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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