# Simplify (2x)/(2x-y) - y/(2x+y) - (4xy)/(4x^2-y^2) ?

Dec 6, 2016

$\frac{2 x - y}{2 x + y}$ Where $y \ne 2 x$

#### Explanation:

First consider the first two terms:

$\frac{2 x}{2 x - y} - \frac{y}{2 x + y} = \frac{2 x \left(2 x + y\right) - y \left(2 x - y\right)}{4 {x}^{2} - {y}^{2}}$

Now combine this result with the third term:

$\frac{2 x \left(2 x + y\right) - y \left(2 x - y\right)}{4 {x}^{2} - {y}^{2}} - \frac{4 x y}{4 {x}^{2} - {y}^{2}}$

$= \frac{4 {x}^{2} + 2 x y - 2 x y + {y}^{2} - 4 x y}{4 {x}^{2} - {y}^{2}}$

$= \frac{4 {x}^{2} + {y}^{2} - 4 x y}{4 {x}^{2} - {y}^{2}}$

Factorise top and bottom:

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

$= \frac{\cancel{\left(2 x - y\right)} \left(2 x - y\right)}{\left(2 x + y\right) \cdot \cancel{\left(2 x - y\right)}}$ Where $2 x - y \ne 0$

$= \frac{2 x - y}{2 x + y}$ Where $y \ne 2 x$