Inequality with Absolute Value on both sides?

Question

My math teacher isn't the best. So can someone please explain how I can solve this problem?

$$|x+3| \ge |6x+9|$$

I tried doing it myself by splitting into two inequalities but my answer was very different.

Answer 1

hint:

square both sides and solve for $x$

Answer 2

Condition 1: For $x\geq-\frac{3}{2}$ $$\begin{align}x+3&\geq6x+9\\ -5x&\geq6\\ x&\leq-\frac{6}{5}\end{align}$$ Merge overlapping interval and we got $$-\frac{3}{2}\leq x\leq-\frac{6}{5}$$

Condition 2: For $-3\leq x < -\frac{3}{2}$ $$\begin{align}x+3&\geq-6x-9\\ 7x&\geq-12\\ x&\geq-\frac{12}{7}\end{align}$$ Merge overlaping interval and we got $$-\frac{12}{7}\leq x \leq -\frac{3}{2}$$

Condition 3: For $x<-3$ $$\begin{align}-x-3&\geq-6x-9\\ 5x&\geq-6\\ x&\geq-\frac{6}{5}\end{align}$$ Merge overlapping interval and we got $$\varnothing$$

Merge all interval from those three conditions and we got $$-\frac{12}{7} \leq x \leq -\frac{6}{5}$$

Answer 3

We have: $$0\geq|6x+9|-|x+3|=\frac{(6x+9)^2-(x+3)^2}{|6x+9|+|x+3|}=\frac{35x^2+102x+72}{|6x+9|+|x+3|}=\frac{(7x+12)(5x+6)}{|6x+9|+|x+3|}$$ and since $$|6x+9|+|x+3|>0,$$ we got the answer: $$\left[-\frac{12}{7},-\frac{6}{5}\right].$$