Question

Solve for x: $\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x};(x\ne 0)$

Answer 1

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

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

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

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

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

$\Rightarrow -ab=x^2+ax+bx$

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

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

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

$\Rightarrow x=-aor-b$