Question

Solve the following equations :
$\frac{x-2}{x+2}+\frac{x+2}{x-2}=\frac{2(x+3)}{x-3}$

Answer 1

The given equations can be written as

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

or $-\frac{4}{x+2}+\frac{4}{x-2}=\frac{12}{x-3}$

or $\frac{-x+2+x+2}{(x+2)(x-2)}$ = $\frac{3}{x-3}$

or 3(x${}^2$ $-4)=4x-12$ or 3$x^2$ $-4x=0$

$\therefore$ x=$0,4/3$