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

Jun 26, 2016

$x = \frac{10}{7} \mathmr{and} x = - 1$

#### Explanation:

When we have an equation with fractions it is possible to get rid of the denominators.

Instead of finding a common denominator and converting all the numerators, find the LCM and multiply each term by the LCM of the denominators. The denominators can cancel.

LCM = $\textcolor{red}{3 \left(x + 1\right) \left(x - 1\right)}$

(color(red)(3(x+1)(x-1))xx2)/(x-1) -(2xxcolor(red)(3(x+1)(x-1)))/3 " " = (4xxcolor(red)(3(x+1)(x-1)))/(x+1)

$\frac{\textcolor{red}{3 \left(x + 1\right) \cancel{\left(x - 1\right)}} \times 2}{\cancel{\left(x - 1\right)}} - \frac{2 \times \textcolor{red}{\cancel{3} \left(x + 1\right) \left(x - 1\right)}}{\cancel{3}} \text{ } = 4 \times \frac{\textcolor{red}{3 \cancel{\left(x + 1\right)} \left(x - 1\right)}}{\cancel{\left(x + 1\right)}}$

$3 \left(x + 1\right) \times 2 - 2 \left(x + 1\right) \left(x - 1\right) = 4 \times 3 \left(x + 1\right) \left(x - 1\right)$

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

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

$14 {x}^{2} - 6 x - 20 = 0 \text{ divide by 2}$

$7 {x}^{2} - 3 x - 10 = 0$

$\left(7 x - 10\right) \left(x + 1\right) = 0 \text{ factorise}$

$\mathmr{if} 7 x - 10 = 0 \text{ then } x = \frac{10}{7}$

$\mathmr{if} x + 1 = 0 \text{ then "x = -1}$