Solve inequality: $-5 < \frac{1}{x} < 0$

I thought about how I can solve this. If I multiply all sides by $x$ I'm afraid I'm removing the answer, cause $\frac{x}{x}=1$. And when $x$ 'leaves' the inequality I'm left with no letter.

How do I get just $x$ in the middle without adding $x$ to other sides or removing $x$?
I then saw that: $\frac{1}{x} = -x$. So can I multiply all sides with $-1$. This also changes the signs. So I'm left with: $5> x > 0$.

Is this correct? If not what did I do wrong?


Since you know that $x\neq 0$ you can multiply by $x$. Just remember that $x<0$. So, multiplying by $x$ reverses the inequalities.

Then you get $-5x>1>0$. This gives $-5x>1$. Now divide by $-5$.

  • $\begingroup$ But doesn't that remove x and get x's to the sides? $\endgroup$ – hiisitmeyoulookingfor Jul 26 '14 at 16:12
  • $\begingroup$ @hiisitmeyoulookingfor I edited the answer with more information. $\endgroup$ – Joe Johnson 126 Jul 26 '14 at 16:14
  • $\begingroup$ I know the answer is 1/-5 and the sign change now, but I don't understand where that 1 comes from and were the 0 is left. $\endgroup$ – hiisitmeyoulookingfor Jul 26 '14 at 16:16
  • $\begingroup$ I get it now, thank you sir. $\endgroup$ – hiisitmeyoulookingfor Jul 26 '14 at 16:19

You don't have to necessarily put $x$ to the center directly. One way to solve such double inequality is to split it to a system of two inequalities:

$$\begin{cases}-5<\frac1x\\ \frac1x<0\end{cases}$$

From the second one you get $x<0$, and from the first one you can obtain $x<-\frac15$ (not forgetting that $x<0$).

Now the answer is intersection of the two intervals: $x\in(-\infty,0)\cap(-\infty,-\frac15)=(-\infty,-\frac15)$.

  • 1
    $\begingroup$ Are you serious? I didn't know you could split the x! This is amazing. I think that makes it really easy! $\endgroup$ – hiisitmeyoulookingfor Jul 26 '14 at 16:21
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    $\begingroup$ @hiisitmeyoulookingfor $a\lt b\lt c$ is really just an abuse of notation for $a\lt b$ and $b\lt c$. ;) $\endgroup$ – Hakim Jul 26 '14 at 16:23
  • $\begingroup$ It makes sense, thanks for explaining! $\endgroup$ – hiisitmeyoulookingfor Jul 26 '14 at 16:24
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    $\begingroup$ I believe you should get $x<-\frac{1}{5}$ from the first inequality, not $x>-\frac{1}{5}$. $\endgroup$ – Jam Jul 26 '14 at 16:29
  • $\begingroup$ @Ruslan You multiply by the $-5$ as well so the sign should switch twice. $\endgroup$ – Jam Jul 26 '14 at 16:31

Note that given $-5\lt \frac 1x \lt 0$, we know that $$\frac 1x < 0 \implies x< 0$$. So when you multiply by $x$ to remove it from the denominator, you need to reverse the directions of the inequalities.

$$-5 \lt \frac 1x \iff -5x \gt 1\iff x \lt -\frac 15 $$ and $$\frac 1x <0 \iff x\lt 0$$ The second inequality we already know, and so the first inequality is the stricter of the two that must be met for both inequalities to hold. $ x\lt \dfrac{-1}{5}$.

  • $\begingroup$ It did reverse them. But doesn't multiplying by x gives me x/x=1 and destroys the inequality. I also get x's on the side then. Or do I see it wrong? $\endgroup$ – hiisitmeyoulookingfor Jul 26 '14 at 16:14
  • $\begingroup$ Ah very clear, thank you sir. $\endgroup$ – hiisitmeyoulookingfor Jul 26 '14 at 16:23
  • $\begingroup$ You're welcome. $\endgroup$ – Namaste Jul 26 '14 at 16:26

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