Algebra Review
Algebra/Calculus Review Answers
ALGEBRA REVIEW
Add as indicated:
1. (2ac) +(–6ac) +(9ac) = 5ac
2. 3x +(–7x) = –4x
3. (–8a) –(–3a) –(2a) = –7a
4. (5x) –(6x) –(7x) = –8x
Add the two expressions in each problem
5. x +2y –8 6. 11m –7n +13
3x –4y +9 3m –8n –21
4x –2y +1 14m –15n –8
Subtract the two expressions in each problem
7. x +2y –8 8. 11m –7n +13
– (3x –4y +9) –(3m –8n –21)
–2x +6y –17 8m +n +34
9. 10a –17b +24c; 13a +14b –16c = –3a –31b +40c
Add the three expressions in each of the problems:
10. 7a –3b +11c; –14a +10b +10c; 8a +8b +13c = a +15b +34c
Combine like terms:
11. 3x +7y –3z +6xc –8y –7z +5 –1 = 3x –y –10z +6xc +4
12. 9xy +3x +4a –5ax +10a –7x +3yx +6xa = 12xy –4x +ax +14a
Remove the symbols of grouping and simplify by combining terms:
13. 3a –(b +c) +(a +b –c) = 4a –2c
14. –[x +(3 –x) –(4 +3x)] = 3x +1
15. –{5a –b –[3b –(c –2b +a) –4a] +c} = –5a +b +[3b –(c –2b +a) –4a] –c
= –5a +b +3b –c +2b -a –4a –c = –10a +6b –2c
Multiply as indicated:
16. (5)(-4)(-2) = 40 17. (3ab)(2a) = 6a2b
18. (6a2b)(3ab2) = 18a3b3 19. (3xy2)(5x2y)(xy) = 15x4y4
20. 3x2y(2xy2 +y) = 6x3y3 +3x2y2
21. –4mn(3 –5m +6mn +3n) = –12mn +20m2n –24m2n2 –12mn2
22. (a +3b)(3a2 +6ab +4b2) = 3a3 +15a2b +22ab2 +12b3
23. (3a +2)(a –2)(2a +1) = (3a2 –6a +2a –4)(2a +1) = 6a3 –8a2 –8a +3a2 –4a –4
= 6a3 –5a2 –12a –4
Divide as indicated:
24. x2/x2 = 1 25. –y3/y3 = –1
26. a8/a5 = a3 27. acz/ac = z
28. 12a4/6a = 2 a3 29. –9c3d4/3c5d3 = –3d/c2
30. 34a3b2/17a2b = 2ab 31. (9x2 –6x3 –3x4)/(-3x2) = –3 + 2x + x2
Perform the indicated operations, giving the results in the lowest term:
32. [pic] = 1/2 33. [pic] = 10/13
34. [pic] = 4/3 35. [pic] = x/12
36. [pic] = [pic] 37. [pic] =[pic]
38. [pic] =[pic] 39. [pic] = [pic]=1/2
Simplify the complex fractions and other expressions;
40. [pic] =[pic]= 1/21 41. [pic] =[pic] =[pic][pic]=[pic]
42. [pic] = [pic] 43. [pic] = [pic]
Solve for x, and check by substitution:
44. 8x –15 = 3x 45. x –4 = 5 +2x
5x = 15 x = –9
x = 3
46. [pic] 47. (3x –1)(x +1) = 3x2
[pic]x = [pic] 3x2 + 3x –x –1 = 3x2
x = 6 x = 1/2
Solve the following systems:
48. – 4x +5y +14 = 0 49. 4xy – 2y –11 = 0
–2(– 2x +2y – 7 = 0) –2(2xy –3y – 4 = 0)
y +28 = 0 4y –3 = 0
y = –28 y = 3/4
–4x +5(–28) +14 = 0 4x(3/4) –2(3/4) –11 = 0
x = –31.5 x = 4.167
50. 3(3x + y – 6 = 0) 51. 3(4x –5y = 9)
4x–3y –21 = 0 –4(3x –4y = 8)
13x –39 = 0 y = –5
x = 3 4x –5(–5) = 9
4(3) –3y –21 = 0 x = –4
y = - 3
Simplify the complex fractions;
52. [pic]= 1.467b/a
53. [pic]= 6.265b/a
54. [pic]= [pic]
55. [pic]= 2[pic]
CALCULUS REVIEW
Differentiate the following functions
56. Y = X2 [pic]= 2X
57 y = 6x3 [pic]=18x2
58. Z = 20Y0.5 [pic]= 10Y–0.5
59. Z =[pic] = 3x1/2 [pic]= 3/2x–1/2
60. y = 30 y’ = 0
61. t = 3x + 6x3 [pic]= 3 + 18x2
62. y = 7x0.5 + 12x1.5 [pic]= 3.5x–0.5 + 18x0.5
63. Y = 6X(6 + 2X2) [pic]= 6X(4X) + 6(6 + 2X2) = 24X2 + 36 + 12X2 = 36 + 36X2
64. y = (2 + 6x)(2 + 5x2) [pic]= (2 + 6x)(10x) + (6)(2 + 5x2) = 12 + 20x + 90x2
65. y = (x2 – 3x3)(x3 + 5) [pic]= (x2 – 3x3)(3x2) + (2x – 9x2)(x3 + 5)
66. y = f(x) = [pic] f’(x) = [pic]=[pic]
67. y = [pic] [pic]= f’(x) =[pic]
Find the maximum and/or minimum values of the following functions. Be sure to state the first and second order conditions.
68. y = 10 + 10x – 0.5x2
FOC y’ = 10 – x = 0 x = 10
SOC y” = –1 Maximum
69. T = – 2000S + 10S2
FOC T’ = –2000 +20S = 0, S = 100
SOC T” = –20 Maximum
70. Z = 145 – 4Y – 0.3Y2
FOC Z’ = –4 – 0.6S = 0, S = – 6.67
SOC Z” = –0.6 Maximum
71. X = – 145 + 4Y + 0.3Y2
FOC X’ = +4 + 0.6Y = 0 S = 6.67
SOC X” = +0.6 Minimum
Find the maximum and/or minimum values and the inflection point of the following functions. Be sure to state the first and second order conditions. (Hint: the roots of the quadratic equation: aX2 + bX + c = 0 are
[pic].)
72. y = 5000 + 500x + 20x2 –0.1x3
FOC y’ = 500 + 40x –0.3x2 = 0
x = -11.51, x = 144.84
SOC y” = 40 –0.6x
y” = 40 –0.6(-11.51) > 0 therefore minimum
y” = 40 –0.6(144.84) < 0 therefore maximum
Inflection point is where y” = 0
y” = 40 –0.6x = 0, x = 6.67
73. z = -2000 +30x +10x2 – 0.3x3
FOC y’ = 30 + 20x –0.9x2 = 0
x = -1.41, x = 23.63
SOC y” = 20 –1.8x
y” = 20 –1.8(-1.41) > 0 therefore minimum
y” = 20 –1.8(23.63) < 0 therefore maximum
Inflection point is where y” = 0
y” = 20 –1.8x = 0, x = 11.11
74. y = 5000x –100x2 –0.5x3
FOC y’ = 5000 –200x –1.5x2 = 0
x = –154.86, x = 21.53
SOC y” = –200 –3x
y” = –200 –3(-154.86) > 0 therefore minimum
y” = –200 –3(21.53) < 0 therefore maximum
Inflection point is where y” = 0
y” = –200 –3x = 0, x = 66.67
75. y = 400x – 25x2 + 0.2x3
FOC y’ = 400 –50x +0.6x2 = 0
x = 8.96, x = 74.37
SOC y” = –50 +1.2x
y” = –50 +1.2(8.96) < 0 therefore maximum
y” = –50 +1.2(74.37) > 0 therefore minimum
Inflection point is where y” = 0
y” = –50 +1.2x = 0, x = 41.67
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