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

Apr 9, 2016

The given equation is invalid.
There are no possible solutions.

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} \frac{2}{x + 3} - \frac{4}{{x}^{2} + 2 x - 3} = \frac{1}{1 - x}$

Noting that
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 2 x - 3 = \left(x + 3\right) \left(x - 1\right)$
the given equation can be re-written as
$\textcolor{w h i t e}{\text{XXX}} \frac{2 \left(x - 1\right)}{{x}^{2} + 2 x - 3} - \frac{4}{{x}^{2} + 2 x - 3} = \frac{\left(- 1\right) \left(x + 3\right)}{{x}^{2} + 2 x + 3}$

Provided $x \ne - 3$ and $\textcolor{red}{x \ne 1}$
then ${x}^{2} + 2 x - 3 \ne 0$

and we have
$\textcolor{w h i t e}{\text{XXX}} 2 \left(x - 1\right) - 4 = - \left(x + 3\right)$

$\textcolor{w h i t e}{\text{XXX}} 2 x - 6 = - x - 3$

$\textcolor{w h i t e}{\text{XXX}} 3 x = 3$

$\textcolor{w h i t e}{\text{XXX}} x = 1$ ...but this is one of the "forbidden values".

Both the terms $\frac{4}{{x}^{2} + 2 x - 3}$ and $\frac{1}{1 - x}$ are undefined when $x = 1$