Question

$2\left(\frac{x-1}{x+3}\right)-7\left(\frac{x+3}{x-2}\right)=5,(x\ne -3,1)$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \text{2}\left(\frac{x-1}{x+3}\right)-7\left(\frac{x+3}{x-1}\right)=5 \\ \mathrm{P}\text{utting}\frac{x-1}{x+3}=y,\mathrm{we}\mathrm{get}: \\ 2y-\frac{7}{y}=5 \\ \Rightarrow \frac{2y^2-7}{y}=5 \\ \Rightarrow 2y^2-7=5y \\ \Rightarrow 2y^2-5y-7=0 \\ \Rightarrow 2y^2-(7-2)y-7=0 \\ \Rightarrow 2y^2-7y+2y-7=0 \\ \Rightarrow y(2y-7)+1(2y-7)=0 \\ \Rightarrow (2y-7)(y+1)=0 \\ \Rightarrow 2y-7=0\text{or}y+1=0 \\ \Rightarrow y=\frac{7}{2}\text{or}y=-1 \\ \text{Case I} \\ \mathrm{If}y=\frac{7}{2},\mathrm{we}\mathrm{get}: \\ \frac{x-1}{x+3}=\frac{7}{2} \\ \Rightarrow \text{2(}x-1)=7(x+3)\left[\mathrm{On}\text{cross multiplying}\right] \\ \Rightarrow \text{2}x-2=7x+21 \\ \Rightarrow -5x=23 \\ \Rightarrow x=\frac{-23}{5} \\ \text{Case II} \\ \mathrm{If}y=-1,\mathrm{we}\mathrm{get}: \\ \frac{x-1}{x+3}\text{=}-\text{1} \\ \Rightarrow x-1=-1(x+3) \\ \Rightarrow x-1=-x-3 \\ \Rightarrow \text{2}x=-2 \\ \Rightarrow x=-1 \\ \text{Hence, the roots of the equation are}\frac{-23}{5}\text{and}-\text{1.} \end{gathered}$$