# How do you solve 2/(x+1) + 5/(x-2)=-2 ?

Oct 27, 2017

$x = \frac{1}{2} \textcolor{w h i t e}{\text{xxxx}}$ or $\textcolor{w h i t e}{\text{xxxx}} x = - 3$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} \frac{2}{x + 1} + \frac{5}{x - 2} = - 2$

Multiply everything (on both sides by $\left(x + 1\right) \left(x - 2\right)$
$\textcolor{w h i t e}{\text{XXX}} 2 \left(x - 2\right) + 5 \left(x + 1\right) = - 2 \left(x + 1\right) \left(x - 2\right)$

$\textcolor{w h i t e}{\text{XXX}} 2 x - 4 + 5 x + 5 = - 2 {x}^{2} + 2 x + 4$

$\textcolor{w h i t e}{\text{XXX}} 7 x + 1 = - 2 {x}^{2} + 2 x + 4$

Subtract $\left(7 x + 1\right)$ from both sides (and reverse the sides)
$\textcolor{w h i t e}{\text{XXX}} - 2 {x}^{2} - 5 x + 3 = 0$

Multiply both sides by $\left(- 1\right)$
$\textcolor{w h i t e}{\text{XXX}} 2 {x}^{2} + 5 x - 3 = 0$

Factor
$\textcolor{w h i t e}{\text{XXX}} \left(2 x - 1\right) \left(x + 3\right) = 0$

$\left.\begin{matrix}\text{Either " & 2x-1=0 & " or } & x + 3 = 0 \\ \null & \rightarrow x = \frac{1}{2} & \null & \rightarrow x = - 3\end{matrix}\right.$