Algebra practice question

Answer: 14

Rewrite |ax − b| ≤ 6 as −6 ≤ ax − b ≤ 6, so b − 6 ≤ ax ≤ b + 6, and (since a > 0) (b − 6)/a ≤ x ≤ (b + 6)/a. Matching endpoints with 3 ≤ x ≤ 9: the center of the interval is (3 + 9)/2 = 6 = b/a, and the half-width is (9 − 3)/2 = 3 = 6/a. From 3 = 6/a, a = 2; then b = 6a = 12, so a + b = 14. Check: |2x − 12| ≤ 6 gives 6 ≤ 2x ≤ 18, i.e., 3 ≤ x ≤ 9. Common wrong answers: 8 comes from reading the center 6 as the value of b (forgetting b = 6a); 3.5 comes from inverting the half-width relation (a = 6/3 misread as a = 3/6 = 1/2, then b = 3); −14 comes from ignoring the condition a > 0 and using a = −2, b = −12, which produces the same solution set but is excluded by the given restriction; 12 comes from stopping after finding b.

Common wrong answers

If you answered 8

You read the center of the interval, 6, as b itself and added it to a = 2. The center gives b/a = 6, so b = 6a = 12 and a + b = 14.

If you answered 12

That's b. You still need to add a = 2 to get a + b = 14.

If you answered 3.5

You inverted the half-width relation: from 3 = 6/a you get a = 2, not a = 6/... = 1/2. With a = 2, b = 12 and a + b = 14.

If you answered -14

a = −2, b = −12 does produce the same solution set, but the problem states a > 0, so use a = 2, b = 12 and a + b = 14.

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