1979 AHSME Problems/Problem 2

Problem 2

For all non-zero real numbers $x$ and $y$ such that $x-y=xy, \frac{1}{x}-\frac{1}{y}$ equals

$\textbf{(A) }\frac{1}{xy}\qquad \textbf{(B) }\frac{1}{x-y}\qquad \textbf{(C) }0\qquad \textbf{(D) }-1\qquad \textbf{(E) }y-x$

Solution 1

Moving all variables to one side of the equation, we can use Simon's Favorite Factoring Trick to factor the equation into $(x+1)(y-1) = -1$ Plugging in $-1$ and $1$ as the $x$ and $y$ sides respectively, we get $x = -2$ and $y = 2$ . Plugging this in to $\frac{1}{x}-\frac{1}{y}$ gives us $\boxed{-1}$ as our final answer.

Solution 2

Notice that we can do $\frac{x+y}{xy} = \frac{xy}{xy}$ . We are left with $\frac{1}{y} - \frac{1}{x} = 1$ . Multiply by $-1$ to achieve $\frac{1}{x} - \frac{1}{y} = \boxed{-1}$ .

1979 AHSME (Problems • Answer Key • Resources)
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1979 AHSME Problems/Problem 2

Contents

Problem 2

Solution 1

Solution 2

See also