Solve the following quadratic equation by factorization:
$\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3}, x≠2, 4$

Given:

Given quadratic equation is $\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3}, x≠2, 4$.

To do:

We have to solve the given quadratic equation.

Solution:

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

$\frac{(x-1)(x-4)+(x-3)(x-2)}{(x-2)(x-4)}=\frac{3\times3+1}{3}$

$\frac{x^2-x-4x+4+x^2-3x-2x+6}{x^2-2x-4x+8}=\frac{9+1}{3}$

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

$3(2)(x^2-5x+5)=10(x^2-6x+8)$

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

$(5-3)x^2+(-30+15)x+40-15=0$

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

To factorise $2x^2-15x+25=0$, we have to find two numbers $m$ and $n$ such that $m+n=-15$ and $mn=2(25)=50$.

If $m=-10$ and $n=-5$, $m+n=-10-5=-15$ and $mn=(-10)\times(-5)=50$.

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

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

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

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

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

$2x=5$ or $x=5$

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

The values of $x$ are  $\frac{5}{2}$ and $5$.