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

Dec 3, 2016

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

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

$x + 3 - 2 {x}^{2} - 2 x = {x}^{2} - 1$

$0 = 3 {x}^{2} + x - 4$

$0 = 3 {x}^{2} - 3 x + 4 x - 4$

$0 = 3 x \left(x - 1\right) + 4 \left(x - 1\right)$

$0 = \left(3 x + 4\right) \left(x - 1\right)$

$x = - \frac{4}{3} \mathmr{and} 1$

However, our restrictions in the original equation are $x \ne \pm 1$, so $x = - \frac{4}{3}$ is the only valid solution.

Hopefully this helps!