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

Jul 18, 2016

Maybe you see that the third fraction can be factorized.

#### Explanation:

$3 - 2 x - {x}^{2} = - \left({x}^{2} + 2 x - 3\right) = - \left(x - 1\right) \left(x + 3\right)$

Which leaves the whole thing as:
$\frac{1}{x - 1} + \frac{2}{x + 3} - \frac{2 x + 2}{\left(x - 1\right) \left(x + 3\right)}$ (mark the $-$ sign)

Next step: make the bottom parts equal:
$\frac{1}{x - 1} \times \frac{x + 3}{x + 3} + \frac{2}{x + 3} \times \frac{x - 1}{x - 1} - \frac{2 x + 2}{\left(x - 1\right) \left(x + 3\right)} =$

$\frac{1 \times \left(x + 3\right)}{\left(x - 1\right) \left(x + 3\right)} + \frac{2 \times \left(x - 1\right)}{\left(x - 1\right) \left(x + 3\right)} - \frac{2 x + 2}{\left(x - 1\right) \left(x + 3\right)} =$

$\frac{\left(x + 3\right) + \left(2 x - 2\right) - \left(2 x + 2\right)}{\left(x - 1\right) \left(x + 3\right)} =$

Put the numbers and the $x$'s together:
$\frac{\left(x + \cancel{2 x} - \cancel{2 x}\right) + \left(3 - 2 - 2\right)}{\left(x - 1\right) \left(x + 3\right)} = \frac{\left(x - 1\right)}{\left(x - 1\right) \left(x + 3\right)} =$

Now cancel:
$\frac{\cancel{\left(x - 1\right)}}{\cancel{\left(x - 1\right)} \left(x + 3\right)} = \frac{1}{x + 3}$

NOTE:
Forbidden solutions are $x = 1 \mathmr{and} x = - 3$ as the numerator of the fraction would be $= 0$
graph{1/(x+3) [-16.02, 16.02, -8, 8.02]}