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

Apr 6, 2018

$x = - 5$ and $x = 2$ are both solutions to this equation.

#### Explanation:

$\frac{2}{x - 1} - \frac{2}{3} = \frac{4}{x + 1}$

Lots of ways to solve this. All of them work! Here's one way.

Divide both sides of the equation by 2.

$\frac{1}{x - 1} - \frac{1}{3} = \frac{2}{x + 1}$

Put the two terms on the left-hand side of the equation under a common denominator.

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

Simplify the numerator of the left-hand side.

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

Invert both sides of the equation.

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

Multiply both sides of the equation by $2 \left(4 - x\right)$.

$6 \left(x - 1\right) = \left(x + 1\right) \left(4 - x\right)$

Expand both sides of the equation using the distributive property.

$6 x - 6 = - {x}^{2} + 3 x + 4$

Put this quadratic equation in standard form.

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

The left-hand side of this equation factors nicely.

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

$x = - 5$ and $x = 2$ are both solutions to this equation.

I bet you can come up with another way to solve this. What I like about THIS way is that you don't have to deal with quadratics AND quotients in the same equation.

Apr 6, 2018

${x}_{1} = - 5$ and ${x}_{2} = 2$

#### Explanation:

$\frac{2}{x - 1} - \frac{2}{3} = \frac{4}{x + 1}$

$\frac{2}{x - 1} - \frac{4}{x + 1} = \frac{2}{3}$

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

$\frac{6 - 2 x}{{x}^{2} - 1} = \frac{2}{3}$

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

$18 - 6 x = 2 {x}^{2} - 2$

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

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

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

So ${x}_{1} = - 5$ and ${x}_{2} = 2$