Question

Solve each of the following quadratic equations:

$\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3},x\ne 2,4$ [CBSE 2010]

Answer 1

$$\begin{gathered} \frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3},x\ne 2,4 \\ \Rightarrow \frac{\left(x-1\right)\left(x-4\right)+\left(x-2\right)\left(x-3\right)}{\left(x-2\right)\left(x-4\right)}=\frac{10}{3} \\ \Rightarrow \frac{x^2-5x+4+x^2-5x+6}{x^2-6x+8}=\frac{10}{3} \\ \Rightarrow \frac{2x^2-10x+10}{x^2-6x+8}=\frac{10}{3} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{x^2-5x+5}{x^2-6x+8}=\frac{5}{3} \\ \Rightarrow 3x^2-15x+15=5x^2-30x+40 \\ \Rightarrow 2x^2-15x+25=0 \\ \Rightarrow 2x^2-10x-5x+25=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 2x\left(x-5\right)-5\left(x-5\right)=0 \\ \Rightarrow \left(x-5\right)\left(2x-5\right)=0 \\ \Rightarrow x-5=0\mathrm{or}2x-5=0 \\ \Rightarrow x=5\mathrm{or}x=\frac{5}{2} \end{gathered}$$
Hence, 5 and $\frac{5}{2}$ are the roots of the given equation.