Question

solve for x:
$\frac{1}{a + b + c } = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$
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Answer 1

Answer:

Given :

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

Now take 1 / x in L.H.S. side :

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

Taking L.C.M and solving it further :

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

$\displaystyle{\dfrac{-1}{(a+b+x)(x)} = \dfrac{1}{ab} }\\\\\\\displaystyle{\dfrac{-ab}{(a+b+x)(x)} = 1}\\\\\\\displaystyle{(a+b+x)(x)+ab=0}$

x² + a x + b x + a b = 0

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

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

x + a = 0  or x + b = 0

x = - a or x = - b .

Therefore , the value of x is - a or - b .