Question

$\left(\frac{4x-3}{2x+1}\right)-10\left(\frac{2x+1}{4x-3}\right)=3,\left(x\ne \frac{-1}{2},\frac{3}{4}\right)$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \left(\frac{4x-3}{2x+1}\right)-10\left(\frac{2x+1}{4x-3}\right)=3 \\ \mathrm{P}\text{utting}\frac{4x-3}{2x+1}=y,\text{we get:} \\ y-\frac{10}{y}=3 \\ \Rightarrow \frac{y^2-10}{y}=3 \\ \Rightarrow y^2-10=3y\left[\mathrm{On}\text{cross multiplying}\right] \\ \Rightarrow y^2-3y-10=0 \\ \Rightarrow y^2-(5-2)y-10=0 \\ \Rightarrow y^2-5y+2y-10=0 \\ \Rightarrow y(y-5)+2(y-5)=0 \\ \Rightarrow (y-5)(y+2)=0 \\ \Rightarrow y-5=0\text{or}y+2=0 \\ \Rightarrow y=5\text{or}y=-2 \\ \text{Case I} \\ \mathrm{If}y=5,\mathrm{we}\mathrm{get}: \\ \frac{4x-3}{2x+1}=5 \\ \Rightarrow \text{4}x-3=5(2x+1)\left[\mathrm{On}\text{cross multiplying}\right] \\ \Rightarrow 4x-3=10x+5 \\ \Rightarrow -6x=8 \\ \Rightarrow -\text{6}x=8 \\ \Rightarrow x=-\frac{{\cancel{8}}^4}{{\cancel{6}}^3} \\ \Rightarrow x=-\frac{4}{3} \\ \text{Case II} \\ \mathrm{If}y=-2,\mathrm{we}\mathrm{get}: \\ \frac{4x-3}{2x+1}\text{=}-2 \\ \begin{array}{l}\Rightarrow 4x-3=-2(2x+1)\\ \Rightarrow \text{4}x-3=-4x-2 \\ \Rightarrow \text{8}x=1 \\ \Rightarrow x=\frac{1}{8} \\ \mathrm{Hence},\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}\mathrm{are}-\frac{4}{3}\text{and}\frac{1}{8}.\end{array} \end{gathered}$$