Solution for College Algebra, 8th edition by Robert F. Blitzer, Chapter 1-8

INSTRUCTOR’S SOLUTIONS MANUAL
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COLLEGE A LGEBRA
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EIGHTH EDITION
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Robert Blitzer
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All Answers Included
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, Chapter P
Fundamental Concepts of Algebra
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Section P.1 6. a. 1− 2
Check Point Exercises Because  1.4, the number inside the
absolute value bars is negative. The absolute
1. 8 + 6(x − 3)2 = 8 + 6(13 − 3)2 value of x when x < 0 is –x. Thus,
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= 8 + 6(10)2 (
1− 2 = − 1− 2 = ) 2 −1
= 8 + 6(100)
= 8 + 600 b.  −3
= 608 Because   3.14, the number inside the
absolute value bars is positive. The absolute
2. a. Since 2016 is 16 years after 2000, substitute 16 value of a positive number is the number itself.
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for x. Thus,
T = −x2 + 361x + 3193  − 3 =  − 3.
= −(16)2 + 361(16) + 3193
= 8713 c.
x
The average cost of tuition and fees at public x
U.S. colleges for the school year ending in Because x  0, x = x.
2016 was $8713.
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x = x =1
b. The formula underestimates the actual answer Thus,
x x
by $65.
7. −4 − (5) = −9 = 9
3. The elements common to {3, 4, 5, 6, 7} and
The distance between –4 and 5 is 9.
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{3, 7, 8, 9} are 3 and 7.
{3, 4, 5, 6, 7} {3, 7,8, 9} = {3, 7}
8. 7(4x2 + 3x) + 2(5x2 + x)
4. The union is the set containing all the elements of = 7(4x2 + 3x) + 2(5x2 + x)
either set.
{3, 4, 5, 6, 7}{3, 7,8, 9} = {3, 4, 5, 6, 7,8, 9} = 28x2 + 21x +10x2 + 2x
= 38x2 + 23x
−9, −1.3, 0, 0.3,  ,
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
5. 9, 10 

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 9. 6 + 4[7 − (x − 2)]
= 6 + 4[7 − x + 2)]
a. Natural numbers: 9 because 9 =3 = 6 + 4[9 − x]
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= 6 + 36 − 4x
b. Whole numbers: 0, 9 = 42 − 4x

Integers: −9, 0,
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c. 9
Concept and Vocabulary Check P.1
d. Rational numbers: −9, −1.3, 0, 0.3, 9
C1. expression

e. Irrational numbers: , 10 C2. b to the nth power; base; exponent
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C3. formula; modeling; models

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f. Real numbers: −9, − 1.3, 0, 0.3, , 9, 10 C4. intersection; A  B
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C5. union; A  B




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, Chapter P Fundamental Concepts of Algebra


C6. natural 6 + 5(8 − 6) = 6 + 5(2)
3 3
10.
C7. whole = 6 + 5(8)
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C8. integers = 6 + 40 = 46

C9. rational 11. 82 − 3(8 − 2) = 64 − 3(6)
= 64 −18 = 46
C10. irrational
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12. 82 − 4 (8 − 3) = 64 − 4 (5) = 64 − 20 = 44
C11. rational; irrational

C12. absolute value; x, −x 5(x + 2) 5(10 + 2)
13. =
2x −14 2(10) −14
C13. b + a ; ba 5(12)
=
6
C14. a + (b + c) ; (ab)c
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= 52
C15. ab + ac = 10

C16. 0; inverse; 0; identity 7(x − 3) 7(9 − 3) 7(6)
14. = = = 7  3 = 21
2x −16 2(9) − 16 2
C17. inverse; 1; identity
2x + 3y
; x = −2, y = 4
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C18. simplified 15.
x +1
C19. a 2(−2) + 3 ( 4 ) −4 +12 8
= = = = −8
−2 +1 −1 −1
Exercise Set P.1
2x + y
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16. ; x = −2 and y = 4
1. 7 + 5(10) = 7 + 50 = 57 xy − 2x
2 (−2) + 4 −4 + 4 0
= = =0
2. 8 + 6 (5) = 8 + 30 = 38 ( −2 )( 4 ) − 2 (−2) −8 + 4 4
3. 6(3) − 8 = 18 − 8 = 10 5 5
17. C= (50 − 32) = (18) = 10
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9 9
4. 8(3) − 4 = 24 − 4 = 20 50F is equivalent to 10C.
5 5 5
5. 82 + 3(8) = 64 + 24 = 88 18. C= (F − 32) = (86 − 32) = (54) = 30
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9 9 9
6. 62 + 5(6) = 36 + 30 = 66 86F is equivalent to 30C.

2 19. h = 4 + 60t −16t 2 = 4 + 60(2) −16(2)2
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7. 7 − 6(7) + 3 = 49 − 42 + 3 = 7 + 3 = 10
= 4 +120 −16(4) = 4 +120 − 64
8. 82 − 7 (8) + 4 = 64 − 56 + 4 = 8 + 4 = 12 = 124 − 64 = 60
Two seconds after it is kicked, the ball’s height is
60 feet.
9. 4 + 5(9 − 7)3 = 4 + 5(2)3
= 4 + 5(8) = 4 + 40 = 44
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