# How do you solve 1/(x-4) + (x-4)/(x-4) = 7 / ( x^2+x-20)?

Jun 19, 2017

Eliminate the denominators by multiplying both sides by $\left(x - 4\right) \left(x + 5\right)$.

#### Explanation:

Given: $\frac{1}{x - 4} + \frac{x - 4}{x - 4} = \frac{7}{{x}^{2} + x - 20}$

Eliminate the denominators by multiplying both sides by $\left(x - 4\right) \left(x + 5\right)$.

$x + 5 + \left(x - 4\right) \left(x + 5\right) = 7$

$x + 5 + {x}^{2} + x - 20 = 7$

${x}^{2} + 2 x - 22 = 0$

$x = \frac{- 2 \pm \sqrt{{2}^{2} - 4 \left(1\right) \left(- 22\right)}}{2 \left(1\right)}$
$x = \frac{- 2 \pm 2 \sqrt{23}}{2}$
$x = - 1 + \sqrt{23} \mathmr{and} x = - 1 - \sqrt{23}$