Solve $ \left|\frac{x}{x+2}\right|\leq 2 $ I am experiencing a little confusion in answering a problem on Absolute Value inequalities which I just started learning. This is the problem: Solve:
$$ 
\left|\frac{x}{x+2}\right|\leq 2
$$
The answer is given to be $x\leq-4$ or $x \geq-1$
This is my attempt to solve the problem:
By dividing, $\left|\frac{x}{x+2}\right|\leq 2$ is equivalent to $\left |1-\frac{2}{x+2}\right|\leq2$ which is also equivalent to $\left |\frac{2}{x+2}-1\right|\leq2$
So, $-2\leq\frac{2}{x+2}-1\leq2$ which is equivalent to $-\frac{1}{2}\leq\frac{1}{x+2}\leq\frac{3}{2}$
Case 1: $x+2>0$. Solving $-\frac{1}{2}\leq\frac{1}{x+2}\leq\frac{3}{2}$, I get $x\geq-4$ and $x\geq-\frac{4}{3}$ which is essentially $x\geq-\frac{4}{3}$.
Case 2: $x+2<0$. Solving $-\frac{1}{2}\times(x+2)\geq{1}\geq\frac{3}{2}\times(x+2)$, I get $x\leq-4$ and $x\leq-\frac{4}{3}$ which is essentially $x\leq-4$.
So, the solutions are: $x\leq-4$ or $x\geq-\frac{4}{3}$.
I couldn't get $x \geq-1$ as a solution. Did I do anything wrong? The book I am using is Schaum's Outlines-Calculus. Another question I would like to ask is that am I using 'and' and 'or' correctly in the above attempt to solve the problem? I have had this problem many times.

Given that, $$\left|\frac{x}{x+2}\right|\le 2$$$$\iff |x|\le 2|x+2| $$
Notice, $x=-2$ & $x=0$ are two critical points on the number line. Now, let's consider the following cases, 
Case 1: If $\color{blue}{x\le -2}$ $$-x\le -2(x+2)\iff x\le -4\ \ \ ({\text{True}})$$
$$\implies \color{red}{x\in(-\infty, -4]}$$ 
Case 2: If $\color{blue}{-2<x<0}$ $$-x\le 2(x+2)\iff x\ge -\frac{4}{3}\ \ \ $$ 
but, $x<0$ hence, $$-\frac{4}{3}<x<0\iff \color{red}{x\in\left[-\frac{4}{3}, 0\right)}$$
Case 3: If $\color{blue}{x\ge 0}$ $$x\le 2(x+2)\iff x\ge -4\ \ \ $$ 
but, $x\ge 0$ hence, $$x\ge 0\iff \color{red}{x\in\left[0, \infty\right)}$$
Hence, the combining the above cases, the complete solution is given as 
$$\color{red}{x\in (-\infty, -4]\cup \left[-\frac{4}{3}, \infty\right) }$$
