Question

solve for x , 1\a+b+c = 1\a+ 1\b+ 1\x . hint a+b+ x not equal to 0​

Answer 1

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

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Here is the answer .Hope its clear

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iamthekingdudefun

iamthekingdudefun Ambitious

ON solving the above equation we have a quadratic equation

(a+b)x^2+(a+b)^2x+ab(a+b)

on solving the above equation by formula -b+root over(b^2-4ac) and -b-root over(b^2-4ac)

we got -a and -b as roots.

According to me this is much easier then others when compared.

Answer 2

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 .