ALGEBRA TEST SOLUTIONS

A/ Factor Completely if Possible

 1) 6a2b2 – 54b4 6b2(a2 – 9b2)6b2(a – 3b)(a + 3b) 2) c2 – a2 – 10ab – 25b2 c2 – (a2 + 10ab + 25b2)c2 – (a + 5b)2 [c – (a + 5b)][c + (a + 5b)][c – a – 5b][c + a + 5b] 3) 81 – m4 (9 – m2)(9 + m2)(3 – m)(3 + m)(9 + m2) 4) 3cx2 + 6cx – 9c3cx2 + 6cx – 9c3c(x2 + 2x – 3)3c(x + 3)(x – 1) 5) 64 – x6 (8 – x3)(8 + x3)(2 – x)(4 + 2x + x2)(2 + x)(4 – 2x + x2) 6) a2b3 – b3 + a2 – 1b3(a2 – 1) + (a2 – 1)(b3 + 1)(a2 – 1)(b + 1)(b2 – b + 1)(a – 1)(a + 1) 7) 3x2 + 11x + 10(3x + 5)(x + 2) 8) 25 – x2 – 6x – 925 – (x2 + 6x + 9)25 – (x + 3)2 [5 – (x + 3)][5 + (x + 3)][5 – x – 3][5 + x + 3] = (2 – x) (8 + x) 9) 7x2 – 27x – 4(7x + 1)(x – 4) 10) 3x3 + 243(x3 + 8)3(x + 2)(x2 – 2x + 4)

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B/ Perform the Indicated Operations. Reduce to Lowest Terms.

1)

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(2)

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(3)

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(4)

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(5)

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 1) 12 + 3(m + 4) = 5m – 2 12 + 3m + 12 = 5m – 224 + 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 + 5n10n + 30 = 7n + 213n = – 9 becomes n = –3 3) 4(x + 2) – 2(5 – x) = x + 84x + 8 – 10 + 2x = x + 86x – 2 = x + 8 becomes 5x = 10 so x = 2 4) 11 – t2 = 2t2 – t – 6 – 3t2 + 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 = – 15x = 7 or x = –8 6) x2 = 2x + 15x2 – 2x – 15 = 0(x – 5)(x + 3) = 0 becomes x = 5 or x = –3 7) 3x2 + 2x = 73x2 + 2x – 7 = 0using the quadratic formula 8) . x2 – 6x + 9 = 4(x – 3)x2 – 10x + 21 = 0 which becomes (x – 7)(x – 3) = 0 so x = 7 or x = 3

 9) 3x + 2y = 16 (eq. 1) 7x + y = 19 (eq. 2) becomes y = 19 – 7x3x + 2(19 – 7x) = 163x + 38 – 14x = 16 becomes 11x = 22x = 2 becomes y = 19 – 7x so y = 5 10) 4r2 + t2 = 25 (eq. 1) 2r + t = 7 (eq. 2) becomes t = 7 – 2r 4r2 + (7 – 2r)2 = 25 4r2 + 49 – 28r + 4r2 = 25 8r2 – 28r + 24 = 0 2r2 – 7r + 6 = 0 (2r – 3)(r – 2) = 0 so r = 3/2 or r = 2both solutions work.

 1) 3x + 2 < 5x – 6–2x < –8 so x > 4 2) x2 + 4x > 12x2 + 4x – 12 > 0(x + 6)(x – 2) > 0x < – 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| > 223x – 5 < – 22 becomes 3x < – 17 so x < –17/33x – 5 > 22 becomes 3x > 27 so x > 9

 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:

let x = the smaller #
x + 4 = the larger #
8x 4(x + 4) = 12
8x 4x 16 = 12
4x = 28
x = 7 and x + 4 = 11
The numbers are 7 and 11.

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:

Since the perimeter = 72, half the perimeter = 36
So length + width = 36
Therefore, length = 36 width.
Let x = the width of the rectangle
so, 36 x = the length
3x = 2(36 x) + 3
3x = 72 2x + 3
5x = 75 becomes x = 15
so 36 – x = 21.
The rectangle is 15 dm. wide and 21 dm. long.

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
Let x = Fred's age now.
so x + 7 = Sam's age now.
x + 6 = 2(x 1)
x = 2x 8
x = 8, so x = 8
x + 7 = 15
Fred is 8 years old and Sam is 15.

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4) The sum of the squares of two consecutive integers is 221. Find 2 solutions for these integers.

Let x = the 1st integer
x + 1 = the 2nd integer
x 2 + (x + 1) 2 = 221
x 2 + x 2 + 2x + 1 = 221
2x 2 + 2x 220 = 0
x 2 + x 110 = 0
(x + 11)(x 10) = 0
so, x = 11 or x = 10
The integers are 11 and 10, or 10 and 11.

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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
Let x = the amount of \$5.50 coffee in the mixture.
80 x = the amount of \$3.00 coffee in the mixture.
5.50 x + 3(80 x) = 320x
5.50 x + 240 3x = 320
2.50 x = 80 becomes x = 32
The mixture should be made up of 32 kg. of \$5.50 coffee and 48 kg. of \$3.00 coffee.

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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
If the 2nd party overtakes the 1st, the 2 parties must have covered the same distance. So, since we have values for rate and time for both parties, and distance = rate × time we can set the distances for the 2 parties equal to each other.
Let x = the late party's rate
so 3x = the late party's distance.
3x = 100 becomes x =
The 2nd party must travel at 33 km/hour to overtake the 1st party in 3 hours.

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