Question

${\displaystyle \frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{31}{3},x\ne 5,7}$ 

 

Answer 1

${\displaystyle \frac{x-4}{x-5}+\frac{x-6}{x-7}=3\frac{1}{3},x\ne 5,7}$ 

⇒${\displaystyle \frac{(x-4)(x-7)+(x-5)(x-6)}{x^2-12x+35}=\frac{10}{3}}$ 

⇒${\displaystyle \frac{x^2-11x+28+x^2-11x+30}{x^2-12x+35}=\frac{10}{3}}$ 

⇒${\displaystyle \frac{2x^2-22x+58}{x^2-12x+35}=\frac{10}{3}}$ 

⇒${\displaystyle \frac{x^2-22x+29}{x^2-12x+35}=\frac{5}{3}}$ 

⇒${\displaystyle \frac{x^2-11x+29}{x^2-12x+35}=\frac{5}{3}}$ 

⇒${\displaystyle 3x^2-33x+87=5x^2-60x+175}$ 

⇒${\displaystyle 2x^2-27x+88=0}$ 

⇒${\displaystyle 2x^2-16x11x+88=0}$ 

⇒${\displaystyle 2x(x-8)-11(x-8)=0}$ 

⇒${\displaystyle (x-8)(2x-11)=0}$ 

⇒${\displaystyle x-8=0 \text{or} 2x-11=0}$ 

⇒${\displaystyle x=8 \text{or} x=\frac{11}{2}}$ 

Hence, 8 and ${\displaystyle \frac{11}{2}}$ are the roots of the given equation.