1972 AHSME Problems/Problem 10

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Problem

For $x$ real, the inequality $1\le |x-2|\le 7$ is equivalent to

$\textbf{(A) }x\le 1\text{ or }x\ge 3\qquad \textbf{(B) }1\le x\le 3\qquad \textbf{(C) }-5\le x\le 9\qquad \\ \textbf{(D) }-5\le x\le 1\text{ or }3\le x\le 9\qquad  \textbf{(E) }-6\le x\le 1\text{ or }3\le x\le 10$

Solution

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

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