Question

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

Answer 1

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

$\Rightarrow \frac{1}{a+b+x}-\frac{1}{a}=\frac{1}{b}+\frac{1}{x}$

$\Rightarrow \frac{a-(a+b+x)}{a(a+b+x)}=\frac{x+b}{bx}$

$\Rightarrow \frac{-(b+x)}{a(a+b+x)}=\frac{(b+x)}{bx}$

$\Rightarrow -bx=a(a+b+x)$

$\Rightarrow a(a+b+x)+bx=0$

$\Rightarrow a^2+ab+ax+bx=0$

$\Rightarrow a(a+b)+x(a+b)=0$

$\Rightarrow (a+b)(a+x)=0$

Either $a+b=0$ or $a+x=0\Rightarrow x=-a$

Also, if $a+b=0\Rightarrow a=-b\Rightarrow x=-(-b)=b$