Question

Solve for x : 1/(a +b+x) = 1/a + 1/b + 1/x , [ a ≠ 0 , b ≠ 0 , x ≠ 0 , x ≠ - (a + b)]

Answer 1

$\textbf{Given equation is}$

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

$\displaystyle\frac{1}{a+b+x}=\frac{bx+ax+ab}{abx}$

$\displaystyle\frac{1}{a+b+x}=\frac{(a+b)x+ab}{abx}$

$\displaystyle\,abx=(x+(a+b))((a+b)x+ab)$

$\displaystyle\,abx=(a+b)x^2+abx+(a+b)^2x+ab(a+b)$

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

$\text{Divide both sides by a+b, we get}$

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

$\displaystyle\,(x+a)(x+b)=0$

$\implies\displaystyle\bf\,x=-a,-b$

$\therefore\textbf{The solution set is \{-a,-b\} }$

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Answer 2

♤YOUR SOLUTION SET FOR THIS QUESTION SHALL BE {-a,-b} ♤