abs((2x-1)/(x-1)) = 3.
I'm trying to figure out what to do and how to finish it since the absolute value is around both (2x+1) and (x-1).
Hi AlphaRock,
$\displaystyle \left|\frac{2x-1}{x-1}\right|=3$
Let's consider case 1.
$\displaystyle \frac{2x-1}{x-1}=3$
$\displaystyle 2x-1=3(x-1)$
$\displaystyle 2x-1=3x-3$
$\displaystyle \boxed{x=2}$
Check this solution in your original absolute value equation.
Now consider case 2.
$\displaystyle \frac{2x-1}{x-1}=-3$
$\displaystyle 2x-1=-3(x-1)$
$\displaystyle 2x-1=-3x+3$
$\displaystyle \boxed{x=\frac{4}{5}}$
Check this solution in your original absolute value equation.