$-1 < \dfrac{2x}{(3x-1)} < 1$ , what are the values of $x$? what I did was 
$-1<\dfrac{2x}{(3x-1)}  =  -3x+1<2x   = -5x<-1   x>1/5 $
and
$\dfrac{2x}{(3x-1)}<1  =  2x<3x-1,\quad    -x<-1,\    x>1 $ 
so intersection is $x>1  $
But, for the first inequality, $-1<2x/(3x-1)  =  ;;  = ;;$ I ended up getting $x>1/5$
but If I pick $x= 0.25$, I get $-2$ which is not inside the boundary
Is there something wrong with calculation ?
 A: Hint. Your first line is not correct, since as written you are not sure that $3x-1>0$. You would better solve
$$
-1< \frac{2x}{3x-1}
$$ like this
$$
0<\frac{2x}{3x-1}+1
$$$$
0<\frac{2x+(3x-1)}{3x-1}
$$ $$
0<\frac{5x-1}{3x-1}
$$ then consider the signs of $5x-1$ and of $3x-1$. 
You may solve
$$
\frac{2x}{3x-1}<1
$$  similarly.
A: $$\dfrac{2x}{3x-1}<1\iff\dfrac{2x}{3x-1}-1<0\iff\dfrac{1-x}{3x-1}<0\iff\dfrac{x-1}{3x-1}>0$$
If $3x-1=0,\dfrac{2x}{3x-1}=?$
Else $$\dfrac{x-1}{3x-1}>0\iff(x-1)(3x-1)>0$$
Now the product of two terms is positive
Hence, either both of them are positive or both negative
A: Another way:  If $-a<y<a$ for real $a,y$  we can write $y^2<a^2$
So here, we have $\left(\dfrac{2x}{3x-1}\right)^2<1\iff4x^2<(3x-1)^2=9x^2-6x+1$
$\iff0<5x^2-6x+1=(5x-1)(x-1)$
Now the product of two terms is positive
Hence, either both of them are positive or both negative
A: Notice, we have the following $$-1<\frac{2x}{3x-1}<1$$ 
It can be divided in LHS & RHS inequalities as follows


*

*Solving LHS inequality $$-1<\frac{2x}{3x-1}$$ $$ \frac{2x}{3x-1}+1>0$$  $$ \frac{5x-1}{3x-1}>0$$ $$x<\frac{1}{5}\ \text{or}\ \ x>\frac{1}{3}$$

*Solving RHS inequality $$\frac{2x}{3x-1}<1$$ $$1- \frac{2x}{3x-1}>0$$  $$ \frac{x-1}{3x-1}>0$$ $$x<\frac{1}{3}\ \text{or}\ \ x>1$$


Taking both the solutions above on the number line & selecting common regions, one gets final solution $$\color{red}{x\in\left(-\infty, \frac{1}{5}\right)\cup(1, \infty)}$$
