2014 AMC 10B Problems/Problem 9

Problem

For real numbers $w$ and $z$ , $\cfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} = 2014.$ What is $\frac{w+z}{w-z}$ ?

$\textbf{(A) }-2014\qquad\textbf{(B) }\frac{-1}{2014}\qquad\textbf{(C) }\frac{1}{2014}\qquad\textbf{(D) }1\qquad\textbf{(E) }2014$

Solution

Multiply numerator and denominator by $w+z$ to get $\frac{z+w}{z-w}=2014$ . Then since $z+w=w+z$ and $w-z=-(z-w)$ , $\frac{w+z}{w-z}=-\frac{z+w}{z-w}=-2014$ , or choice $\boxed{A}$ .

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2014 AMC 10B Problems/Problem 9

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