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Difference between revisions of "1978 AHSME Problems/Problem 3"

(Created page with "== Problem 3 == For all non-zero numbers <math>x</math> and <math>y</math> such that <math>x = 1/y</math>, <math>\left(x-\frac{1}{x}\right)\left(y+\frac{1}{y}\right)</math> e...")
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Revision as of 15:41, 20 January 2020

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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