Question

Find $x$ in terms of $a,b$ and $c$
$\frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c}$ where $x\ne \{a,b,c\}$

Answer 1

$\frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c}$

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

$\Rightarrow \frac{ax-ab+bx-ab}{x^2-bx-ax+ab}=\frac{2c}{x-c}$

$\Rightarrow \frac{ax+bx-2ab}{x^2-bx-ax+ab}=\frac{2c}{x-c}$

$(ax+bx-2ab)(x-c)=2c(x^2-bx-ax+ab)$

$\Rightarrow a{x}^2+b{x}^2-2abx-2acx-bcx+2abc=2c{x}^2-2bcx-2acx+2abc$

$\Rightarrow a{x}^2+b{x}^2-2c{x}^2-acx+2acx-bcx+2bcx+2abc-2abc=0$

$\Rightarrow a{x}^2+b{x}^2-2c{x}^2+acx+bcx=0$

$\Rightarrow x^2(a+b-x)+x(ac+bc)=0$

$\Rightarrow x\left[\right(a+b+c-2c)x+(ac+bc\left)\right]=0$

$\Rightarrow (a+b-2c)x+(ac+bc)=0$

$x(a+b-2c)=-(ax+bc)$

$x=\frac{-(ac+bc)}{a+b-2c}$