Hunting The Variables!

Algebra Level 2

Given that;

x y + x = y x + y \dfrac xy + x = \dfrac yx + y

for x y x \neq y , find the value of:

1 x + 1 y \dfrac 1x + \dfrac 1y


The answer is -1.

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1 solution

Sravanth C.
Mar 26, 2016

We can rewrite the expression as:

x y + x = y x + y x y y x = ( y x ) \begin{aligned} \dfrac xy + x &= \dfrac yx + y\\ \dfrac xy - \dfrac yx &=(y - x)\\ \end{aligned}

Now we can solve this expression to get our desired result:

x 2 y 2 x y = ( y x ) ( x + y ) ( x y ) x y = ( y x ) ( x + y ) x y = 1 1 x + 1 y = 1 \begin{aligned} \dfrac{x^2 - y^2}{xy} &= (y - x)\\ \dfrac{(x+ y)(x-y)}{xy} &= (y - x)\\ \dfrac{(x+ y)}{xy} &= -1\\ \dfrac 1x + \dfrac 1y &= -1\\ \end{aligned}

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