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

Given: The eqation:$\frac{1}{( x-1)( x-2)} +\frac{1}{( x-2)( x-3)} =\frac{2}{3} \ ,\ x\
eq 1,2,3$

To do: To find $x=?$

The given equation is

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

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

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

$\Rightarrow \frac{2x-4}{x^{3} -3x^{2} -3x^{2} +9x+2x-6} =\frac{2}{3}$

$3( 2x-4) =2( x^{3} -3x^{2} -3x^{2} +9x+2x-6)$

$6x-12=2x^{3} -12x^{2} +22x-12$

$2x^{3} -12x^{2} +16x=0$

$2x( x^{2} -6x+8) =0$

$x^{2} -6x+8=0$

$x^{2} -4x-2x+8=0$

$x( x-4) -2( x-4) =0$

$( x-2)( x-4) =0$

If $x-2=0$

$x=2$

If $x-4=0$

$x=4$

Thus The given equation has two solution $x=2,\ 4$