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

Jun 23, 2016

Put on an equivalent denominator.

#### Explanation:

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

We can now eliminate the denominators and solve.

$6 x + 6 - 2 \left({x}^{2} - 1\right) = 4 \left(3 x - 3\right)$

$6 x + 6 - 2 {x}^{2} + 2 = 12 x - 12$

$0 = 2 {x}^{2} + 6 x - 20$

$0 = 2 {x}^{2} - 4 x + 10 x - 20$

$0 = 2 x \left(x - 2\right) + 10 \left(x - 2\right)$

$0 = \left(2 x + 10\right) \left(x - 2\right)$

$x = - 5 \mathmr{and} 2$

None of these solutions are extraneous; they don't make the denominator $0$.

The solution set is $\left\{x = - 5 , 2\right\}$.

Hopefully this helps!