Question

Solve the following quadratic equations by factorization:

$\frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3};x\ne 3,5$

Answer 1

$$\begin{gathered} \frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3} \\ \Rightarrow \frac{x-2}{x-3}-\frac{10}{3}=-\frac{x-4}{x-5} \\ \Rightarrow \frac{3\left(x-2\right)-10\left(x-3\right)}{3\left(x-3\right)}=-\frac{x-4}{x-5} \\ \Rightarrow \frac{3x-6-10x+30}{3x-9}=-\frac{x-4}{x-5} \\ \Rightarrow -\frac{7x-24}{3x-9}=-\frac{x-4}{x-5} \\ \Rightarrow \left(7x-24\right)\left(x-5\right)=\left(3x-9\right)\left(x-4\right) \\ \Rightarrow 7x^2-59x+120=3x^2-21x+36 \\ \Rightarrow 4x^2-38x+84=0 \\ \Rightarrow 2x^2-19x+42=0 \\ \Rightarrow 2x^2-12x-7x+42=0 \\ \Rightarrow 2x(x-6)-7(x-6)=0 \\ \Rightarrow (2x-7)(x-6)=0 \\ \Rightarrow 2x-7=0\mathrm{or}x-6=0 \\ \Rightarrow x=\frac{7}{2}\mathrm{or}x=6 \end{gathered}$$

Hence, the factors are 6 and $\frac{7}{2}$.