Question

Solve for x=1/a+b+x=1/a+a/b+1/x

Answer 1

$\dfrac{1}{a+b+x}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x}\\ \\ \\ \implies \dfrac{1}{a+b+x}-\dfrac{1}{x}=\dfrac{1}{a}+\dfrac{1}{b}\\ \\ \\ \implies \dfrac{x-a-b-x}{x(a+b+x)}=\dfrac{a+b}{ab}\\ \\ \\ \implies\dfrac{-(a+b)}{x(a+b+x)}=\dfrac{a+b}{ab}\\ \\ \\ \implies \dfrac{(-1)}{x(a+b+x)}=\dfrac{1}{ab}\\ \\ \\ \implies x(a+b+x)=-ab\\ \\ \\ \implies x(a+b+x)+ab=0\\ \\ \\ \implies x^{2}+(a+b)x +ab =0\\ \\ \\(Applying\:identity:-)\\ \\ \boxed{\boxed{(x+y)(x+z)=x^{2}+(y+z)x+yz}}\\\\ \\ \implies (x+a)(x+b) = 0$

$\rule{200}{2}$

$\rule{200}{2}$

On comparing the quantities we get :-

(x+a)=0

x =(-a)

(x+b)=0

x=(-b)

$\rule{200}{2}$

Answer 2

please see the attachment