1979 AHSME Problems/Problem 2

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Solution

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 $1/x-1/y$ gives us $\boxed{-1}$ as our final answer.

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