Simplify the inequality $2<\frac{10}{x}<3$ to an inequality for $x$. I'm not sure what to google search for this.  Let's say I have the expression:
$$ 2 &lt; \frac{10}{x} &lt; 3 $$
We need to see if x is positive or negative.  If positive, we have:
$$ 2x &lt; 10 &lt; 3x $$
If negative we have:
$$ 2x &gt; 10 &gt; 3x $$
It can't be negative since $2x$ can't be greater than $10$.  So $x$ must be positive.
Now, what do I do with $ 2x &lt; 10 &lt; 3x $ to get $x$ in the middle so that it is $x$ not $1/x$.  I am not sure what rule or term to google for.
Thanks!
Note that the inequality $2x&lt;10&lt;3x$ is actually two inequalities:
$$2x&lt;10\qquad\text{ and }\qquad 3x&gt;10.$$
Dividing them by $2$ and $3$, respectively, shows that
$$x&lt;\frac{10}{2}\qquad\text{ and }\qquad x&gt;\frac{10}{3},$$
which can be written more concisely as $\tfrac{10}{3}&lt;x&lt;\tfrac{10}{2}.$

More abstractly, for the function $f(x)=\tfrac1x$ we can write the inequalities as
$$f(\tfrac12)&lt;f(\tfrac{x}{10})&lt;f(\tfrac13).$$
Because $f(x)$ is positive if and only if $x$ is positive, we see that also $\tfrac{x}{10}$ is positive, so $x$ is positive. Moreover, because $f$ is strictly decreasing on the positive numbers, it follows that
$$\tfrac12&gt;\tfrac{x}{10}&gt;\tfrac13,$$
yielding the bounds $\tfrac{10}{3}&lt;x&lt;\tfrac{10}{2}$.
Solution without cases.
We need to solve $$\frac{10-3x}{x}&lt;0$$ and $$\frac{2x-10}{x}&lt;0,$$ which gives
$$x\in(-\infty,0)\cup\left(\frac{10}{3},+\infty\right)$$ and $$x\in(0,5),$$ which gives the answer:$$\left(\frac{10}{3},5\right).$$