# How do you solve 15/(x^2-1)=4/(x-1)-3/(x+1)?

Mar 30, 2018

$x = 8$

#### Explanation:

We can combine like fractions:

$\frac{4}{x - 1} - \frac{3}{x + 1} = \frac{4 \left(x + 1\right)}{\left(x + 1\right) \left(x - 1\right)} - \frac{3 \left(x - 1\right)}{\left(x - 1\right) \left(x + 1\right)}$
$= \frac{4 x + 4 - 3 x + 3}{{x}^{2} - 1} = \frac{x + 7}{{x}^{2} + 1}$

Setting this equal to the first equation, we get
$\frac{15}{{x}^{2} - 1} = \frac{x + 7}{{x}^{2} - 1}$

Let $x \ne \pm 1$ so there aren't infinities we will deal with.

This means that
$x + 7 = 15$
$x = 8$

If we plug in $x = 8$ to the original, we can check:

$\frac{15}{{8}^{2} - 1} = \frac{4}{8 - 1} - \frac{3}{8 + 1}$
$\frac{15}{63} = \frac{4}{7} - \frac{3}{9} = \frac{4}{7} - \frac{1}{3}$
$\frac{5}{21} = \frac{12}{21} - \frac{7}{21} = \frac{5}{21.}$
As we hope.