1978 AHSME Problems/Problem 3

Problem 3

For all non-zero numbers $x$ and $y$ such that $x = 1/y$ , $\left(x-\frac{1}{x}\right)\left(y+\frac{1}{y}\right)$ equals

$\textbf{(A) }2x^2\qquad \textbf{(B) }2y^2\qquad \textbf{(C) }x^2+y^2\qquad \textbf{(D) }x^2-y^2\qquad \textbf{(E) }y^2-x^2$

Solution 1

Using substitution, we can substitute y into the equation in the first parentheses. Therefore, we'll get $\left(x-\frac{1}{\frac{1}{y}}\right)\left(y+\frac{1}{y}\right)$

Because $x = \frac{1}{y}$ , we can also see that $y = \frac{1}{x}$ . Using substitution again, we can substitute x into the second equation getting $\left(x-\frac{1}{\frac{1}{y}}\right)\left(y+\frac{1}{\frac{1}{x}}\right)$

Simplifying, we get $(x-y)(y+x)$ . Multiplying, we get $\boxed{\textbf{(D) }x^2-y^2}$ ~awin

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1978 AHSME Problems/Problem 3

Problem 3

Solution 1