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

Solution 1:

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 $-5$.

Solution 2:

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: $x\in(-\infty,0)\cap(-\infty,-\frac15)=(-\infty,-\frac15)$.

Solution 3:

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}$.