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.

So you have
$$
\lvert x \rvert \leq 2\lvert x+2\rvert.
$$
If $x \geq 0$, then $x + 2 > 0$. And then the equation reads $x \leq 2x + 4$ implying that $x \geq -4 $. This in all gives you the solutions $x \geq 0$.
If $x\in [-2, 0)$, then the equation becomes $-x \leq 2x + 4$. So $-4\leq 3x$. So $\frac{-4}{3} \leq x$. In all you get $x\in [-4/3, 0)$.
Now the last case is where $x <-2$. Then $-x \leq -2x - 4$. So $4 \leq -x$. So $-4\geq x$. This gives the set of solutions $(-\infty, -4]$.
Putting it all together gives you the closed interval:
$$
(-\infty, -4] \cup [-4/3,\infty)
$$
It looks like your solution is correct.
