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.
 A: Great start. Right off the bat, we have to assume $x\neq 2$, otherwise the expression doesn't make sense. 
For $0 < \frac{1}{x-2}$, we just need for the right term to be positive, i.e. for $x-2>0$, and so $x>2$.
For the next inequality, we can assume $x-2>0$ from the last part. Then we have $5 < x-2$ and so $x > 7$. Taking the intersection of $(2, \infty)$ and $(7, \infty)$, the latter set is our solution. 
A: The right inequality yields $x-2>5$ or $x>7$ (provided $x-2>0$; if not, the first inequality is not satisfied, and we can ignore this case). When the left inequality is reciprocated, $\frac10$ should be interpreted as $\infty$, so $\frac10=\infty>x-2$, which is true for all values of x and thus can be dropped. Hence the range of x values satisfying the inequality is $x>7$.
(Technically this operation is not exactly justified, but the range of values satisfying $\epsilon<\frac1x$ or $\frac1\epsilon>x$ increases to the right without bound as $\epsilon\rightarrow0$.)
A: $$\frac { 1 }{ x-2 } -\frac { 1 }{ 5 } <0\\ \frac { 5-x+2 }{ 5\left( x-2 \right)  } <0\\ \frac { 7-x }{ \left( x-2 \right)  } <0\\ \frac { \left( 7-x \right) \left( x-2 \right)  }{ { \left( x-2 \right)  }^{ 2 } } <0\\ \left( 7-x \right) \left( x-2 \right) <0\\ x\in \left( -\infty ;2 \right) \cup \left( 7;+\infty  \right) $$
on the other hand $0<\frac { 5 }{ x-2 } \Rightarrow x>2$ so the answer will be $$\left( -\infty ;2 \right) \cup \left( 7;+\infty  \right) \cap \left( 2;+\infty  \right) \Rightarrow \left( 7;+\infty  \right) \\ \\ $$
A: 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?
