Question

Solve for $x$.

If $\frac{1}{a}+\frac{1}{b}+\frac{1}{x}=\frac{1}{a+b+x};$where $a\ne 0,b\ne 0,x\ne 0.$

Answer 1

Given,

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

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

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

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

$\Rightarrow -[a+b]\left[ab\right]=[a+b][x^2+ax+bx]$

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

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

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

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

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

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

$x+b=0$ and $x+a=0$

$x=-b$ and $x=-a$.