Question

Solve the following quadratic equations by factorization:

$\frac{2}{x+1}+\frac{3}{2(x-2)}=\frac{23}{5x};x\ne 0,-1,2$

Answer 1

$$\begin{gathered} \frac{2}{x+1}+\frac{3}{2(x-2)}=\frac{23}{5x} \\ \Rightarrow \frac{4(x-2)+3(x+1)}{2(x-2)(x+1)}=\frac{23}{5x} \\ \Rightarrow \frac{4x-8+3x+3}{2(x^2+x-2x-2)}=\frac{23}{5x} \\ \Rightarrow \frac{7x-5}{2x^2-2x-4}=\frac{23}{5x} \\ \Rightarrow 5x\left(7x-5\right)=23\left(2x^2-2x-4\right) \\ \Rightarrow 35x^2-25x=46x^2-46x-92 \\ \Rightarrow 46x^2-35x^2-46x+25x-92=0 \\ \Rightarrow 11x^2-21x-92=0 \\ \Rightarrow 11x^2-44x+23x-92=0 \\ \Rightarrow 11x(x-4)+23(x-4)=0 \\ \Rightarrow (11x+23)(x-4)=0 \\ \Rightarrow 11x+23=0\mathrm{or}x-4=0 \\ \Rightarrow x=-\frac{23}{11}\mathrm{or}x=4 \end{gathered}$$

Hence, the factors are 4 and $-\frac{23}{11}$.