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Unformatted text preview: very term by 20.
4 − 30 −26 = 15 15 Exercise 5
1. (A) (a + b) (a – b) = a2 – b2 1 1 2 2 3 4 = a − b 1 = a2 − b2 12 2. (C)
a Multiply every term by x2. 2. 1 = 2 ax ax (D) (a – b)2 = a2 – 2ab + b2 = 40 3. (C)
y− x y+ x Multiply every term by xy. Substituting 8 for ab, we have
a 2 − 16 + b 2 = 40 a 2 + b 2 = 56 4. (A)
x+y x Multiply every term by xy. 3. (C) (a + b) (a – b) = a2 – b2 5. (E)
4+t 2 Multiply every term by t . 2 2 8(a − b) = 24 (a − b) = 3 4. 5. (A) (D) x2 + 4x – 45 = (x + 9) (x – 5)
1 1 1 1 1 1 − + = − c d c d c2 d 2 (5)(3) = 12 − 12 c d 1 1 15 = 2 − 2 c d www.petersons.com Factoring and Algebraic Fractions 169 Retest
1. 2. 3. (B) (A) (B)
= 4 n + n 5n n 1 = ==n 10 10 2 2 x 3 x − 3y −= y1 y
x 2 + 2x − 8 3 ⋅ 4+ x 2− x ( x + 4)( x − 2) 3 ⋅ 4+ x 2− x 3( x − 2) 3( x − 2) Divide x + 4. 2 − x = −1( x − 2) = −3 4. 5. (D) 5a 3 5a 3 5a 3 125a 9 ⋅ ⋅ = b b b b3 (E) Multiply every term by x.
3x − 1 y 6. (B) Multiply every term by x2.
3x 2 − x 3 7. (A) a2 – b2 = (a + b)(a – b) = 100 Substituting 25 for a + b, we have 25(a – b) = 100 a–b=4 8. 9. (D) (A) x2 – 8x – 20 = (x – 10)(x + 2)
1 1 1 1 1 1 a − b a + b = 2 − 2 a b 1 (6)(5) = 2 − 12 a b 1 1 30 = 2 − 2 a b 10. (E) (x – y)2 = x2 – 2xy + y2 = 30 Substituting 17 for xy, we have
x 2 − 34 + y 2 = 30 x 2 + y 2 = 64 www.petersons.com Problem Solving in Algebra 12 DIAGNOSTIC TEST
Directions: Work out each problem. Circle the letter that appears before your answer. Answers are at the end of the chapter. 1. Find three consecutive odd integers such that the sum of the first two is four times the third. (A) 3, 5, 7 (B) –3, –1, 1 (C) –11, –9, –7 (D) –7, –5, –3 (E) 9, 11, 13 Find the shortest side of a triangle whose perimeter is 64, if the ratio of two of its sides is 4 : 3 and the third side is 20 less than the sum of the other two. (A) 6 (B) 18 (C) 20 (D) 22 (E) 24 A purse contains 16 coins in dimes and quarters. If the value of the coins is...
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