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.
 A: Just use the observation that, on the domain of the inequation ($x\ne 2$), 
$$\dfrac{\lvert A\rvert}{\lvert B\rvert}<1\iff\lvert A\rvert< \lvert B\rvert\iff A^2<B^2.$$
After some simplification this yields
$$3x^2+8x-3<0.$$
The Rational roots theorem yields $x=-3,\dfrac13$, hence the solutions are
$$-3<x<\dfrac13$$
A: I suggest first simplifying so you don't have $x$ in both the numerator and denominator:
$\dfrac{2x+1}{x-2}\ =\ \dfrac{2x-4+5}{x-2}\ =\ 2+\dfrac5{x-2}$
Now try and solve
$$-1\ <\ 2+\dfrac5{x-2}\ <\ 1$$
A: $|\frac {2x+1}{x-2}|<1$
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 $(x-2) < 0$
$x\in(-3 , \frac 13)$
A: Here's another way:
$$\left|\frac{2x+1}{x-2}\right|<1\implies \frac{|2x+1|}{|x-2|}<1$$
$$\implies (2x+1)^2<(x-2)^2$$
$$\implies(2x+1+x-2)(2x+1-x+2)<0$$
$$\implies(3x-1)(x+3)<0$$
$$\implies -3<x<\frac 13$$
