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. :-/