Solving the inequality $−5 < \frac 1 x \le 1$ I can solve for $\frac 1 x \le 1$, but I cannot solve for $−5 < \frac 1 x$. Please help!
 A: My advice: draw the graph of $y = \frac1x$, and look at the places where the $y$-coordinate is between $-5$ and $1$.
This should give you an answer and help you visualize things.  Then if you want to prove the answer algebraically, split into cases $x > 0$ and $x < 0$ as suggested in the comments.  (What about the case $x = 0$?)
A: $$-5<\frac1x\le1\stackrel{\text{mult. by}\;x^2}\iff-5x^2<x\le x^2\iff\begin{cases}-5x^2<x\iff x(5x+1)>0\\{}\\\text{and}\\{}\\x\le x^2\iff x(x-1)\ge 0\end{cases}$$
Checking both parabolas we got above in the rightmost inequalities, we get:
$$\begin{align*}x(5x+1)>0&\iff x<-\frac15\;\;\vee\;\;x>0\\
x(x-1)\ge 0&\iff x\le 0\;\;\vee\;\;x\ge 1\end{align*}$$
The solution to the above "and" inequalities system is $\;x\le-\frac15\;\;\vee\;\;x\ge 1\;$
Observe carefully the first multiplication: this slick trick makes things simpler since then we don't have to distinguish between positive or negative multiplicatives: $\,x^2\,$ is always non-negative.
A: Note that $\frac1x$ will never be zero for any real $x$. Hence, $$-5<\frac1x\le 1\qquad\Longleftrightarrow\qquad-5<\frac1x<0\quad\text{or}\quad0<\frac1x\le 1.$$ In both cases, we can simply multiply by $x,$ bearing in mind the sign of $x$ in each case.
