Question

How do you solve this this?

Answer 1

First of all, we have to observe that the expression makes sense only if

$x\neq\pm 1$

otherwise, you'd have zero denominators, which are not allowed. So, assuming $x\neq\pm 1$, let's sum the two fractions at the left hand side:

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

Again, since we're assuming $x\neq\pm 1$, we can multiply both sides by $x^2-1$ to get

$x^2+3x-2=2 \iff x^2+3x-4=0$

This equation has solutions $x=-4$ and $x=1$, but we can only accept the first, since 1 is a forbidden value.

So, the only solutions to this equation is $x=-4$