# Solve $\dfrac{x}{x-2}>2$ by first rewriting it in the form $\dfrac{P(x)}{Q(x)}>0$

Edit: So then is this the correct final solution? $x<4,(\infty,4), x\ne2$

I am asked to do this:

Solve $\dfrac{x}{x-2}>2$ by first rewriting it in the form $\dfrac{P(x)}{Q(x)}>0$

$$\dfrac{x}{x-2}>2$$ $$\dfrac{x}{x-2}-\dfrac{2(x-2)}{1(x-2)}>0$$ $$\dfrac{x-2x+4}{x-2}>0$$ $$\dfrac{-1(-x+4)}{x-2}>0$$ $$\dfrac{x-4}{x-2}<0(x-2)$$ $$x-4<0$$ $$x<4$$

• Advice in the future: As a sanity check, try $x=0$. It is greater than $-4$, but $x/(x-2)=0 \le 2$. This means something went wrong. See the answers below for where your mistake is. Commented Jul 9, 2014 at 18:54
• See this solution to the problem
– L.K.
Commented Jul 9, 2014 at 18:59
• would you accept a solution? Commented Sep 2, 2014 at 10:46

Your method is correct. You can perform the following operations:

$$\dfrac{x}{x-2}>2$$ $$\dfrac{x}{x-2}-\dfrac{2(x-2)}{1(x-2)}>0$$

However, your error occurs on the second line:

$$x - 2(x - 2) = x -2x + 4$$

• corrected this. Commented Jul 9, 2014 at 18:56
• @nitrous2 that was the only problem with your question. Everything else was fine. Please upvote. Commented Jul 9, 2014 at 18:57
• So the interval for this inequality would be $(4,\infty)$? Commented Jul 9, 2014 at 19:01

Your method not is correct because: if

$$\frac{4-x}{x-2}>0\equiv-1\cdot\frac{(4-x)}{x-2}<(-1)\cdot0\equiv\frac{x-4}{x-2}<0$$

then $(x-4)>0$ and $(x-2)<0$, or $(x-4)<0$ and $(x-2)>0$

solutions $2<x<4$

Check Second Line$$\cdots-2(x-2)=\cdots-2x+4$$