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

See Also

1978 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png