Solve inequality: $-5 < \frac{1}{x} < 0$ 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?
 A: 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 $-5$.
A: 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: $x\in(-\infty,0)\cap(-\infty,-\frac15)=(-\infty,-\frac15)$.
A: 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}$. 
