Solve the following quadratic equation by factorization:
$\frac{a}{x-b}+\frac{b}{x-a}=2$

$\frac{a}{x-b}+\frac{b}{x-a}=2$
$$\begin{gathered} \Rightarrow \frac{ax-a^2+bx-b^2}{\left(x-a\right)\left(x-b\right)}=2 \\ \Rightarrow ax-a^2+bx-b^2=2x^2-2bx-2ax+2ab \\ \Rightarrow 2x^2-2bx-2ax+2ab-ax+a^2-bx+b^2=0 \\ \Rightarrow 2x^2+x\left[-2b-2a-a-b\right]+a^2+b^2+2ab=0 \\ \Rightarrow 2x^2-3x\left[a+b\right]+{\left(a+b\right)}^2=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 2x^2-2\left(a+b\right)x-\left(a+b\right)x+{\left(a+b\right)}^2=0 \\ \Rightarrow 2x\left[x-\left(a+b\right)\right]-\left(a+b\right)\left[x-\left(a+b\right)\right]=0 \\ \Rightarrow \left[2x-\left(a+b\right)\right]\left[x-\left(a+b\right)\right]=0 \end{gathered}$$
So, the value of x will be
$x=\frac{a+b}{2},a+b$