Solve the following quadratic equations by factorization:

We have been given

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

$3\left(x^{2}-5 x+4+x^{2}-5 x+6\right)=10\left(x^{2}-6 x+8\right)$

$4 x^{2}-30 x+50=0$

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

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

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

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

Therefore,

$2 x-5=0$

$2 x=5$

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

or,

$x-5=0$

$x=5$

Hence, $x=\frac{5}{2}$ or $x=5$.