Question

$\frac{a}{(x-b)}+\frac{b}{(x-a)}=2,(x\ne b,a)$

Answer 1

$$\begin{gathered} \begin{array}{l}\frac{a}{(x-b)}+\frac{b}{(x-a)}=2\\ \begin{array}{l}\Rightarrow [\frac{a}{(x-b)}-1]+[\frac{b}{(x-a)}-1]=0\\ \Rightarrow \frac{a-(x-b)}{x-b}+\frac{b-(x-a)}{x-a}=0\end{array}\end{array} \\ \Rightarrow \frac{a-x+b}{x-b}+\frac{a-x+b}{x-a}=0 \\ \Rightarrow (a-x+b)[\frac{1}{(x-b)}+\frac{1}{(x-a)}]=0 \\ \Rightarrow (a-x+b)\left[\frac{(x-a)+(x-b)}{(x-b)(x-a)}\right]=0 \\ \Rightarrow (a-x+b)\left[\frac{2x-(a+b)}{(x-b)(x-a)}\right]=0 \\ \Rightarrow (a-x+b)[2x-(a+b\left)\right]=0 \\ \Rightarrow a-x+b=0\text{or}2x-(a+b)=0 \\ \Rightarrow x=a+b\text{or}x=\frac{a+b}{2} \\ \text{Hence, the roots of the equation are}(a+b)\text{and}\left(\frac{a+b}{2}\right)\text{.} \end{gathered}$$