Question

Solve the following quadratic equations by factorization:

$3\left(\frac{3x-1}{2x+3}\right)-2\left(\frac{2x+3}{3x-1}\right)=5;x\ne \frac{1}{3},-\frac{3}{2}$

Answer 1

$$\begin{gathered} 3\left(\frac{3x-1}{2x+3}\right)-2\left(\frac{2x+3}{3x-1}\right)=5 \\ \Rightarrow \frac{3(3x-1{)}^2-2{\left(2x+3\right)}^2}{\left(2x+3\right)\left(3x-1\right)}=5 \\ \Rightarrow \frac{3\left(9x^2+1-6x\right)-2\left(4x^2+9+12x\right)}{6x^2-2x+9x-3}=5 \\ \Rightarrow \frac{27x^2+3-18x-8x^2-18-24x}{6x^2+7x-3}=5 \\ \Rightarrow 19x^2-42x-15=5\left(6x^2+7x-3\right) \\ \Rightarrow 19x^2-42x-15=30x^2+35x-15 \\ \Rightarrow 30x^2-19x^2+35x+42x-15+15=0 \\ \Rightarrow 11x^2+77x=0 \\ \Rightarrow x^2+7x=0 \\ \Rightarrow x\left(x+7\right)=0 \\ \Rightarrow x=0\mathrm{or}x+7=0 \\ \Rightarrow x=0\mathrm{or}x=-7 \end{gathered}$$

Hence, the factors are 0 and −7.