Question

Solve the following quadratic equations by factorization:

$\frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1},x\ne -1,\frac{1}{3}$

Answer 1

$$\begin{gathered} \frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1} \\ \Rightarrow \frac{6-\left(x+1\right)}{2(x+1)}=\frac{2}{3x-1} \\ \Rightarrow \frac{6-x-1}{2x+2}=\frac{2}{3x-1} \\ \Rightarrow \left(5-x\right)\left(3x-1\right)=2\left(2x+2\right) \\ \Rightarrow 15x-5-3x^2+x=4x+4 \\ \Rightarrow -3x^2+16x-5-4x-4=0 \\ \Rightarrow -3x^2+12x-9=0 \\ \Rightarrow 3x^2-12x+9=0 \\ \Rightarrow x^2-4x+3=0 \\ \Rightarrow x^2-3x-x+3=0 \\ \Rightarrow x(x-3)-1(x-3)=0 \\ \Rightarrow (x-1)(x-3)=0 \\ \Rightarrow x-1=0\mathrm{or}x-3=0 \\ \Rightarrow x=1\mathrm{or}x=3 \end{gathered}$$

Hence, the factors are 3 and 1.