# Difference between revisions of "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 the numerator and denominator of the LHS by $wz$ 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 $\boxed{A}$.

~savannahsolver