# 1964 AHSME Problems/Problem 4

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## Problem

The expression $$\frac{P+Q}{P-Q}-\frac{P-Q}{P+Q}$$

where $P=x+y$ and $Q=x-y$, is equivalent to: $\textbf{(A)}\ \frac{x^2-y^2}{xy}\qquad \textbf{(B)}\ \frac{x^2-y^2}{2xy}\qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{x^2+y^2}{xy}\qquad \textbf{(E)}\ \frac{x^2+y^2}{2xy}$

## Solution 1

Simplifying before substituting by obtaining a common denominator gives: $\frac{(P+Q)^2 - (P - Q)^2}{(P-Q)(P+Q)}$ $\frac{(P^2 + 2PQ + Q^2) - (P^2 - 2PQ + Q^2)}{P^2 - Q^2}$ $\frac{4PQ}{P^2 - Q^2}$

Substituting gives: $\frac{4(x+y)(x-y)}{(x+y)^2 - (x - y)^2}$ $\frac{4(x^2 - y^2)}{(x^2 + 2xy + y^2) - (x^2 - 2xy + y^2)}$ $\frac{4(x^2 - y^2)}{4xy}$ $\frac{x^2 - y^2}{xy}$ $\boxed{\textbf{(A)}}$

## Solution 2

Substituting and then simplifying, noting that $P+Q = 2x$ and $P - Q = 2y$, gives: $\frac{2x}{2y} - \frac{2y}{2x}$ $\frac{x^2}{xy} - \frac{y^2}{xy}$ $\frac{x^2 - y^2}{xy}$ $\boxed{\textbf{(A)}}$

## Solution 3

If it's true for all values of $x$ and $y$, then it's true for a specific value of $x$ and $y$. Letting $x=5$ and $y = 4$ gives $P = 9$ and $Q = 1$. The fraction becomes: $\frac{9 + 1}{9 - 1} - \frac{9 - 1}{9 + 1}$ $\frac{10}{8} - \frac{8}{10}$ $\frac{5}{4} - \frac{4}{5}$ $\frac{25}{20} - \frac{16}{20}$ $\frac{9}{20}$

Plugging in $(x, y) = (5, 4)$ into the answer choices gives $\frac{9}{20}, \frac{9}{40}, 1, \frac{41}{20}, \frac{41}{40}$. Therefore, the answer is $\boxed{\textbf{(A)}}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 