Question
Solve:
$\left(\frac{2 x-3}{x-1}\right)-4\left(\frac{x-1}{2 x-3}\right)=3$

Answer

$\left(\frac{2 x-3}{x-1}\right)-4\left(\frac{x-1}{2 x-3}\right)=3$
Let $\frac{2 x -3}{ x -1}= y$
Then $y -\frac{4}{ y }=3$
$\Rightarrow \frac{y^2-4}{y}=3$
$\Rightarrow y^2 - 4 = 3y$
$\Rightarrow y^2 - 3y - 4 = 0$
$\Rightarrow y^2 - 4y + y - 4 = 0$
$\Rightarrow y( y -4) +1 ( y -4) = 0$
$\Rightarrow ( y -4) (y + 1) = 0$
$If y -4 = 0$ or $y + 1 = 0$
Then $y = 4$ or $y = -1$
$\Rightarrow \frac{2 x-3}{x-1}=4$ or $\frac{2 x-3}{x-1}=-1$
$\Rightarrow 4x -4 = 2x -3$ or $2x -3 = -x +1$
$\Rightarrow 2x = 1$ or $3x = 4$
$\Rightarrow x =\frac{1}{2}$ or $x =\frac{4}{3}=1 \frac{1}{3}$

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