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.
 A: One thing you can do here is square to get rid of the absolute values and make it easier to solve, obtaining the equivalent statement
$${(2x - 3)^2 \over (2x + 3)^2} < 1$$
The denominator is nonnegative so you can multiply it through, obtaining
$$(2x - 3)^2 < (2x + 3)^2$$
This can be rewritten as
$$4x^2 - 12x + 9 < 4x^2 + 12x + 9$$
This simplifies to just
$$x > 0$$
Note that since there was originally a $(2x + 3)$ in the denominator, if $x = -{3 \over 2}$ were in the above solution we would have had to exclude it. But it wasn't, so we don't have to worry about it.
A: 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.  
A: I assume you mean
$$-1 < \frac{2x+3}{2x-3} < 1 \, ?$$
That's a reasonable way to go. At this point, you should break it up into two separate inequalities ($-1<\frac{2x+3}{2x-3}$, $\frac{2x+3}{2x-3} < 1$) and solve each one individually. The original inequality will be true when both of the new inequalities are true.
The other thing you could do: Since $|ab|=|a||b|$ and $|2x-3|$ is always positive, the inequality you started with is equivalent to
$|2x-3| < |2x+3|$; divide this by 2 to get $|x-3/2| < |x-(-3/2)|$. That is, you want those $x$ which are closer to the point 3/2 than to the point -3/2...
A: First, eliminate $x = -3/2$.  Now multiply and divide by 2 to get
$$|x - 3/2| = d(x, 3/2) < |x + 3/2| = d(x, -3/2).$$
The locus of points where the distances are equal is a point. Now choose sides.
Don't forget to omit the verboten point $x = -3/2$ if necessary.
