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.
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.
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.
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...
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.
