Solve for $x$:

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

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:

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

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

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

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

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

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

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

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

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=2, b=-15$ and $c=25$.

Therefore, the roots of the given equation are

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-15)\pm \sqrt{(-15)^2-4(2)(25)}}{2(2)}$ 

$x=\frac{15\pm \sqrt{225-200}}{4}$ 

$x=\frac{15\pm \sqrt{25}}{4}$ 

$x=\frac{15\pm 5)}{4}$ 

$x=\frac{15+5}{4}$ or $x=\frac{15-5}{4}$

$x=\frac{20}{4}$ or $x=\frac{10}{4}$

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

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