# How do you express ((y^2-xy)/(x^2+xy)-xy+y^2)(x/(x-y))+(y/(x+y)) as a single fraction?

May 26, 2017

$\left(\frac{{y}^{2} - x y}{{x}^{2} + x y} - x y + {y}^{2}\right) \left(\frac{x}{x - y}\right) + \left(\frac{y}{x + y}\right) = - x y$

#### Explanation:

$\left(\frac{{y}^{2} - x y}{{x}^{2} + x y} - x y + {y}^{2}\right) \left(\frac{x}{x - y}\right) + \left(\frac{y}{x + y}\right)$

$= \left(\frac{- y \left(x - y\right)}{x \left(x + y\right)} - y \left(x - y\right)\right) \left(\frac{x}{x - y}\right) + \left(\frac{y}{x + y}\right)$

$= \left(\frac{- y \left(x - y\right)}{x \left(x + y\right)} + \frac{- y \left(x - y\right) x \left(x + y\right)}{x \left(x + y\right)}\right) \left(\frac{x}{x - y}\right) + \left(\frac{y}{x + y}\right)$

$= \left(\frac{- y \left(x - y\right) \left(1 + x \left(x + y\right)\right)}{x \left(x + y\right)}\right) \left(\frac{x}{x - y}\right) + \left(\frac{y}{x + y}\right)$

$= \left(\frac{- y \cancel{\left(x - y\right)} \left(1 + x \left(x + y\right)\right)}{\cancel{x} \left(x + y\right)}\right) \left(\frac{\cancel{x}}{\cancel{\left(x - y\right)}}\right) + \left(\frac{y}{x + y}\right)$

$= \frac{- y \left(1 + x \left(x + y\right)\right)}{\left(x + y\right)} + \left(\frac{y}{x + y}\right)$

$= - \frac{y}{\left(x + y\right)} - \frac{y x \left(x + y\right)}{\left(x + y\right)} + \left(\frac{y}{x + y}\right)$

$= - x y$