# 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. :-/

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For case 2, you should not change the denominator to $-(x-2)$. Instead you proceed as follows: $$\dfrac{2x}{x-2} > 1$$ $$2x < x-2$$ You change the inequality, as you know the denominator is negative. However, you do not introduce a negative sign. You wil have $x<-2$ as desired.

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That easy, huh? I got to remember this... I think I was still influenced under the spell of absolute values since yesterday. I need to separate them in my mind! Thanks for a quick answer. Problem solved. :) – matfor Sep 9 '12 at 11:59

Another common approach is to reduce your inequality to something of the form $\displaystyle\frac{N(x)}{D(x)}\lesseqgtr 0$.
You then have to study the sign of the numerator and the sign of the denominator, i.e. solving some non fractional inequalities. Finally, you can study the sign of the ratio of $N(x)$ and $D(x)$ and answer properly to the problem.

In your example, you obtain the equivalent inequality $\displaystyle \frac{x+2}{x-2} > 0$. After having ruled out the value $x=2$, you solve:

• $x+2 > 0 \Longrightarrow x>-2$, i.e. your numerator is positive iff $x>-2$;
• $x-2 > 0 \Longrightarrow x>2$, i.e. your denominator is positive iff $x>2$.

The conclusion follows, since

• in the interval $(-\infty,-2)$ both $x+2$ and $x-2$ are negative, so their ratio is positive;
• in the interval $(-2,2)$ numerator and denominator have opposite sign, so their ratio is negative;
• in the interval $(2,\infty)$ both $x+2$ and $x-2$ are positive, so their ratio is positive.
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Note also that the method of multiplying through by the non-negative $(x-2)^2$ - as advertised in a wrongly worked through post on which I was trying to comment before it was deleted - does work as follows $$2x(x-2)>(x-2)^2$$ which gives $$x^2-4>0$$ ie $$(x+2)(x-2)>0$$ which is equivalent to the advertised answer.

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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:

$$\frac{2x}{x-2} > 1 \iff x \in (-\infty,-2) \, \cup \, (2,\infty) \, .$$

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Others have done a good job of locating your mistake. But no one's posted this frequently seen other method, so I'll do that.

You have $$\frac{2x}{x-2}>1.$$ This becomes $$\frac{2x}{x-2}-1>0.$$ The common denominator is $x-2$: $$\frac{2x}{x-2} - \frac{x-2}{x-2} > 0.$$ Simplify: $$\frac{x+2}{x-2}>0.$$ A quotient is positive if and only if the numerator and denominator are either both positive or both negative. They're both positive when $x>2$ and $x>-2$, thus when $x>2$. They're both negative when $x<2$ and $x<-2$, thus when $x<-2$. So the solution is $x>2$ or $x<-2$.

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