# How to simplify this .?

## $\frac{4 {x}^{2} - {y}^{2}}{12 {x}^{2} - 4 x y - {y}^{2}}$

##### 1 Answer
Jul 31, 2017

$\frac{4 {x}^{2} - {y}^{2}}{12 {x}^{2} - 4 x y - {y}^{2}} = \frac{2 x + y}{6 x + y}$

with exclusion $2 x \ne y$

#### Explanation:

Given:

$\frac{4 {x}^{2} - {y}^{2}}{12 {x}^{2} - 4 x y - {y}^{2}}$

We can factor both numerator and denominator and then cancel any common factor.

The numerator is a difference of squares, so factors as:

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

To factor the denominator find a pair of factors of $12$ which differ by $4$. The pair $6 , 2$ works, so use that to split the middle term and factor by grouping:

$12 {x}^{2} - 4 x y - {y}^{2} = \left(12 {x}^{2} - 6 x y\right) + \left(2 x y - {y}^{2}\right)$

$\textcolor{w h i t e}{12 {x}^{2} - 4 x y - {y}^{2}} = 6 x \left(2 x - y\right) + y \left(2 x - y\right)$

$\textcolor{w h i t e}{12 {x}^{2} - 4 x y - {y}^{2}} = \left(6 x + y\right) \left(2 x - y\right)$

So we find:

$\frac{4 {x}^{2} - {y}^{2}}{12 {x}^{2} - 4 x y - {y}^{2}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(2 x - y\right)}}} \left(2 x + y\right)}{\left(6 x + y\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(2 x - y\right)}}}}$

$\textcolor{w h i t e}{\frac{4 {x}^{2} - {y}^{2}}{12 {x}^{2} - 4 x y - {y}^{2}}} = \frac{2 x + y}{6 x + y}$

with exclusion $2 x \ne y$