Question

Solve the quadratic equation $\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3}$

Answer 1

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

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

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

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

$\Rightarrow 6x^2-30x+30=10x^2-60x+80$

$\Rightarrow 6x^2-30x+30-10x^2+60x-80=0$

$\Rightarrow -4x^2-30x-50=0$

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

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

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

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

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

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