Question

Solve each of the following quadratic equations:

$\frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1},x\ne -1,\frac{1}{3}$ [CBSE 2014]

Answer 1

$$\begin{gathered} \frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1},x\ne -1,\frac{1}{3} \\ \Rightarrow \frac{3}{x+1}-\frac{2}{3x-1}=\frac{1}{2} \\ \Rightarrow \frac{9x-3-2x-2}{\left(x+1\right)\left(3x-1\right)}=\frac{1}{2} \\ \Rightarrow \frac{7x-5}{3x^2+2x-1}=\frac{1}{2} \end{gathered}$$
$$\begin{gathered} \Rightarrow 3x^2+2x-1=14x-10\left(\mathrm{Cross}\mathrm{multiplication}\right) \\ \Rightarrow 3x^2-12x+9=0 \\ \Rightarrow x^2-4x+3=0 \\ \Rightarrow x^2-3x-x+3=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x\left(x-3\right)-1\left(x-3\right)=0 \\ \Rightarrow \left(x-3\right)\left(x-1\right)=0 \\ \Rightarrow x-3=0\mathrm{or}x-1=0 \\ \Rightarrow x=3\mathrm{or}x=1 \end{gathered}$$
Hence, 1 and 3 are the roots of the given equation.