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.

• Did you mean $|x-2|\leq 1$? – Lucas Corrêa Dec 29 '18 at 2:51
• Yes, that's what I meant. – user503154 Dec 29 '18 at 2:53

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 $$x=-9$$, $$|-9-2| = |-11| = 11 > 1$$
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 $$x\in [-9,-7]\cup [1,3]$$: