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

Jul 3, 2016

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

#### Explanation:

Make a common denominator of all the fractions. The obvious choice is to make every denominator $3 \left(x - 1\right) \left(x + 1\right)$. We can do this because multiplying the top and bottom of a fraction does not change the fraction itself.

So, we multiply top and bottom of each fraction by the appropriate part of the required denominator:

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

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

Because everything is over a common denominator, we can ignore it and focus on the numerators.

$6 \left(x + 1\right) - 2 \left(x - 1\right) \left(x + 1\right) = 12 \left(x - 1\right)$

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

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

We've arrived at a simple quadratic, can solve by factorisation but for completeness we'll use the quadratic formula.

Have a = 2, b = 6, c = -20.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- 6 \pm \sqrt{36 + 160}}{4} = \frac{- 6 \pm \sqrt{196}}{4} = \frac{- 6 \pm 14}{4}$

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