Question

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

Answer 1

$\dfrac{1}{a + b + x} = \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{x}$

$\dfrac{1}{a + b + x} - \dfrac{1}{x} = \dfrac{1}{a} + \dfrac{1}{b}$

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

$\dfrac{ - \cancel{ ( a + b)}}{ax + bx + {x}^{2} } = \dfrac{ \cancel{(a + b)}}{ab}$

$- ab = {x}^{2} + ax + bx$

${x}^{2} + ax + bx + ab = 0$

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

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

$\implies \boxed{ x = - a, - b}$

Answer 2

Answer

We have,

$\dashrightarrow \sf \dfrac{1}{(a + b + x)} = \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{x}$

$\dashrightarrow \sf \dfrac{1}{(a + b + x)} - \dfrac{1}{x} = \dfrac{1}{a} + \dfrac{1}{b}$

$\dashrightarrow \sf \dfrac{x - (a + b + x)}{x (a + b + x)} = \dfrac{b + a}{ab}$

$\dashrightarrow \sf \dfrac{- (a + b)}{x (a + b + x)} = \dfrac{a + b}{ab}$

On dividing both sides by (a + b)

$\sf \dashrightarrow \dfrac{-1}{x (a + b + x)} = \dfrac{1}{ab}$

By cross multiplication,

$\dashrightarrow \sf x (a + b + x) = -ab$

$\dashrightarrow \sf {x}^{2} + ax + bx + ab = 0$

$\dashrightarrow \sf x (x + a) + b (x + a) = 0$

$\dashrightarrow \sf (x + a) (x + b) = 0$

$\dashrightarrow \sf x + a = 0 \: \: or \: \: x + b = 0$

$\sf \dashrightarrow x = -a \: \: or \: \: x = -b$

Thus, -a and -b are the roots of the given equation.