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

May 16, 2018

It has no solutions.

#### Explanation:

Sum the two fractions of right hand side:

$\setminus \frac{x}{x + 2} - \setminus \frac{2}{2 - x} = \setminus \frac{x}{x + 2} + \setminus \frac{2}{x - 2}$

$= \setminus \frac{x \left(x - 2\right) + 2 \left(x + 2\right)}{\left(x + 2\right) \left(x - 2\right)} = \setminus \frac{{x}^{2} - 2 x + 2 x + 4}{{x}^{2} - 4} = \setminus \frac{{x}^{2} + 4}{{x}^{2} - 4}$

Now the equation becomes

$\setminus \frac{{x}^{2}}{{x}^{2} - 4} = \setminus \frac{{x}^{2} + 4}{{x}^{2} - 4}$

Since the denominators are the same (and are never zero), the equation holds if and only if it holds between the numerators:

$\setminus \cancel{{x}^{2}} = \setminus \cancel{{x}^{2}} - 4 \setminus \implies 0 = - 4$

which is clearly impossible.