question_answer

Find x in terms of a, b and c: \[\frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c},x\ne a,b,c\]

Answer:

We have, \[\frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c},x\ne a,b,c\]

\[a(x-b)(x-c)+b(x-a)(x-c)=2c(x-a)(x-b)\]

\[a({{x}^{2}}-bx-cx+bc)+b({{x}^{2}}-ax-cx+ac)=2c({{x}^{2}}-ax-bx+ab)\]

\[a{{x}^{2}}-abx-acx+abc+b{{x}^{2}}-abx-bcx+abc=2c{{x}^{2}}-2acx-2bcx+2abc\]

\[a{{x}^{2}}+b{{x}^{2}}-2abx-acx-bcx+2abc=2c{{x}^{2}}-2acx-2bcx+2abc\]

\[a{{x}^{2}}+b{{x}^{2}}-2c{{x}^{2}}-2abx-acx-bcx+2acx+2bcx=0\]

\[(a+b-2c){{x}^{2}}+(-2ab+ac+bc)x=0\]

\[x[(a+b-2c)x+(ac+bc-2ab)]=0\]

\[x=0,-\frac{(ac+bc-2ab)}{a+b-2c}\]