# How do you solve (x+1)/(x-3) = 4 - 12/(x^2-2x-3)?

May 31, 2015

Given $\frac{x + 1}{x - 3} = 4 - \frac{12}{{x}^{2} - 2 x - 3}$
and
assuming that ${x}^{2} - 2 x - 3 \ne 0$ (otherwise we have an undefined division)

Note that x^2-2x-3= (x-3)((x+1)

So we can re-write the equation with a common denominators as
$\textcolor{w h i t e}{\text{XXXXX}}$$\frac{\left(x + 1\right) \left(x + 1\right)}{{x}^{2} - 2 x - 3} = \frac{4 \left({x}^{2} - 2 x - 3\right) - 12}{{x}^{2} - 2 x - 3}$
and since ${x}^{2} - 2 x - 3 \ne 0$
$\textcolor{w h i t e}{\text{XXXXX}}$$\left(\left(x + 1\right) \left(x + 1\right)\right) = \left(4 \left({x}^{2} - 2 x - 3\right) - 12\right)$

Simplifying
$\textcolor{w h i t e}{\text{XXXXX}}$${x}^{2} + 2 x + 1 = 4 {x}^{2} - 8 x - 24$

$\textcolor{w h i t e}{\text{XXXXX}}$$3 {x}^{2} - 10 x - 25 = 0$

$\textcolor{w h i t e}{\text{XXXXX}}$$\left(3 x + 5\right) \left(x - 5\right) = 0$

$x = - \frac{5}{3}$ or $x = 5$