# How do you add or subtract (4xy)/(x^2-y^2) + ( x-y)/(x+y)?

Jun 28, 2018

See a solution process below:

#### Explanation:

To add or subtract fractions they must be over a common denominator.

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

Therefore we need to multiply the fraction on the right by $\left(x - y\right) \left(x - y\right)$, a form of $1$, to get both fractions over a common denominator:

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

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

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

Now, we can add the numerators of the two fractions over their common denominator:

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

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

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

$\frac{{x}^{2} + 2 x y + {y}^{2}}{{x}^{2} - {y}^{2}} \implies$

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

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + y\right)}}} \left(x + y\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + y\right)}}} \left(x - y\right)} \implies$

$\frac{x + y}{x - y}$