<s>[INST] How to solve this inequality with absolute value: $ \frac{\left|x-3\right|}{\left|x+2\right|}\le 3 $ Good morning to everyone. I have an inequality that I don't know how to solve: $$ \frac{\left|x-3\right|}{\left|x+2\right|}\le 3 $$ I tried to solve it in this way: $$ \frac{\left|x-3\right|}{\left|x+2\right|}\le 3 \rightarrow \frac{\left|x-3\right|}{\left|x+2\right|} - 3 \le 0 \rightarrow \frac{\left|x-3\right|-3\left|x+2\right|}{\left|x+2\right|}\le 0 \rightarrow $$
Case 1:
$$ \left|x+2\right| > 0 \wedge \left|x-3\right|-3\left|x+2\right| \le 0 $$
Case 1 a) $$ x+2 >0 \rightarrow x>-2$$
Case 1 b) $$ -x-2 <0 \rightarrow x>-2$$
Case 1 c) $$ \left|x-3\right|-3\left|x+2\right| \le 0 \rightarrow \left|x-3\right| \le 3\left|x+2\right| \rightarrow x-3 \le x+2 \rightarrow $$
The solution are all real numbers
Case 1 d)$$ -x+3 < x+2 \rightarrow x>\frac{1}{2} $$
Case 1 e) $$ x-3 \le -x-2 \rightarrow x \le \frac {1}{2} $$
Case 1 f) $$ x-3 < x+2  $$
The solution: all real numbers
Case 2: $$ \left|x+2\right| \le 0 \wedge \left|x-3\right|-3\left|x+2\right| > 0 $$
It'll have the same solutions because are the same inequalities. Therefore $x$ belongs to $(\frac{1}{2}, \infty)$ But my answer sheet shows that it belongs to $(-\infty, -\frac{9}{4}) \wedge (-\frac{3}{4},\infty)$
 [/INST] Simply write the homographic function in canonical form:
$$\biggl\lvert\frac{x-3}{x+2}\biggr\rvert=\biggl\lvert\frac{(x+2)-5}{x+2}\biggr\rvert=\biggl\lvert1-\frac{5}{x+2}\biggr\rvert.$$
The inequation means the distance between $1$ and $\dfrac{5}{x+2}$ is at most $3$, so it's equivalent to
$$1-3=-2\le \frac{5}{x+2}\le 1+3=4.$$
As $\dfrac{5}{x+2}$ can't be $0$, we have two cases


*

*$-2\le \dfrac{5}{x+2}<0\iff x+2\le-\dfrac52\iff x\le-\dfrac92$,

*$0<\dfrac{5}{x+2}\le 4\iff x+2\ge\dfrac54\iff x\ge-\dfrac34$.

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