Question

factorise a/x-a +b/x-b = 2c/x-c

Answer 1

$\textbf{Given:}$

$\dfrac{a}{x-a}+\dfrac{b}{x-b}=\dfrac{2c}{x-c}$

$\textbf{To find:}$

$\text{Factors of \dfrac{a}{x-a}+\dfrac{b}{x-b}=\dfrac{2c}{x-c} }$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{a}{x-a}+\dfrac{b}{x-b}=\dfrac{2c}{x-c}$

$\dfrac{a(x-b)+b(x-a)}{(x-a)(x-b)}=\dfrac{2c}{x-c}$

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

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

$((a+b)x-2ab)(x-c)=2c(x^2-(a+b)x+ab)$

$(a+b)x^2-2ab\,x-c(a+b)x+2abc=2c\,x^2-2c(a+b)x+2abc$

$(a+b)x^2-2ab\,x-c(a+b)x=2c\,x^2-2c(a+b)x$

$\text{Rearranging terms we get}$

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

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

$(a+b-2c)x^2-(2ab-ac-bc)x=0$

$x((a+b-2c)x-(2ab-ac-bc))=0$

$\implies\,x=0\;\text{(or)}\;x=\dfrac{2ab-ac-bc}{a+b-2c}$

$\textbf{Answer:}$

$\textbf{Factors of \dfrac{a}{x-a}+\dfrac{b}{x-b}=\dfrac{2c}{x-c} are x and (a+b-2c)x-(2ab-ac-bc) }$

$\textbf{Solution set is}\bf\;\{0,\dfrac{2ab-ac-bc}{a+b-2c}\}$

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