# How do you solve the system  [(x)/(x+5)]- [(5)/(5-x)] = (x+31)/(x^2-25)?

Jan 30, 2016

Multiply numerators and denominators to get a common denominator ${x}^{2} - 25$ then equate the numerators to get a quadratic, hence $x = 3$ or $x = - 2$

#### Explanation:

$\frac{x}{x + 5} - \frac{5}{5 - x}$

$= \frac{x}{x + 5} + \frac{5}{x - 5}$

$= \frac{x \left(x - 5\right)}{\left(x - 5\right) \left(x + 5\right)} + \frac{5 \left(x + 5\right)}{\left(x - 5\right) \left(x + 5\right)}$

$= \frac{{x}^{2} - \textcolor{red}{\cancel{\textcolor{b l a c k}{5 x}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{5 x}}} + 25}{{x}^{2} - 25}$

$= \frac{{x}^{2} + 25}{{x}^{2} - 25}$

So we want to solve ${x}^{2} + 25 = x + 31$ with exclusion $x \ne \pm 5$

Subtract $x + 31$ from both sides to get:

$0 = {x}^{2} - x - 6 = \left(x - 3\right) \left(x + 2\right)$

So $x = 3$ or $x = - 2$