Is the converse of this implication true?

Question

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.

Answer 1

For the first question: suppose that $|x-2|\leq 1$. So, $$|x+k| = |(x-2) + (k+2)| \leq |x-2| + |k+2| \leq 1 + |k+2|.$$

For the second: $$4 \leq |x+3| \leq 6 \Longrightarrow x \in [1,3]\cup[-9,-7]$$ $$5 \leq |x+4| \leq 7 \Longrightarrow x \in [1,3]\cup[-11,-9]$$ then if $x$ satisfies both, $x \in [1,3]\cup\{-9\}$. Take $x=-9$, $$|-9-2| = |-11| = 11 > 1$$

Answer 2

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 $x\in [-9,-7]\cup [1,3]$: