2013 AMC 12A Problems/Problem 8

Problem

Given that $x$ and $y$ are distinct nonzero real numbers such that $x+\tfrac{2}{x} = y + \tfrac{2}{y}$, what is $xy$?

$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad$

Solution 1

$x+\tfrac{2}{x}= y+\tfrac{2}{y}$

Since $x\not=y$, we may assume that $x=\frac{2}{y}$ and/or, equivalently, $y=\frac{2}{x}$.

Cross multiply in either equation, giving us $xy=2$.

$\boxed{\textbf{(D) }{2}}$

Solution 2

$x+\tfrac{2}{x}= y+\tfrac{2}{y}$

$x-y+\frac{2}{x}-\frac{2}{y} = 0$

$(x-y)+2(\frac{y-x}{xy}) = 0$

$(x-y)(1-\frac{2}{xy})=0$

Since $x\not=y$

$1 = \frac{2}{xy}$

$xy = 2$

Video Solution

https://youtu.be/CCjcMVtkVaQ ~sugar_rush

See also

2013 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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
All AMC 12 Problems and Solutions

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