Question

Solve each of the following quadratic equations:

$\frac{x-4}{x-5}+\frac{x-6}{x-7}=3\frac{1}{3},x\ne 5,7$ [CBSE 2014]

Answer 1

$$\begin{gathered} \frac{x-4}{x-5}+\frac{x-6}{x-7}=3\frac{1}{3},x\ne 5,7 \\ \Rightarrow \frac{\left(x-4\right)\left(x-7\right)+\left(x-5\right)\left(x-6\right)}{\left(x-5\right)\left(x-7\right)}=\frac{10}{3} \\ \Rightarrow \frac{x^2-11x+28+x^2-11x+30}{x^2-12x+35}=\frac{10}{3} \\ \Rightarrow \frac{2x^2-22x+58}{x^2-12x+35}=\frac{10}{3} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{x^2-11x+29}{x^2-12x+35}=\frac{5}{3} \\ \Rightarrow 3x^2-33x+87=5x^2-60x+175 \\ \Rightarrow 2x^2-27x+88=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 2x^2-16x-11x+88=0 \\ \Rightarrow 2x\left(x-8\right)-11\left(x-8\right)=0 \\ \Rightarrow \left(x-8\right)\left(2x-11\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x-8=0\mathrm{or}2x-11=0 \\ \Rightarrow x=8\mathrm{or}x=\frac{11}{2} \end{gathered}$$
Hence, 8 and $\frac{11}{2}$ are the roots of the given equation.