Question

Solve the following equations:
$\frac{2x-3}{x-1}+1=\frac{6x-x^2-6}{x-1}$

Answer 1

We have

$\frac{2x-3}{x-1}+1=\frac{6x-x^2-6}{x-1}$

$\Rightarrow \frac{6x-x^2-6}{x-1}$-$\frac{2x-3}{x-1}=1$

$\Rightarrow \frac{6x-x^2-6-2x+3}{x-1}=1$

$\Rightarrow -x^2+4x-3=x-1$

$\Rightarrow x^2-3x+2=0$

$\Rightarrow x^2-x-2x+2=0$

$\Rightarrow x(x-1)-2(x-1)=0$

$\Rightarrow (x-1)(x-2)=0$

$\therefore x=1,2$