# 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 $x\not=y$

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

$xy = 2$

## Solution 3

Let $A = x + \frac{2}{x} = y + \frac{2}{y}.$ Consider the equation $$u + \frac{2}{u} = A.$$ Reorganizing, we see that $u$ satisfies $$u^2 - Au + 2 = 0.$$ Notice that there can be at most two distinct values of $u$ which satisfy this equation, and $x$ and $y$ are two distinct possible values for $u.$ Therefore, $x$ and $y$ are roots of this quadratic, and by Vieta’s formulas we see that $xy$ thereby must equal $\boxed{2}.$

~ Professor-Mom

## Solution 4

$$x + \frac{2}{x} = y + \frac{2}{y}.$$

Multiply both sides by xy to get

$$x^2y + 2y = y^2x +2x$$

Rearrange to get

$$x^2y - y^2x = 2x - 2y$$

Factor out $xy$ on the left side and $2$ on the right side to get

$$xy(x-y) = 2(x-y)$$

Divide by $(x-y)$ {You can do this since x and y are distinct} to get

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

~ pi_is_3.14

## Video Solution

https://youtu.be/CCjcMVtkVaQ ~sugar_rush