the solution set of $\left | \frac{2x - 3}{2x + 3} \right |< 1$ what is the solution set of $\left | \frac{2x - 3}{2x + 3} \right |< 1$ ?
I solved it by first assuming:  $-1 < \frac{2x - 3}{2x + 3 } < 1$
ended with: $x > 0 > -3/2$
Is that a correct approach?
And how to derive the solution set from the last inequality?
Is it  $(-\frac{3}{2}, \infty)$ or $(0, \infty )$ ?
Thanks in advance.

Equivalently, you are indeed trying to solve the
inequalities 
$$-1\lt \frac{2x-3}{2x+3}\lt 1,$$
so it is reasonable to start by saying so. Then there are two cases to consider, $2x+3>0$ and $2x+3\lt 0$. In the first case we obtain the equivalent inequalities 
$$-(2x+3) \lt 2x-3\lt 2x+3.$$ 
The inequality on the right always holds. The inequality on the left holds iff $-2x-3\lt 2x-3$, that is, iff $x\gt 0$.
In the case $2x+3 \lt 0$, multiplying through by $2x+3$ switches the direction of the inequalities, so we want
$$-(2x+3) \gt 2x-3 \gt 2x+3.$$
But the inequality $2x-3\gt 2x+3$ can never hold. We conclude that our inequality holds precisely if $x$ is in the interval $(0,\infty)$.
I would recommend that for more complicated problems you consider the following sort of approach.  The expression $\frac{2x-3}{2x+3}$ changes sign "at" $x=-3/2$ and at $x=3/2$. Consider the three cases (i) $x\lt -3/2$, (ii) $-3/2\lt x\lt 3/2$, and (iii) $x \gt 3/2$ separately.  
