Answer
{$\frac{2}{3}$}
Work Step by Step
First, we find the LCM of the denominators of the two fractions on the left-hand side of the equation in order to develop a single fraction:
$\frac{x}{x-2}+\frac{4}{x+2}=1$
$\frac{x(x+2)+4(x-2)}{(x-2)(x+2)}=1$
$\frac{x^{2}+2x+4x-8}{(x-2)(x+2)}=1$
$\frac{x^{2}+6x-8}{(x-2)(x+2)}=1$
Now, we cross multiply the two sides:
$\frac{x^{2}+6x-8}{(x-2)(x+2)}=1$
$x^{2}+6x-8=(x-2)(x+2)$
$x^{2}+6x-8=x^{2}-2^{2}$
$x^{2}+6x-8=x^{2}-4$
$6x-8=-4$
$6x=-4+8$
$6x=4$
$x=\frac{4}{6}$
$x=\frac{2}{3}$
Therefore, the solution set is {$\frac{2}{3}$}.
