Question

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

Answer 1

We have,

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

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

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

$\frac{-(a+b)}{ax+bx+x^2}=\frac{a+b}{ab}$

Therefore,

$\Rightarrow x^2+(a+b)x=-ab$

$\Rightarrow x^2+(a+b)x+ab=0$

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

$x=-a$ or $x=-b$

Hence, this is the answer.