Question

Solve for $x$:
$\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}=\frac{2}{3}$
where $x\ne 1,2,3$

Answer 1

Given

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

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

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

$\frac{2x-4}{(x-1)(x-3)}=\frac{2(x-2)}{3}$

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

$3=(x-1)(x-3)$

$x^2-x-3x+3=3$

$x^2-4x=0$

$x(x-4)=0$

$x=0,4$