How to solve this inequality with absolute value: $ \frac{\left|x-3\right|}{\left|x+2\right|}\le 3 $

Question

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 $(-\infty, -\frac{9}{4}) \wedge (-\frac{3}{4},\infty)$

Answer 1

hint: $x \neq -2$, and $|x-3| \leq 3|x+2|$, and square both sides.

Answer 2

If $|y|\le a$ for real $y, a\ge0,$

we need $-a\le y\le a$

So, $-3\le\dfrac{x-3}{x+2}\le3$

Now $\dfrac{x-3}{x+2}\le3\iff\dfrac{x-3}{x+2}-3\le0\iff\dfrac{2x+9}{x+3}\ge0$

$\iff x\le-\dfrac92$ or $x>-3$

can you take it from here?

Answer 3

Hint:

For $x \ne -2$ the inequality is equivalent to: $$ |x-3| \le 3|x+2| $$ so since the arguments of the absolute values change sign at $x=-2$ and $x=3$ the inequality is equivalent to the systems: $$ \begin{cases} x<-2\\ 3-x\le 3(-x-2) \end{cases} \quad \lor \quad \begin{cases} -2<x\le3\\ 3-x\le 3(x+2) \end{cases} \quad \lor \quad \begin{cases} x\ge 3\\ x-3\le 3(-x-2) \end{cases} $$

Answer 4

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

• $-2\le \dfrac{5}{x+2}<0\iff x+2\le-\dfrac52\iff x\le-\dfrac92$,
• $0<\dfrac{5}{x+2}\le 4\iff x+2\ge\dfrac54\iff x\ge-\dfrac34$.