Solve each of the following quadratic equations:

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

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

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

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

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

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

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

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

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

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

$\Rightarrow x-5=0$ or $2 x-5=0$

$\Rightarrow x=5$ or $x=\frac{5}{2}$

Hence, 5 and $\frac{5}{2}$ are the roots of the given equation.