Question

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

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \text{2}\left(\frac{2x-1}{x+3}\right)-3\left(\frac{x+3}{2x-1}\right)=5 \\ \mathrm{P}\text{utting}\frac{2x-1}{x+3}=y,\text{we get:} \\ 2y-\frac{3}{y}=5 \\ \Rightarrow \frac{2y^2-3}{y}=5 \\ \Rightarrow 2y^2-3=5y\left[\mathrm{On}\text{cross multiplying}\right] \\ \Rightarrow 2y^2-5y-3=0 \\ \Rightarrow 2y^2-(6-1)y-3=0 \\ \Rightarrow 2y^2-6y+y-3=0 \\ \Rightarrow 2y(y-3)+1(y-3)=0 \\ \Rightarrow (y-3)(2y+1)=0 \\ \Rightarrow y-3=0\text{or 2}y+1=0 \\ \Rightarrow y=3\text{or}y=-\frac{1}{2} \\ \text{Case I} \\ \mathrm{If}y=3,\mathrm{we}\mathrm{get}: \\ \frac{2x-1}{x+3}=3 \\ \Rightarrow \text{2}x-1=3(x+3)\left[\mathrm{On}\text{cross multiplying}\right] \\ \Rightarrow \text{2}x-1=3x+9 \\ \Rightarrow -x=10 \\ \Rightarrow x=-10 \\ \text{Case II} \\ \mathrm{If}y=-\frac{1}{2},\mathrm{we}\mathrm{get}: \\ \Rightarrow \frac{2x-1}{x+3}\text{=}-\frac{1}{2} \\ \Rightarrow \text{2}(2x-1)=-1(x+3) \\ \Rightarrow \text{4}x-2=-x-3 \\ \Rightarrow \text{5}x=-1 \\ \Rightarrow x=-\frac{1}{5} \\ \text{Hence, the roots of the equation are}-\text{10 and}-\frac{1}{5}\text{.} \end{gathered}$$