Question

Solve the following quadratic equations by factorization:

$\frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3};x\ne 5,7$

Answer 1

$$\begin{gathered} \frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3} \\ \Rightarrow \frac{\left(x-4\right)\left(x-7\right)+\left(x-6\right)\left(x-5\right)}{\left(x-5\right)\left(x-7\right)}=\frac{10}{3} \\ \Rightarrow \frac{x^2-11x+28+x^2-11x+30}{x^2-12x+35}=\frac{10}{3} \\ \Rightarrow \frac{2x^2-22x+58}{x^2-12x+35}=\frac{10}{3} \\ \Rightarrow 3\left(2x^2-22x+58\right)=10\left(x^2-12x+35\right) \\ \Rightarrow 6x^2-66x+174=10x^2-120x+350 \\ \Rightarrow 4x^2-54x+176=0 \\ \Rightarrow 2x^2-27x+88=0 \\ \Rightarrow 2x^2-11x-16x+88=0 \\ \Rightarrow x\left(2x-11\right)-8\left(2x-11\right)=0 \\ \Rightarrow \left(x-8\right)\left(2x-11\right)=0 \\ \Rightarrow x-8=0\mathrm{or}2x-11=0 \\ \Rightarrow x=8\mathrm{or}x=\frac{11}{2} \end{gathered}$$

Hence, the factors are 8 and $\frac{11}{2}$.