Solve |2x + 3| ≤ 9.

An absolute value inequality of the form $|A|\le9$ means $-9\le A\le9$. So $-9\le2x+3\le9$. Subtract $3$ from all parts: $-12\le2x\le6$. Divide by $2$: $-6\le x\le3$, choice A. Test $x=3$: $|9|=9\le9$, true. Test $x=-6$: $|-9|=9\le9$, true. Choice B ($-3\le x\le6$) has the endpoints mixed up, and $x=6$ gives $|15|>9$. Choices C and D each describe only one side and allow values far outside the range, such as $x=-20$ in choice C. The solution set is a single interval because the inequality uses less than or equal to.