Solving $\frac{2x -1}{1 - x} \ge \frac{3x + 2}{1 - 2x}$ Can someone please give me a tip on how to solve the inequality in the title? I know that $x \ne 1$ and $x \ne \frac{1}{2}$, and I also know that I cannot cross-multiply since I don't know the sign of the factors. What else should I try?
 A: $$\frac{2x -1}{1 - x} \ge \frac{3x + 2}{1 - 2x} \iff \frac{2x - 1}{1-x} - \frac{3x + 2}{1-2x} \geq 0 \\ \\ \iff \frac{(1-2x)(2x-1) - (1-x)(3x+2)}{(1-x)(1-2x)}\geq 0$$
Now, expand the numerator, and then factor if possible. 
$$ \frac{(1-2x)(2x-1) - (1-x)(3x+2)}{(1-x)(1-2x)}\geq 0 \iff \frac{-(4x^2 - 4x+1) + 3x^2 -x -2}{(x-1)(2x-1)}\geq 0\\ \\ \iff \frac {-x^2 +3x -3}{(x - 1)(2x-1)} = \frac{(x^2 - 3x +3)}{(1-x)(2x-1)}\geq 0$$
Examine the signs of the factors on the left that yield a positive result, and determine which values of $x$ satisfy the inequality.
A: Move everything to one side so that you can compare against $0$. You will need to do the fraction math to combine both terms into one fraction. Simplify, including factoring if possible, the numerator but leave the denominator factored. 
A: First of all, you should know which point are impossible. As you said, $x\ne 1$ and $x\ne\frac{1}{2}$. Then, You should change inequality to form $f(x)\geq0$. So,  $\frac{2x -1}{1 - x} - \frac{3x + 2}{1 - 2x}\geq0$ and it implies that $\frac{(1-2x)(2x-1) - (1-x)(3x+2)}{(1-x)(1-2x)}\geq 0$. Now you have two cases:
Case (I): $(1-2x)(2x-1) - (1-x)(3x+2)\geq 0$ and $(1-x)(1-2x)> 0$.
Case(II): $(1-2x)(2x-1) - (1-x)(3x+2)\leq 0$ and $(1-x)(1-2x)< 0$.
After you should determine which values of $x$ satisfy these cases.
