# 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 $(0, \infty )$ ?

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The solution set is $(0,\infty)$. Of course $0\gt -3/2$, but that has nothing to do with conditions on $x$. –  André Nicolas Jun 16 '12 at 2:17
thank you André –  FaMu Jun 20 '12 at 2:00

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.

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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...

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just edit it. I mean $\frac{2x - 3}{2x + 3}$ –  FaMu Jun 16 '12 at 1:54

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.

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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.

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