<s>[INST] Absolute value rational inequality I just stumbled upon this particular question and cannot answer
$$\left\lvert \frac{2x+1}{x-2} \right\rvert<1$$
I know that there are some rules in order to answer this Problem 
$\frac{2x+1}{x-2} $ if only the  $\frac{2x+1}{x-2}≥ 0 $ 
and
$\frac{(-)2x+1}{x-2}$ if only the $\frac{2x+1}{x-2}< 0 $ 
but I just can't find the answer because everything what i've learnt and saw on YouTube mixed up, and I can't tell which one is correct.
Can anybody give me an Explanation?
*Ps, I'm sorry, I'm not really familiar with MathJax
btw, I have seen another Problem on this site which is also Absolute value rational inequality, but I don't really understand.
 [/INST] $|\frac {2x+1}{x-2}|<1$
To get rid of the absolute value:
$-1< \frac {2x+1}{x-2}<1$
When you mulitiply through by $x-2$ it is going to flip the direction of the inequalities if  $x-2 < 0$
Suppose (x-2) > 0
$2-x < 2x+1 <x-2$ and $(x-2) > 0$
$2-x < 2x+1$ and $2x+1 <x-2$ and $(x-2) > 0$
$\frac 13 < x$ and $x <-3$ and $x > 2$
not compatible!
Suppose $(x-2) < 0$
$2-x > 2x+1 >x-2$ and $(x-2) < 0$
$2-x > 2x+1$ and $2x+1 > x-2$ and $(x-2) < 0$
$\frac 13 > x$ and $x > -3$ and $(x-2) < 0$
$x\in(-3 , \frac 13)$
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