ALGEBRA TEST SOLUTIONS 
A/ Factor Completely if Possible
1) 6a^{2}b^{2} – 54b^{4} 6b^{2}(a^{2} – 9b^{2}) 6b^{2}(a – 3b)(a + 3b) 
2) c^{2} – a^{2} – 10ab – 25b^{2} c^{2} – (a^{2} + 10ab + 25b^{2}) c^{2} – (a + 5b)^{2} [c – (a + 5b)][c + (a + 5b)] [c – a – 5b][c + a + 5b] 
3) 81 – m^{4} (9 – m^{2})(9 + m^{2}) (3 – m)(3 + m)(9 + m^{2}) 
4) 3cx^{2} + 6cx – 9c 3cx^{2} + 6cx – 9c 3c(x^{2} + 2x – 3) 3c(x + 3)(x – 1) 
5) 64 – x^{6} (8 – x^{3})(8 + x^{3}) (2 – x)(4 + 2x + x^{2})(2 + x)(4 – 2x + x^{2}) 
6) a^{2}b^{3} – b^{3} + a^{2} – 1 b^{3}(a^{2} – 1) + (a^{2} – 1) (b^{3} + 1)(a^{2} – 1) (b + 1)(b^{2} – b + 1)(a – 1)(a + 1) 
7) 3x^{2} + 11x + 10 (3x + 5)(x + 2) 
8) 25 – x^{2} – 6x – 9 25 – (x^{2} + 6x + 9) 25 – (x + 3)^{2} [5 – (x + 3)][5 + (x + 3)] [5 – x – 3][5 + x + 3] = (2 – x) (8 + x) 
9) 7x^{2} – 27x – 4 (7x + 1)(x – 4) 
10) 3x^{3} + 24 3(x^{3} + 8) 3(x + 2)(x^{2} – 2x + 4) 
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B/ Perform the Indicated Operations. Reduce to Lowest Terms.
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(2)
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(3)
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(4)
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(5)
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C/ Solve These Equations and Check Your Answer(s).
1) 12 + 3(m + 4) = 5m – 2 12 + 3m + 12 = 5m – 2 24 + 3m = 5m – 2 – 2m = – 26 becomes m = 13 
2) 8(n + 3) + (6 + 2n) = 2(n – 2) + 5(5 + n) 8n + 24 + 6 + 2n = 2n – 4 + 25 + 5n 10n + 30 = 7n + 21 3n = – 9 becomes n = –3 
3) 4(x + 2) – 2(5 – x) = x + 8 4x + 8 – 10 + 2x = x + 8 6x – 2 = x + 8 becomes 5x = 10 
4)
11 – t^{2} = 2t^{2} – t – 6 – 3t^{2} + 8t + 3 14 = 7t becomes t = 2 but t = 2 makes the last fraction's denominator = 0. There are no solutions. 
5) 2x + 1 = 15 or 2x + 1 = – 15 x = 7 or x = –8 
6) x^{2} = 2x + 15 x^{2} – 2x – 15 = 0 (x – 5)(x + 3) = 0 becomes x = 5 or x = –3 
7) 3x^{2} + 2x = 7 3x^{2} + 2x – 7 = 0 using the quadratic formula 
8) .
x^{2} – 6x + 9 = 4(x – 3) x^{2} – 10x + 21 = 0 which becomes (x – 7)(x – 3) = 0 
9) 3x + 2y = 16 _{(eq. 1)} 7x + y = 19 _{(eq. 2)} 3x + 2(19 – 7x) = 16 3x + 38 – 14x = 16 becomes 11x = 22 x = 2 becomes y = 19 – 7x so y = 5 
10) 4r^{2} + t^{2} = 25 _{(eq. 1)} 2r + t = 7 _{(eq. 2)} becomes t = 7 – 2r 4r^{2} + (7 – 2r)^{2} = 25 4r^{2} + 49 – 28r + 4r^{2} = 25 8r^{2} – 28r + 24 = 0 2r^{2} – 7r + 6 = 0 (2r – 3)(r – 2) = 0 so r = 3/2 or r = 2 both solutions work. 
D/ Solve These Inequalities and Check Your Answers.
1) 3x + 2 < 5x – 6 –2x < –8 so x > 4 
2) x^{2} + 4x > 12 x^{2} + 4x – 12 > 0 (x + 6)(x – 2) > 0 x < – 6 or x > 2 
3) 3(3x + 2) < 2(5x – 4) 9x + 6 < 10x – 8 – x < – 14 becomes x > 14 
4) –1 < x < 13/5 and x > 5 
5)  3x – 5 > 22 3x – 5 < – 22 becomes 3x < – 17 so x < –17/3 3x – 5 > 22 becomes 3x > 27 so x > 9 
E/ Simplify these Radical Expressions:
1) =

2) = = =

3) =

d) 
e)  f) 
g)  h) 
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F/ Rationalize these denominators
1)

2)

3)

4)

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G/ Simplify. Leave no negative exponents
1)  2)

3)  4)

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H/ Solve these word problems:
(remember to make a "let" statement and a table or diagram when helpful.)
1) The larger of two numbers exceeds the smaller by 4. Eight times the smaller decreased by four times the larger is 12. What are the two numbers?
solution:
2) The perimeter of a rectangle is 72 dm. Three times the width exceeds twice the length by 3 cm. Find the length and width of the rectangle.
Solution:
3) Sam is presently 7 years older than Fred. One year ago, Sam was twice as old as Fred was then. How old is each now?
person  age now  age 1 year ago 
Fred  x  x – 1 
Sam  x + 7  x + 7 – 1 
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4) The sum of the squares of two consecutive integers is 221. Find 2 solutions for these integers.
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5) A grocer wishes to make a mixture of 80 Kg. of coffee to sell at $4.00 per Kg. He has some coffee that sells for $5.50/kg and cheaper coffee that sells for $3.00/kg. How many kilograms of each type of coffee must he put in the mixture?
item  amount  unit price($)  value($) 
$5.50 coffee  x kg  5.50  5.50 x 
$3.00 coffee  (80 – x) kg  3.00  3.00(80 – x) 
mixture  80kg  4.00  320x 
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6) A bicycle party travels at 25 km/h. At what speed must a second bicycle party travel in order to overtake the first party in 3 hours if they start from the same place 1 hour later than the first party?
party  distance  rate  time 
early party  100 km.  25 km/h  4 hours 
late party  3x km  x km/h  3 hours 
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