<s>[INST] 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.
 [/INST] $$\frac { 1 }{ x-2 } -\frac { 1 }{ 5 } <0\\ \frac { 5-x+2 }{ 5\left( x-2 \right)  } <0\\ \frac { 7-x }{ \left( x-2 \right)  } <0\\ \frac { \left( 7-x \right) \left( x-2 \right)  }{ { \left( x-2 \right)  }^{ 2 } } <0\\ \left( 7-x \right) \left( x-2 \right) <0\\ x\in \left( -\infty ;2 \right) \cup \left( 7;+\infty  \right) $$
on the other hand $0<\frac { 5 }{ x-2 } \Rightarrow x>2$ so the answer will be $$\left( -\infty ;2 \right) \cup \left( 7;+\infty  \right) \cap \left( 2;+\infty  \right) \Rightarrow \left( 7;+\infty  \right) \\ \\ $$
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