Question

Solve
$\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3},x\ne 2,4$

Answer 1

$\frac{x-1}{x-2}+\frac{x-3}{x-4}=\frac{10}{3}$

$\Rightarrow \frac{(x-1)(x-4)+(x-3)(x-2)}{(x-2)(x-4)}=\frac{10}{3}$

$\Rightarrow \frac{x^2-5x+4+x^2-5x+6}{x^2-6x+8}=\frac{10}{3}$

$\Rightarrow \frac{2x^2-10x+10}{x^2-6x+8}=\frac{10}{3}$

$\Rightarrow \frac{2(x^2-5x+5)}{x^2-6x+8}=\frac{10}{3}$

$\Rightarrow \frac{(x^2-5x+5)}{x^2-6x+8}=\frac{5}{3}$

$\Rightarrow 3x^2-15x+15=5x^2-30x+40$

$\Rightarrow 5x^2-30x+40-3x^2+15x-15=0$

$\Rightarrow 2x^2-15x+25=0$

$\Rightarrow 2x^2-10x-5x+25=0$

$\Rightarrow 2x(x-5)-5(x-5)=0$

$\Rightarrow (x-5)(2x-5)=0$

$\therefore x=5,\frac{5}{2}$