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:
$$ 2 < \frac{10}{x} < 3 $$
We need to see if x is positive or negative.  If positive, we have:
$$ 2x < 10 < 3x $$
If negative we have:
$$ 2x > 10 > 3x $$
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 < 10 < 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!
 A: Note that the inequality $2x<10<3x$ is actually two inequalities:
$$2x<10\qquad\text{ and }\qquad 3x>10.$$
Dividing them by $2$ and $3$, respectively, shows that
$$x<\frac{10}{2}\qquad\text{ and }\qquad x>\frac{10}{3},$$
which can be written more concisely as $\tfrac{10}{3}<x<\tfrac{10}{2}.$

More abstractly, for the function $f(x)=\tfrac1x$ we can write the inequalities as
$$f(\tfrac12)<f(\tfrac{x}{10})<f(\tfrac13).$$
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
$$\tfrac12>\tfrac{x}{10}>\tfrac13,$$
yielding the bounds $\tfrac{10}{3}<x<\tfrac{10}{2}$.
A: Since $$\frac{10}x>0 \implies x>0$$
then
$$2 < \frac{10}{x} < 3 \iff 2x<10 \quad \land \quad3x>10 $$
that is
$$ x<5 \quad \land \quad x>\frac{10}3 $$
or $x\in \left(\frac{10}3,5\right)$.
A: Note that $x$ can't be negative since then $10/x$ will be negative and won't lie between $2,3$. You have to inequalities $2x<10$ and $10<3x$ and you are looking for values of $x$ such that both inequalities are satisfied simultaneously. The first gives $x<5$, the second gives $x>10/3$ and since you require $x$ to satisfy both, you take the common values in these two ranges i.e. $\frac{10}3<x<5$.
A: Break it into two problems.
You have $2 < \frac{10} x$ and $x > 0$.
So $2x < 10$ and $x < 5$.
ANd you have $\frac {10}x < 3$ so $10 < 3x$ and $x > \frac {10}3$.
So $\frac {10}3 < x < 5$.
.....
Actually if we know for positive numbers that if $a < b$ then $\frac 1b < \frac 1a$ (which is easy to prove if you can't take it for a given) we can do this in one fell swoop:
$2 < \frac {10}x < 3 \implies$
$\frac 12 > \frac x{10} > \frac 13\implies $
$10\frac 12 > x > 10\frac 13 \implies$
$\frac {10}3 < x < 5$.
A: Solution without cases.
We need to solve $$\frac{10-3x}{x}<0$$ and $$\frac{2x-10}{x}<0,$$ which gives
$$x\in(-\infty,0)\cup\left(\frac{10}{3},+\infty\right)$$ and $$x\in(0,5),$$ which gives the answer:$$\left(\frac{10}{3},5\right).$$
