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

Aug 18, 2016

In an equation with fractions, we can get rid of the denominators by multiplying each term by the LCD (LCM of denominators).
IN this case the LCM is $\textcolor{red}{3 \left(x - 1\right) \left(x + 1\right)}$

(color(red)(3(cancel(x-1))(x+1))xx2)/cancel(x-1) - (color(red)(cancel3(x-1)(x+1))xx2)/cancel3 =(color(red)(3(x-1)cancel(x+1))xx4)/(cancel(x+1)

After cancelling the denominators, this leads to a simpler equation.

$6 \left(x + 1\right) - 2 \textcolor{b l u e}{\left(x - 1\right) \left(x + 1\right)} = 12 \left(x - 1\right) \text{ } \textcolor{b l u e}{D O T S}$

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

$6 x + 6 - 2 {x}^{2} + 2 = 12 x - 12 \text{ make =0}$

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

${x}^{2} + 3 x - 10 = 0 \text{ factorise}$

$\left(x + 5\right) \left(x - 2\right) = 0$

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