Question

Solve for $x:\frac{1}{(a+b+x)}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x},[a\ne 0,b\ne 0,x\ne 0,x\ne -(a+b\left)\right]$

Answer 1

Given, $\frac{1}{(a+b+x)}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}$

$\Rightarrow \frac{1}{a+b+x}=\frac{bx+ax+ab}{abx}$

$\Rightarrow abx=(a+b+x)(ax+bx+ab)$

$\Rightarrow abx=a^{2}x+abx+a^{2}b+abx+b^{2}x+a{b}^2+a{x}^2+b{x}^2+abx$

$\Rightarrow 0=2x[ab+a^2+b^2]+ab[a+b]$

$\Rightarrow 2x[ab+a^2+b^2]=-ab[a+b]$

$\Rightarrow x=\frac{-ab[a+b]}{2[ab+a^2+b^2]}$