .   145 10 145 . 145 100

Section 1.4 Solutions - Formulas 1.

I. V  r 2 h ; V  hr 2 ;

II. V  r 2 h ; V  r 2 h ; 2.

 2 V V V  hr  r2 ; r2  ; ;  h h h  h

V

r

2



r2  

V

h

; r

V

h

V V  r 2 h h ; h 2 ; r 2 r  r 2

I. 2 x  2 y  3 x  y   5 ; 2x  2 y  3x  3y  5 ; 2x  3x  2 y  3x  3x   3 y  5 ;  x  2 y  0  3y  5 ;  x  2 y  3 y  5 ;  x  2 y  2 y   3 y  2 y  5 ;  x  0  y  5 ;  x  y  5 ;

y5 x y  5  ; x ; x  y  5 1 1 1

II. 2 x  2 y  3 x  y   5 ; 2 x  2 x   2 y  3x  2 x   3 y  5 ; 0  2 y  x  3 y  5 ; 2 y  x  3 y  5 ; 2 y  3 y  x  3 y  3 y   5 ;  y  x  0  5 ;  y  x  5 ;

y x 5 x5  ; y ; y  x  5 1 1 1

C C 2  r C  r ; r ; 2 2  2 2

3.

C  2r ;

4.

I. d  rt ;

5.

I. y  b 

d d d rt   ; t ; t  r r r r

II. d  rt ;

d d rt d  ; r ; r  t t t t

2 1 2 2 2 1 1 1 2 1 x  b ; y  b  b  x  b  b ; y  b  b  x  0 ; y  b  0.67b  x ; y  1.67b  x 3 3 3 3 3 3 3 3 3 3

; 3   y  1.67b  3  II. y  b 

1 x ; 3y  5.01b  x ; x  3 y  5.01b 3

2 1 2 2 2 1 1 1 2 1 x  b ; y  b  b  x  b  b ; y  b  b  x  0 ; y  b  0.67b  x ; y  1.67b  x 3 3 3 3 3 3 3 3 3 3

; y  y  1.67b 

0.33 x  y   b 0.33x  y 1 1 1.67 x  y ; 0  1.67b  x  y ; 167  ; b . b  0.33x  y ;  3 3 1.67 1.67 1.67

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Linear Equations and Inequalities

6.

y

Solutions

y abc abc  ; ; 3  y  1  a  b  c ; 3y  a  b  c ; 3y  a   a  a   b  c ; 3y  a  0  b  c 3 1 3

; 3y  a  b  c ; 3y  a  b   b  b  c ; 3y  a  b  0  c ; 3y  a  b  c ;

3 y  a  b c  1 1

; 3y  a  b  c ; 3y  a  b  c ; c  3 y  a  b 7.

I. m  II. m  ;

8.

yb m yb  ; ; m  x   y  b 1 ; mx  y  b ; mx  b  y  b  b ; mx  b  y ; y  mx  b x 1 x yb m yb  ; ; m  x   y  b 1 ; mx  y  b ; mx  y  y  y  b ; mx  y  0  b ; mx  y  b x 1 x

mx  y  b mx  y b   ; mx  y  b ; b  mx  y ;  1 1 1 1

3V 3V 3V r 2 h 1 r 2 h V r 2 h  I. V  r 2 h ; V  ; ; 3  V  1  r 2 h ; 3V  r 2 h ; ; 2  ;   2  2 2 1 3 3 3 r h r h r h r h









 2 3V 1 3V  hr r 2 h V r 2 h   r2 II. V  r 2 h ; V  ; ; 3  V  1  r 2 h ; 3V  r 2 h ; 3V  hr 2 ; ;  1 3 h 3 h  h 3 ; r2 

3V ; h

r2  

3V 3V ; r h h

3V 3V 1 3V  r 2 h r 2 h V r 2 h  h ; h 2 III. V  r 2 h ; V  ; ; 3  V  1  r 2 h ; 3V  r 2 h ; ;  2  2 2 1 3 3 3 r r r  r



9.

I. E  mc 2 ; II. E  mc 2 ;

2 E E E mc   c2 ; c2  ; ;  m m m m 

E c

2



mc 2 c

2

;

E c

2

m ; m

c2  



E E ; c m m

E c2

10. I. y  2 x  3 y   3  5 y  x ; y  2x  3y  3  5 y  x ;  y  3 y   2 x  3  5 y  x ; 4 y  2x  3  5 y  x ; 4 y  4 y   2 x  3  5 y  4 y  x ; 0  2x  3  y  x ; 2x  3  y  x ; 2x  3  3  y  x  3 ; 2x  0  y  x  3 ; 2x  y  x  3 ; 2x  x  y  x  x  3 ;  x  y  3 ;

x y  3  ; x   y  3 ; x  3  y 1 1

II. y  2 x  3 y   3  5 y  x ; y  2x  3y  3  5 y  x ;  y  3 y   2 x  3  5 y  x ; 4 y  2x  3  5 y  x ; 4 y  5 y   2 x  3  5 y  5 y  x ;  y  2x  3  0  x ;  y  2x  3   x ;  y   2 x  2 x   3   x  2 x ; y  0  3  x ; y  3 x ; y  3 3 x  3 ; y  0  x  3 ; y  x  3 ;

y x 3 x3  ; y ; y  3 x 1 1 1

Section 1.5 Case I Solutions - Addition and Subtraction of Linear Inequalities 1.

x  10  12 ; x  10  10  12  10 ; x  0  12  10 ; x  22

2.

3   u  8 ; 3  u   u  u  8 ; 3  u  0  8 ; 3  u  8 ; 3  3  u  8  3 ; 0  u  11 ; u  11

3.

8  x 5 ; 8 x   x  x 5 ; 8 x  05 ; 8 x 5 ; 88 x 58 ; 0 x  3 ; x   3

4.

.  32 .  w   2.8  32 . ; 0 w6 ; w  6 32 .  w   2.8 ; 32

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Linear Equations and Inequalities

Solutions

2  3  2  0.65 2 62 8  0.65 ; t   0.65 ; t  2.67  0.65 ; t  2.02 ; 0.65  0.65  t  ; 0t  3 3 3 3

5.

0.65  t  2

6.

s

7.

1  3  1  0.9  0.8 31 4 1 0.8  w  1  0.9 ; 0.8  0.8  w   01 . ; w   01 . ; w  133 ; 0 w .  01 . ; w  1.43 3 3 3 3

8.

1

1  7  2   h  2  8  3 72 16  3 9 19 9 19 2 3  h2 ;  h ;  ;  h ;   h h h 7 8 7 8 7 8 7 8 7 8

; 

19  7  9  8 9 19 9 9 19 9 133  72 205 9 19 h 0  ; h ;   h ;    h ; h ; h ; h  3.66 87 7 8 7 8 7 7 8 7 56 56

9.

3 3 1  5  3 3 53 3 8 3 83 3 3 11  ; s0 1 ; s     ; s  ; s ; s ; s  2.2 5 5 5 5 5 5 5 5 5 5 5 5

y  1.25   2

2  4  3  1.25 83 11 3  1.25 ; y    1.25 ; y   2.75  125 ; y  1.25  1.25   ; y 0 . 4 4 4 4

; y   1.5

1  3  2   2  5  2  2 5 12  7   2  5 2 2 2 5 12 2 3  2 10  2 2   ; x   ; x  10. x  1  2  ; x  ; x 3 5 7 3 5 7 3 5 7 3 5 7 3 5 7 ; x

 74  3  5  35 5 84  10 5 74 222  175 47 5 5 74 5   ; x0 ; x  ; x   ; x ; x ; x  0.45 35  3 3 35 3 35 3 3 35 3 105 105 Section 1.5 Case II Solutions - Multiplication and Division of Linear Inequalities 2 1 2 2 1 2 1 ;  4 y    ; y   ; y ; y   0.167 3 4 12 3 4 3 4

1.

4y  

2.



3.

2 2 2 2 2 2 1 2  1 2 2 1 2   2h ;  2h   2h  2h ;  2h  0 ;   2h  0  ; 0  2h   ; 2h   ;  2 h    ; h   5 5 5 5 5 5 5 5 2 5 2 5  2

1  3  2 2 3 2 5 3 5  3 5 2 2 2 3 2 2 5 x 1 ;  x  ;  x ;  x  ;   x   ; x  ; x   ; x   2.5 3 3     3 3 3 3 3 3 2 3 3 2 3 2 2

; h

1 5

Another way of solving the problem is:

2 1 2 h 1 2 2 1   2h ;    h ;   h or h   ;   5 5 2 2 10 5 5

4.

w w w   3 ; 7    3  7 ;   21 ; w   21 7 7 1

5.

w w w   5 ; 2    5  2 ;   10 ; w  10 2 2 1

6.

1  5  1 8  1 6 4 9 64 24 1 1  2  4  1 5 1 9 6 4 u 2 u  1 ; u ; ; u ;  u   ; u ; u ; u   0.53 4 5 4 5 4 5 4 5 9 4 5 9 59 45

7.

2 x  2

2  4  3  1 11  1  1  4 83 1 15 3 11 1 11  4  1 ; 2 x   ; 2 x   ; 2 x  ; 2x  ; 2 x  ; 2 x  4 1 4 1 4 4 1 4 1 4 4

15 1 15  1 1 15 1 15  2 x   ; x   ; x   ; x ; x  1.875 2 4 2 4 2 42 8 24  x 3 .28 2.4 2.4 24  100 2400 10  3.28x  2.4 ; ; x ; x ; x ; x ; x  0.732 328  3 .28 3.28 3.28 10  328 3280 100 1 1  y   2 ;   4 y   2  4 ;  y   8 ; y  8 4 4

; 

8.

9.

Hamilton Education Guides

74

Linear Equations and Inequalities

Solutions

5  3  1   2  4  3 x 15  1 8  3 16 11 16 11 1 3 11 11 16 11  x ;  x ;  x x x ;  x0 10. 5   2 x ; ; 3 4 3 4 3 4 3 4 3 4 4 4 3 4  16 4 11 16 4 16  4 64 16 16 11 16 11 16 11   x  0 x  x  ; 0 x   ; ; ; x ; x ; x  1.94   4 3 3 4 3 4 3 4 3 11 3 11 3  11 33

;

Section 1.5 Case III Solutions - Mixed Operations Involving Linear Inequalities 1.

2x  9  9x  20 ; 2x  9x  9  9x  9x  20 ; 11x  9  0  20 ; 11x  9   20 ; 11x  9  9   20  9 ; 11x  0   11 ; 11x   11 ;

2.

   x  11 11  ; x 1   11 11

15x  3  20x ; 15x  20x  3  20x  20x ; 5x  3  0 ; 5x  3  3  0  3 ; 5x  0   3 ; 5x   3 ; ; x

5 x 3  5 5

3 ; x  0.6 5 4 x 5 1 5  ; x   ; x   1 ; x  1.25 4 4 4 4

3.

4x  5  10 ; 4x  5  5  10  5 ; 4x  0  5 ; 4 x  5 ;

4.

12t  4  4t  8 ; 12t  4t  4  4t  4t  8 ; 16t  4  0  8 ; 16t  4   8 ; 16t  4  4   8  4 ; 16t  0   12 ; 16t   12 ;

5.

  t 12 3 16 12  ; t ; t  ; t  0.75   16 16 16 4

4w  5  8w  17 ; 4w  8w  5  8w  8w  17 ; 12w  5  0  17 ; 12w  5  17 ; 12w  5  5  17  5 ; 12w  0  22 ; 12w  22 ;

6.

10 y  4  4 y  12 ; 10 y  4 y  4  4 y  4 y  12 ; 6 y  4  0  12 ; 6 y  4   12 ; 6 y  4  4   12  4 ; 6 y  0   8 ; 6y   8 ;

;

y 109 y y 24  85 y   7.27 ; 3   3  7.27 ; y  21.81 ;  ; 3 15 3 15 3 3

3

2  3  1 15  4 6  1 19 7 19 7 4 1 3  5  4 7 7 19 7 19 7 t t2 ; t t ;  t  ;  t0 ;  t ; ; 5 3 5 3 5 3 5 3 3 3 5 3 5 3 5 3

;

9.

6 y 8 4 1   ; y   ; y   1 ; y   1.33 6 6 3 3

8  3  17  5 y 2 3 y 5  3  2 1  5  3 y 15  2 5  3 y 17 8 y 17 17 8 17 y  0 5 1 ;      ;     ; ; ; 3 5 3 53 3 3 5 3 3 3 5 3 3 5 3 3 3 5 3

7.

8.

 w 12 22 22 5 11  ; w ; w ; w   1 ; w   183 .  12 12 12 6 6

19  3   7  5  t 53

;

57  35 22 7 t ;  t ; 1  t ; 147 .  t or t  1.47 . A second way to solve this problem is as follows: 15 15 15

3

2  3  1 15  4 4 1 4 1 4 1 3  5  4 6 1 4 1 t t2 ; 3 tt t 2 ; 3 t 02 ; 3 t2 ; t ; 5 3 5 3 5 3 5 3 5 3 5 3

;

 7  5  19  3 19 7 19 19 7 19 35  57 7 19 22 22 t ;  t   ; 0t  ; t  ; t  ; t   ; t ; t  1.47 35 5 5 3 5 3 5 15 15 15 5 3

2  5  3 10  3 13 3 .  w   2.6 ; 3.4  w  w  w  ; 3.4  w   0  ; 3.4  w   ; 34 5 5 5 5  w 0.8  .  34 .  w   2.6  34 . ; 0  w  0.8 ; w  08 . ; ; 34 ; w   0.8 1 1 3.4  w  2

. x  0.35 ; 0.48x  15 . x  25 .  15 . x  15 . x  035 . ; 102 . x  25 .  0  035 . ; 102 . x  25 .   035 . 10. 0.48x  2.5  15

. x  25 .  035 .   035 .  035 . ; 102 . x  285 .  0 ; 102 . x  2.85  2.85  0  285 . ; 102 . x  0   285 . ; 102 285  .   2.85 285  100 285 81 102 2.85 100 x . x   2.85 ; ; 102 ; x ; x ; x ; x ; x2 ; x  2.794 102 .       102 102 102 . 100 102 102 . 102 100

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About the Author Dan Hamilton received his B.S. degree in Electrical Engineering from Oklahoma State University and Master's degree, also in Electrical Engineering from the University of Texas at Austin. He has taught a number of math and engineering courses as a visiting lecturer at the University of Oklahoma, Department of Mathematics, and as a faculty member at Rose State College, Department of Engineering Technology, at Midwest City, Oklahoma. He is currently working in the field of ae rospace technology and has published several books and numerous technical papers.

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