Solutions for College Algebra 6th edition by Mark Dugopolski, All Chapters 1-8

College Algebra – 6th Edition


SOLUTIONS
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MANUAL
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Mark Dugopolski
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Complete Solutions Manual for Instructors and
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Students
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© Mark Dugopolski
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All rights reserved. Reproduction or distribution without permission is prohibited.
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, Table of Contents
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Chapter P...……………………………………………………………………………………1

Chapter 1……………………………………………………………………………………..34

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

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 − .
7. False, since |x| = −8 has no solution. 2 2
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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
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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
√
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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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5
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3 The solution set is − .
14. Since −2x = −3, the solution set is . 8
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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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, 1.1 Linear, Rational, and Absolute Value Equations 35



28. Subtract 5a from both sides of 5a = 6a to 40. Multiply by 60x.
get 0 = a. The latter equation is conditional
whose solution set is {0}. 12 − 15 + 20 = −17x
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17 = −17x
29. Note, 5x = 4x is equivalent to x = 0. The
latter equation is conditional whose solution A conditional equation with solution set {−1}.
set is {0}.
41. Multiply by 3(z − 3).
30. Note, 4(y − 1) = 4y − 4 is true by the
3(z + 2) = −5(z − 3)
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distributive law. The equation is an identity
and the solution set is R. 3z + 6 = −5z + 15
8z = 9
31. Equivalently, we get 2x + 6 = 3x − 3 or 9 =  
x. The latter equation is conditional whose 9
A conditional equation with solution set .
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solution set is {9}. 8

32. Equivalently, we obtain 2x + 2 = 3x + 2 or 42. Multiply by (x − 4).
0 = x. The latter equation is conditional 2x − 3 = 5
whose solution set is {0}.
2x = 8
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33. Using the distributive property, we find x = 4
3x − 18 = 3x + 18 Since division by zero is not allowed, x = 4
−18 = 18. does not satisfy the original equation. We have
an inconsistent equation and so the solution
The equation is inconsistent and the solution
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set is ∅.
set is ∅.
43. Multiplying by (x − 3)(x + 3).
34. Since 5x = 5x + 1 or 0 = 1, the equation is
inconsistent and the solution set is ∅. (x + 3) − (x − 3) = 6
35. An identity and the solution set is {x|x 6= 0}. 6 = 6
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36. An identity and the solution set is {x|x 6= −2}. An identity with solution set
{x|x 6= 3, x 6= −3}.
37. Multiplying by 2(w − 1), we get
44. Multiply by (x + 1)(x − 1).
1 1 1
− =
w − 1 2w − 2 2w − 2 4(x + 1) − 9(x − 1) = 3
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2 − 1 = 1. 4x + 4 − 9x + 9 = 3
An identity and the solution set is {w|w 6= 1} −5x = −10
38. Multiply by x(x − 3). A conditional equation with solution set {2}.
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(x − 3) + x = 9 45. Multiply by (y − 3).
2x = 12
4(y − 3) + 6 = 2y
A conditional equation with solution set {6}. 4y − 6 = 2y
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39. Multiply by 6x. y =3

6−2 = 3+1 Since division by zero is not allowed, y = 3
does not satisfy the original equation. We have
4 = 4
an inconsistent equation and so the solution
An identity with solution set {x|x 6= 0}. set is ∅.

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