Simplify the inequality $2<\frac{10}{x}<3$ to an inequality for $x$.

Question

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!

Answer 1

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 $\tfrac{10}{3}<x<\tfrac{10}{2}.$

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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 $\tfrac{10}{3}<x<\tfrac{10}{2}$.

Answer 2

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 $x\in \left(\frac{10}3,5\right)$.

Answer 3

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. $\frac{10}3<x<5$.

Answer 4

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 $

$10\frac 12 > x > 10\frac 13 \implies$

$\frac {10}3 < x < 5$.

Answer 5

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:$$\left(\frac{10}{3},5\right).$$