Question

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

Answer 1

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

Putting ${\displaystyle \frac{4x-3}{2x+1}=y}$, We get: 

${\displaystyle y-\frac{10}{y}=3}$ 

⇒${\displaystyle \frac{y^2-10}{y}=3}$ 

⇒${\displaystyle y^2-10=3y}$              [On cross multiplying] 

⇒${\displaystyle y^2-3y-10=0}$ 

⇒${\displaystyle y^2-(5-2)y-10=0}$ 

⇒${\displaystyle y^2-5y+2y-10=0}$ 

⇒${\displaystyle y(y-5)+2(y-5)=0}$ 

⇒${\displaystyle (y-5)(y+2)=0}$ 

⇒${\displaystyle y-5=0 \text{or} y+2=0}$ 

⇒${\displaystyle y=5 \text{or} y=-2}$ 

Case I : 

If y=5, We get: 

${\displaystyle \frac{4x-3}{2x+1}=5}$  

⇒${\displaystyle 4x-3=5(2x+1)}$                      [On cross multiplying] 

⇒${\displaystyle 4x-3=10x+5}$ 

⇒${\displaystyle -6x=8}$ 

⇒${\displaystyle -6x=8}$ 

⇒${\displaystyle x=\frac{8}{6}}$ 

⇒${\displaystyle x=-\frac{4}{3}}$ 

Case II 

If y=-2, We get : 

${\displaystyle \frac{4x-3}{2x+1}=2}$ 

⇒${\displaystyle 4x-3=-2(2x+1)}$ 

⇒${\displaystyle 4x-3=-4x-2}$ 

⇒${\displaystyle 8x=1}$ 

⇒${\displaystyle x=\frac{1}{8}}$ 

Hence, the roots of the equation are ${\displaystyle -\frac{4}{3}}$ and ${\displaystyle \frac{1}{8}}$