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

May 25, 2018

${x}_{1} = 0$ and ${x}_{2} = 2$

#### Explanation:

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

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

$\frac{4 x - 2}{{x}^{2} - 1} = 2$

$4 x - 2 = 2 \cdot \left({x}^{2} - 1\right)$

${x}^{2} - 1 = 2 x - 1$

${x}^{2} - 2 x = 0$

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

Hence ${x}_{1} = 0$ and ${x}_{2} = 2$

May 25, 2018

$x = 0$, $x = 2$

#### Explanation:

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

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

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

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

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

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

$\Rightarrow 4 x - 2 = 2 {x}^{2} - 2 x + 2 x - 2$

$\Rightarrow 4 x - 2 = 2 {x}^{2} - 2$

$\Rightarrow 2 {x}^{2} - 4 x = 0$

$\Rightarrow 2 x \left(x - 2\right) = 0$

$\Rightarrow x - 2 = 0 \to x = 2$

$\Rightarrow 2 x = 0 \to x = 0$