I'm trying to solve
$$\frac{2x}{x-2}>1$$
but I can't seem to get the correct answer. I'm doing something wrong but I don't know what; that is why I'm asking. This is what I've got:
$$\frac{2x}{x-2}>1$$
Since we do not know if the denominator is positive or negative, we can't multiply both sides with the expression $x-2$. Instead we solve for 2 cases (and $x \not= 2$):
Case 1: $x-2 > 0$, so $x > 2$.
$$\frac{2x}{x-2}>1$$ $$2x > x-2$$ $$x > -2$$
Case 2: $x-2 < 0$, so $x < 2$.
$$\frac{2x}{-(x-2)}>1$$ $$2x < 2-x$$ $$x < \frac{2}{3}$$
For case 1 we get the inequality $x > 2$ and $x > -2$. This simplifies to $x > 2$.
For case 2 we get the inequality $x < 2$ and $x < \frac{2}{3}$. This simplifies to $x < \frac{2}{3}$.
Combining our 2 cases we get the answer: $x < \frac{2}{3}$ or $x > 2$.
Which unfortunately is wrong! The answer should be $x < -2$ or $x > 2$.
I'm guessing case 2 is flawed. I have tried different ways doing it, but I never got the correct answer. In my answer above, I wrote case 2 as I wrote it during my first attempt at this. As I said though, I did try other ways too. :-/