Question

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

Answer 1

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

On solving this we get

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

$16x^2+9-24x-40x^2-10-40x=3(8x^2-2x-3)$

$16x^2-40x^2-24x^2-24x-40x+6x+9-10+9=0$

$-48x^2-58x+8=0$

$48x^2+58x-8=0$

$\Rightarrow 48x^2+58x-8=0$ On dividing by $2$

$24x^2+29x-4=0$

$x=\frac{-29\pm \sqrt{1225}}{48}=\frac{-29\pm 35}{48}$

$x=\frac{6}{48},\frac{-64}{48}$

Hence values of x are $(\frac{1}{8},\frac{-2}{3})$.