Given that (a - b)^2a^2 - b^2 = 16, what is the value of a + ba - b =

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PS · Problem Solving

sub-505

Algebra

Given that $\frac{(a - b)^2}{a^2 - b^2} = 16$ , what is the value of $\frac{a + b}{a - b} =$

Answer Choices

Correct answer marked below

1/16

Correct

B

1/4

C

1

D

4

E

16

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Official Explanation

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Starting with

\frac{(a - b)^2}{a^2 - b^2} = 16

, we can rewrite it as

\frac{(a - b)(a - b)}{(a + b)(a - b)} = 16

.…

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Given that $\frac{(a - b)^2}{a^2 - b^2} = 16$ , what is the value of $\frac{a + b}{a - b} =$

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Official Explanation

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Given that (a−b)2a2−b2=16\frac{(a - b)^2}{a^2 - b^2} = 16a2−b2(a−b)2​=16, what is the value of a+ba−b=\frac{a + b}{a - b} =a−ba+b​=

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Official Explanation

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What it tests

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Given that $\frac{(a - b)^2}{a^2 - b^2} = 16$ , what is the value of $\frac{a + b}{a - b} =$