Instructor's Solution Manual for College Algebra 6th Edition by Mark Dugopolski, All Chapters Included

INSTRUCTOR’S
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SOLUTIONS MANUAL
EDGAR REYES
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Southeastern Louisiana University
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C OLLEGE A LGEBRA
SIXTH EDITION
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Mark Dugopolski
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Southeastern Louisiana University
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Reproduced by Pearson from electronic files supplied by the author.

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ISBN-13: 978-0-321-91666-2
ISBN-10: 0-321-91666-2
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www.pearsonhighered.com

, Table of Contents
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Chapter P...……………………………………………………………………………………1
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Chapter 1……………………………………………………………………………………..34

Chapter 2…………………………………………………………………………………….121

Chapter 3…………………………………………………………………………………….179

Chapter 4…………………………………………………………………………………….255
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Chapter 5…………………………………………………………………………………….306

Chapter 6…………………………………………………………………………………….388

Chapter 7…………………………………………………………………………………….458
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Chapter 8…………………………………………………………………………………….501
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, 34 Chapter 1 Equations, Inequalities, and Modeling


 
For Thought 1
17. Since 14x = 7, the solution set is .
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1. True, since 5(1) = 6 − 1.
18. Since −2x = 2, the solution set is {−1}.
2. True, since x = 3 is the solution to both
equations. 19. Since 7 + 3x = 4x − 4, the solution set is {11}.
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3. False, −2 20. Since −3x + 15 = 4 − 2x, the solution set
√ is not a solution of the first equation is {11}.
since −2 is not a real number.
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4. True, since x − x = 0. 21. Since x = − · 18, the solution set is {−24}.
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5. False, x = 0 is the solution. 6. True 3

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22. Since x = · (−9), the solution set is − .
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7. False, since |x| = −8 has no solution. 2 2
x 23. Multiplying by 6 we get
8. False, is undefined at x = 5.
x−5
3x − 30 = −72 − 4x
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9. False, since we should multiply by − . 7x = −42.
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10. False, 0 · x + 1 = 0 has no solution. The solution set is {−6}.

24. Multiplying by 4 we obtain
1.1 Exercises
x − 12 = 2x + 12
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1. equation −24 = x.
2. linear
The solution set is {−24}.
3. equivalent
25. Multiply both sides of the equation by 12.
4. solution set
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18x + 4 = 3x − 2
5. identity
15x = −6
6. inconsistent equation 2
x = − .
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7. conditional equation  
2
8. extraneous root The solution set is − .
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9. No, since 2(3) − 4 = 2 6= 9. 10. Yes 26. Multiply both sides of the equation by 30.
11. Yes, since (−4)2 = 16.
15x + 6x = 5x − 10
√
12. No, since 16 6= −4. 16x = −10
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x = − .
 
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13. Since 3x = 5, the solution set is . 8
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3 5
14. Since −2x = −3, the solution set is . The solution set is − .
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15. Since −3x = 6, the solution set is {−2}. 27. Note, 3(x − 6) = 3x − 18 is true by the
distributive law. It is an identity and the
16. Since 5x = −10, the solution set is {−2}. solution set is R.

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