Help solving $1 < \frac{x + 3}{x - 2} < 2$ I worked a lot of inequalities here in MSE and that greatly helped me.
I've seen a similar inequality [here] [1], but the one I have today is significantly different, in that I'll end up with a division by 0, which is not possible.
[1] [Simple inequality
$$1 < \frac{x + 3}{x - 2} < 2$$
$$\implies 1 < \frac{x - 2 + 2 + 3}{x - 2}$$
$$\implies 1 < 1 + \frac{5}{x - 2} < 2$$
$$\implies 0 <  \frac{5}{x - 2} < 1$$
$$\implies 0 <  \frac{1}{x - 2} < \frac{1}{5}$$
I can't continue without falling into the division by 0. So my many thanks for your help.

Break it into two separate inequalities when you get to here:
$$ 0 < \frac{5}{x-2} < 1$$
First let's consider $0 < \dfrac{5}{x-2}$.  Since $5 > 0$, this inequality is satisfied when $x-2 > 0$, i.e., when $x > 2$.
Now let's consider $\dfrac{5}{x-2} < 1$.  Subtract $1$ from both sides and get a common denominator to get $$\dfrac{5}{x-2} - \dfrac{x-2}{x-2} < 0.$$
Combine and we have $$ \dfrac{5 - (x-2)}{x-2} < 0.$$
Simplify to get $$\frac{7-x}{x-2} < 0.$$
This inequality is satisfied when either of the following two conditions hold:


*

*$7-x > 0$ and $x-2 < 0$

*$7-x < 0$ and $x-2 > 0$


Can you take it from here?
