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Find *x* in terms of *a*, *b* and *c*: `a/(x-a)+b/(x-b)=(2c)/(x-c)`

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#### Solution

Consider the given equation:

`a/(x-a)+b/(x-b)=(2c)/(x-c)`

Taking the LCM and then cross multiplying, we get

`(a(x-b)+b(x-a))/((x-a)(x-b))=(2c)/(x-c)`

⇒(x−c)[a(x−b)+b(x−a)]=2c(x−a)(x−b)

⇒(x−c)[ax−ab+bx−ab]=2c(x2−bx−ax+ab)

⇒ax^{2}−2abx+bx^{2}−acx+2abc−bcx=2cx^{2}−2bcx−2acx+2abc

⇒ax^{2}+bx^{2}−2cx^{2}=2abx−acx−bcx

⇒(a+b−2c)x^{2}=x(2ab−ac−bc)

⇒(a+b−2c)x^{2}−x(2ab−ac−bc)=0

⇒x[(a+b−2c)x−(2ab−ac−bc)]=0

⇒x=0 or (a+b−2c)x−(2ab−ac−bc)=0

⇒x=0 or (a+b−2c)x=(2ab−ac−bc)

`=>x = 0 `

Thus, the two roots of the given equation are *x = *0 and `(2ab-ac-bc)/(a+b-2c)`

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