Is the converse of this implication true? If we have,
\begin{align} 
|x-2|\leq {1} \iff & -1\leq{x-2}\leq1\\ \iff & 4\leq {x+3}\leq6 \\ \iff & 5\leq {x+4}\leq7.\end{align} 
Then In particular,
$ |x-2|\leq {1} \implies  4\leq {|x+3|}\leq6$ and $5\leq {|x+4|}\leq7.$
Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?
$4\leq {|x+3|}\leq6$ and $5\leq {|x+4|}\leq7 \implies |x-2|\leq {1}$.
This is a question in my textbook. I feel like the implication can be reversed. 


Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?

Note that:
$$\begin{align}4\le x+3\le 6 &\Rightarrow 4\le |x+3|\le 6, \ \ \text{but} \\
4\le |x+3|\le 6 &\not\Rightarrow 4\le x+3\le 6, \end{align}$$
because $4\le x+3\le 6$ has a solution $x\in [1,3]$:

whereas $4\le|x+3|\le 6$ has a solution $x\in [-9,-7]\cup [1,3]$:

