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

1 Answer
May 16, 2018

Answer:

It has no solutions.

Explanation:

Sum the two fractions of right hand side:

#\frac{x}{x+2}-\frac{2}{2-x} = \frac{x}{x+2}+\frac{2}{x-2}#

#=\frac{x(x-2)+2(x+2)}{(x+2)(x-2)} = \frac{x^2-2x+2x+4}{x^2-4}= \frac{x^2+4}{x^2-4}#

Now the equation becomes

#\frac{x^2}{x^2-4} = \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:

#\cancel(x^2) = \cancel(x^2)-4 \implies 0=-4#

which is clearly impossible.