Question

(1 ÷ x- 1)+(2÷ x+2)=(3÷2) solve and check for extraneous solutions

Answer 1

$\frac{1}{x-1}+ \frac{2}{x+2}= \frac{3}{2}$
$\frac{1(x+2)}{(x-1)(x+2)}+ \frac{2(x-1)}{(x+2)(x-1)}= \frac{3(x-1)(x+2)}{2}$
$\frac{x+2+2(x-1)}{(x-1)(x+2)}= \frac{3(x-1)(x+2)}{2}$
$\frac{x+2+2x-1}{(x-1)(x+2)}= \frac{3(x-1)(x+2)}{2}$
$\frac{3x+1}{(x-1)(x+2)}= \frac{3(x^2+x-2)}{2}$
$\frac{}{(x-1)(x+2)}= \frac{3x^2+3x-6)}{2}$
$\frac{3x+1}{(x^2+x-2)}= \frac{3x^2+3x-6)}{2}$
$\frac{2(3x+1)}{2(x^2+x-2)}= \frac{3x^2+3x-6)(x^2+x-2)}{2(x^2+x-2)}$
$\frac{2(3x+1)}{2(x^2+x-2)}-\frac{3x^2+3x-6)(x^2+x-2)}{2(x^2+x-2)}=0$
$\frac{2(3x+1)-3x^2+3x-6}{2(x^2+x-2)}=0$

this will be continued