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 $\boxed{\textbf{(A)}}$