Q) Solve the following quadratic equations by factorization method: $\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x},a+b\ne0$

$\frac{1}{a+b+x}-\frac{1}{x}=\frac{1}{a}+\frac{1}{b}$ $\frac{x-(a+b+x)}{x(a+b+x)}=\frac{a+b}{ab}$ $\frac{-(a+b)}{x(a+b+x)}=\frac{a+b}{ab}$ $-ab(a+b)=(a+b)x(a+b+x)$ $-ab\cancel{(a+b)}=\cancel{(a+b)}x(a+b+x)$ $-ab=x(a+b+x)$ $x(a+b+x)+ab=0$ $ax+bx+x^2+ab=0$ $x^2+ax+bx+ab=0$ $x(x+a)+b(x+a)=0$ $(x+a)(x+b)=0$ $x+a=0\implies x=-a, ~or,$ $x+b=0\implies x=-b$ Please do press "Heart" button if you liked the solution.