# How do you solve (6/(x^2-1)) - (1/(x+1)) = 1/2?

Apr 6, 2016

$x \approx - 4.464 \text{ or } 2.464$ to 3 decimal paces

#### Explanation:

${x}^{2} - 1$ can be written as ${x}^{2} - {1}^{2}$

Compare this to ${a}^{2} + {b}^{2} = \left(a + b\right) \left(a - b\right)$

We can use this to our advantage

Write the given equation as:

$\frac{6}{{x}^{2} - {1}^{2}} - \frac{1}{x + 1} = \frac{1}{2}$

By substitution this is

$\frac{6}{\left(x + 1\right) \left(x - 1\right)} - \frac{1}{x + 1} = \frac{1}{2}$

$\frac{6 - \left(x - 1\right)}{\left(x + 1\right) \left(x - 1\right)} = \frac{1}{2}$

Multiply both sides by 2

$\frac{12 - 2 \left(x - 1\right)}{\left(x + 1\right) \left(x - 1\right)} = 1$

$12 - 2 \left(x - 1\right) = \left(x + 1\right) \left(x - 1\right)$

$12 - 2 x + 2 = {x}^{2} - 1$

${x}^{2} + 2 x - 11 = 0$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now solve as a quadratic -

Completing the square method (skipping many steps)

${\left(x + 1\right)}^{2} - 12 = 0$

$x + 1 = \pm \sqrt{12}$

$x = - 1 \pm 2 \sqrt{3}$

$x \approx - 4.464 \text{ or } 2.464$ to 3 decimal paces