Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Find the answer

13. Given the inequality $|a x-3| \leqslant b$ has the solution set $\left[-\frac{1}{2}, \frac{7}{2}\right]$ . Then $a+b=$ $\qquad$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

13.6.

From $|a x-3| \leqslant b$ , we get $3-b \leqslant a x \leqslant 3+b$ . It is easy to see that $a \neq 0$ , so $\frac{3-b}{a}+\frac{3+b}{a}=-\frac{1}{2}+\frac{7}{2}$ . Solving this, we get $a=2$ .
Therefore, $\frac{3-b}{a}=-\frac{1}{2}$ . Solving this, we get $b=4$ . Hence, $a+b=6$ .

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