# 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 $\boxed{A}$.

## Solution 2

Muliply both sides by $\left(\frac{1}{w}-\frac{1}{z}\right)$ to get $\frac{1}{w}+\frac{1}{z}=2014\left(\frac{1}{w}-\frac{1}{z}\right)$. Then, add $2014\cdot\frac{1}{z}$ to both sides and subtract $\frac{1}{w}$ from both sides to get $2015\cdot\frac{1}{z}=2013\cdot\frac{1}{w}$. Then, we can plug in the most simple values for z and w ($2015$ and $2013$, respectively), and find $\frac{2013+2015}{2013-2015}=\frac{2(2014)}{-2}=-2014$, or answer choice $\boxed{A}$.

## Solution 3

Let $a = \frac{1}{w}$ and $b = \frac{1}{z}$. To find values for a and b, we can try $a+b = 2014$ and $a-b=1$. However, that leaves us with a fractional solution, so scaling it by 2, we get $a+b = 4028$ and $a-b=2$. Solving by adding the equations together, we get $b = 2015$ and $a = 2013$. Now, substituting back in, we get $w = \frac{1}{2015}$ and $z = \frac{1}{2013}$. Now, putting this into the desired equation with $n = 2015 \cdot 2013$ (since it will cancel out), we get $\frac{\frac{2013+2015}{n}}{\frac{2013-2015}{n}}$. Dividing, we get $\frac{4028}{-2} = \boxed{\textbf{(A)}\ -2014}$.

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## Video Solution (CREATIVE THINKING)

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