Помогите решить уравнение!

$\frac{x}{x+1}+ \frac{x}{x-1}=2 \frac{2}{3}$
$\frac{x(x-1)+x(x+1)}{(x+1)(x-1)}=\frac{2*3+2}{3}$
$\frac{x(x-1+x+1)}{(x+1)(x-1)}=\frac{8}{3}$
$\frac{x^2}{(x+1)(x-1)}-\frac{4}{3}=0$
$\frac{3x^2-4(x+1)(x-1)}{3(x+1)(x-1)}=0$
$\frac{3x^2-4(x^2-1)}{3(x+1)(x-1)}=0$

$\left \{ {{3x^2-4x^2+4=0} \atop {x \neq 1,and,x \neq -1}} \right. ; \left \{ {{-x^2+2^2=0} \atop {x \neq 1,and,x \neq -1}} \right. ; \left \{ {{x^2-2^2=0} \atop {x \neq 1,and,x \neq -1}} \right. ; \left \{ {{(x-2)(x+2)=0} \atop {x \neq 1,and,x \neq -1}} \right. ;$
$\left \{ {{x=2,or,x=-2} \atop {x \neq 1,and,x \neq -1}} \right.$

$x=2,or,x=-2$

Ответ: 2; -2.