For $x$ real, the inequality $1\le |x-2|\le 7$ is equivalent to
$\textbf{(A) }x\le 1\text{ or }x\ge 3\qquad \textbf{(B) }1\le x\le 3\qquad \textbf{(C) }-5\le x\le 9\qquad \\ \textbf{(D) }-5\le x\le 1\text{ or }3\le x\le 9\qquad  \textbf{(E) }-6\le x\le 1\text{ or }3\le x\le 10$
We can split the inequality into two smaller inequalities and solve them individually.
\[|x-2| \ge 1 \quad \rightarrow \quad x \ge 3 \quad \text{and} \quad x \le 1\]
\[|x-2| \le 7 \quad \rightarrow \quad x \le 9 \quad \text{and} \quad x \ge -5\]
Combining these inequalities, we get $x \in [-5, 1] \cup [3, 9].$
The answer is $\textbf{(D)}.$
-edited by coolmath34