Question

Solve the following equations :
$\frac{x-ab}{a+b}+\frac{x-ac}{a+c}+\frac{x-bc}{b+c}$ = a+ b + c

Answer 1

The equation can be written as
$(\frac{x-ab}{a+b}-c)+(\frac{x-ac}{a+c}-b)+(\frac{x-bc}{b+c}-a)$ = 0
or $\frac{x-ab-bc-ca}{a+b}+\frac{x-ab-bc-ca}{a+c}+\frac{x-ab-bc-ca}{b+c}$ =0
$(x-ab-bc-ca)(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c})$ =0
Assuming $(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c})$ $\ne$ 0,
$x=ab+bc+ca$ as the root of the equation. If, however,
$(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c})$ = 0, the given equation turns into an identity which holds for every value of x.