Solving inequality $\frac{2x}{x-2}>1$ 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. :-/

When solving fractional inequalities, the first step is to remove the denominator. We could multiply through by $x-2$ but then we would need to think about the sign of $x-2$ and how the $x-2 < 0$ and $x - 2 > 0$ cases effect things. Instead, we could multiply through by $(x-2)^2$ because $(x-2)^2 \ge 0.$ Doing this gives:
$$ 2x(x-2) > (x-2)^2 \iff 2x^2 - 4 > x^2 - 4x + 4 \iff (x-2)(x+2) > 0 .$$
We see that $(x-2)(x+2) > 0$ if and only if $x-2$ and $x+2$ have the same signs. Clearly $x = -2$ and $x = 2$ are important values. In the region $(-\infty,-2)$ we have $x\pm 2<0$. In the region $(-2,2)$ we have $x-2 < 0$ and $x+2 > 0$, in the region $(2,\infty)$ we have $x \pm 2 > 0$. It follows that:
$$ \frac{2x}{x-2} > 1 \iff x \in (-\infty,-2) \, \cup \, (2,\infty)  \, . $$
