Beginning and Intermediate Algebra, Trobey - Downloadable Solutions Manual (Revised)

Practice Final Examination
1 3 13 13 65 52 117 17 1 1
1. 3 + 2 = + = + = =5 12. x − 4 = x +1
4 5 4 5 20 20 20 20 3 2
1
x = −5
⎛ 1 ⎞⎛ 2 ⎞ ⎛ 7 ⎞⎛ 8 ⎞ 7 2 ⋅ 4 28 1 6
2. ⎜1 ⎟ ⎜ 2 ⎟ = ⎜ ⎟ ⎜ ⎟ = × = =3
⎝ ⎠⎝
6 3 ⎠ ⎝ ⎠⎝ ⎠
6 3 2 ⋅ 3 3 9 9 x = −30
13. −4 x + 3 y = 7
15 3 15 8 3 ⋅ 5 4 ⋅ 2 10 3 y = 4x + 7
3. ÷ = × = × = = 10
4 8 4 3 4 3 1 4x + 7
y=
3
4. 1.63
× 3.05 14. 5 x + 3 − (4 x − 2) ≤ 6 x − 8
815 5x + 3 − 4x + 2 ≤ 6x − 8
4 890 −5 x ≤ −13
4.9715 13
x≥ = 2.6
5
20 0000
5. 0.0006 ∧ 120.0000 ∧
12
15. P = 2L + 2W = 1760
0
L + W = 880
12 ÷ 0.0006 = 200,000 2W − 200 + W = 880
3W = 1080
6. 7% of 64,000 = 0.07 × 64,000 = 4480
W = 360
L = 2W − 200 = 520
7. (4 − 3)2 + 9 ÷ (−3) + 4 = 12 + 3 ÷ (−3) + 4 The width is 360 meters and the length is
= 1 + (−1) + 4 520 meters.
=4
d 6
2 2 16. r = = =3
8. 5a − 2ab − 3a − 6a − 8ab + 2a 2 2
4 4 4
= (5a − 6a) + (−2ab − 8ab) + (−3a 2 + 2a 2 ) V = πr 3 ≈ (3.14)(3)3 = (3.14)(27) = 113.04
3 3 3
= −a − 10ab − a 2 The volume is approximately 113.04 cubic feet.
9. −2 x + 3y{7 − 2[ x − (4 x + y)]} 17. 7x − 2y = −14
= −2 x + 3y[7 − 2( x − 4 x − y)] Let x = 0.
= −2 x + 3y[7 − 2(−3x − y)] 7(0) − 2 y = −14
= −2 x + 3 y ( 7 + 6 x + 2 y ) y=7
= −2 x + 21y + 18 xy + 6 y 2 Let y = 0.
7 x − 2(0) = −14
10. 2 x 2 − 3xy − 4 y, x = −2, y = 3 x = −2

2(−2)2 − 3(−2)(3) − 4(3) = 8 + 18 − 12 = 14 x y
0 7
9
11. F = C + 32
5 −2 0
9
F = (−35) + 32
5
F = −31
The temperature is −31°F.

364 Copyright © 2017 Pearson Education, Inc.

,ISM: Beginning and Intermediate Algebra Practice Final Examination


1 2
24. x + y = 1 (1)
2 3
1
x + y = −1 (2)
3
Multiply equation (1) by 6 and equation (2) by 3
to clear fractions.
3x + 4 y = 6 (3)
x + 3 y = −3 (4)
Multiply equation (4) by −3 and add to equation
(3).
3x + 4 y = 6
18. 3x − 4y ≤ 6 −3 x − 9 y = 9
Graph 3x − 4y = 6 with a solid line. −5 y = 15
Test point: (0, 0) y = −3
3(0) − 4(0) ≤ 6
Replace y with −3 in equation (4).
0 ≤ 6, True x + 3(−3) = −3
Shade the region containing (0, 0). x − 9 = −3
x=6
The solution is (6, −3).
25. 4 x − 3 y = 12 (1)
3x − 4 y = 2 (2)
Multiply equation (1) by 3 and equation (2) by
−4.
12 x − 9 y = 36
−12 x + 16 y = −8
7 y = 28
y −y −3 − 5 8 y=4
19. m = 2 1 = =
x2 − x1 −2 − 1 3 Replace y with 4 in equation (1).
4 x − 3(4) = 12
2 2 4 x − 12 = 12
20. y = − x + 4; m = − 4 x = 24
3 3
x=6
2
The slope of a parallel line is m = − . The solution is (6, 4).
3
26. Solve the system.
2 2 x + 3y − z = 16 (1)
21. f ( x) = 3x − 4 x − 3
x − y + 3z = −9 (2)
f (3) = 3(32 ) − 4(3) − 3 5 x + 2 y − z = 15 (3)
f (3) = 27 − 12 − 3
Multiply equation (1) by 3 and add to equation
f (3) = 12 (2).
6 x + 9 y − 3z = 48
22. f ( x) = 3x 2 − 4 x − 3 x − y + 3z = −9
f (−2) = 3((−2) 2 ) − 4(−2) − 3 7 x + 8y = 39 (4)
f (−2) = 12 + 8 − 3 Multiply equation (3) by 3 and add to equation
f (−2) = 17 (2).
x − y + 3 z = −9
23. {(2, −7), (−1, 1), (3, 2)} 15 x + 6 y − 3z = 45
Domain: {2, −1, 3}
16 x + 5 y = 36 (5)
Range: {−7, 1, 2}
Multiply equation (4) by 5, equation (5) by −8,
and add the results.
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,Practice Final Examination ISM: Beginning and Intermediate Algebra


35 x + 40 y = 195 29. (−3x 2 y )(−6 x3 y 4 ) = −3(−6) x 2+3 y1+ 4 = 18 x5 y 5
−128 x − 40 y = −288
−93x = −93
30. (3 x − 1)(2 x + 5) = 6 x 2 + 15 x − 2 x − 5
x =1
= 6 x 2 + 13 x − 5
Replace x with 1 in equation (4).
7(1) + 8 y = 39
x+4
7 + 8 y = 39
8 y = 32 31. x + 3 x 2 + 7 x + 12
y=4 x 2 + 3x
Replace x with 1 and y with 4 in equation (1). 4 x + 12
2(1) + 3(4) − z = 16 4 x + 12
2 + 12 − z = 16 0
14 − z = 16 ( x 2 + 7 x + 12) ÷ ( x + 3) = x + 4
−z = 2
z = −2
The solution is (1, 4, −2). 32. 9 x 2 − 30 x + 25 = (3 x)2 − 2(3 x)(5) + 52
= (3 x − 5) 2
27. y + z = 2 (1)
x + + z = 5 (2) 33. x3 + 2 x 2 − 4 x − 8 = x 2 ( x + 2) − 4( x + 2)
x+ y = 5 (3)
= ( x + 2)( x 2 − 4)
Subtract equation (3) from equation (2).
x +z=5 = ( x + 2)( x 2 − 22 )
−x − y = −5 = ( x + 2)( x + 2)( x − 2)
− y + z = 0 (4)
Add equations (1) and (4). 34. 2 x3 + 15 x 2 − 8 x = x(2 x 2 + 15 x − 8)
y+z =2 = x(2 x − 1)( x + 8)
−y + z = 0
2z = 2 35. x 2 + 15 x + 54 = 0
z =1 ( x + 9)( x + 6) = 0
Replace z with 1 in equation (1). x +9 = 0 x +6 = 0
y +1 = 2 x = −9 x = −6
y =1
Replace z with 1 in equation (2). 9 x3 − x x (9 x 2 − 1)
36. =
x +1 = 5 3 x 2 − 8 x − 3 (3 x + 1)( x − 3)
x=4 x (3 x + 1)(3x − 1)
=
The solution is (4, 1, 1). (3x + 1)( x − 3)
x (3 x − 1)
28. 3y ≥ 8x − 12 2x + 3y ≤ −6 =
Test point: (0, 0) Test point: (0, 0) ( x − 3)
3(0) ≥ 8(0) − 12 2(0) + 3(0) ≤ −6
0 ≥ −12, true 0 ≤ −6, false x2 − 9 x 2 − 3x
37. ÷
2 x 2 + 7 x + 3 2 x 2 + 11x + 5
( x + 3)( x − 3) (2 x + 1)( x + 5)
= ⋅
(2 x + 1)( x + 3) x ( x − 3)
x +5
=
x




366 Copyright © 2017 Pearson Education, Inc.

, ISM: Beginning and Intermediate Algebra Practice Final Examination


3x 2
38. −
x + 5 x 2 + 7 x + 10
3x ( x + 2) 2
= −
( x + 5)( x + 2) ( x + 5)( x + 2)
3x ( x + 2) − 2
=
( x + 5)( x + 2)
3x 2 + 6 x − 2
=
( x + 5)( x + 2)


39.
3 +2
2 x +1
=
(
(2 x + 1)(2 x − 1) 2 x3+1 + 2 )
1 − 22
4 x −1 (
(2 x + 1)(2 x − 1) 1 − (2 x +1)(2
2
x −1) )
3(2 x − 1) + 2(2 x + 1)(2 x − 1)
=
(2 x + 1)(2 x − 1) − 2
6 x − 3 + 2(4 x 2 − 1)
=
4 x2 − 1 − 2
6 x − 3 + 8x2 − 2
=
4 x2 − 3
8 x2 + 6 x − 5
=
4 x2 − 3

x −1 2 4
40. = +
x −4 2 x + 2 x − 2
x −1 2 4
= +
( x + 2)( x − 2) x + 2 x − 2
⎛ x −1 ⎞ 2 4
( x + 2)( x − 2) ⎜ ⎟ = ( x + 2)( x − 2) ⋅ + ( x + 2)( x − 2) ⋅
⎝ ( x + 2)( x − 1) ⎠ x+2 x−2
x − 1 = 2( x − 2) + 4( x + 2)
x − 1 = 2x − 4 + 4x + 8
5 x = −5
x = −1

5 x −4 y −2 5 −4 −( −1/2) −2−3
41. x= y
−1/2 3 15
15 x y
1
= x −7/2 y −5
3
1
=
−7/2 5
3x y

42. 3 40 x 4 y 7 = 3 8 x3 y 6 ⋅ 5 xy
= 3 8 x3 y 6 ⋅ 3 5 xy
= 2 xy 2 3 5 xy

43. 5 2 − 3 50 + 4 98 = 5 2 − 3 25 ⋅ 2 + 4 49 ⋅ 2
= 5 2 − 15 2 + 28 2
= 18 2

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