For how many integer values of x is |x − 3| + |x + 1| ≤ 6?
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Answer & Explanation
Correct answer
C
Method 1 (region analysis): the sum |x − 3| + |x + 1| equals the total distance from x to 3 and to −1. For −1 ≤ x ≤ 3 the sum is the fixed gap 4, which is ≤ 6. For x > 3 the sum is 2x − 2, and 2x − 2 ≤ 6 gives x ≤ 4. For x < −1 the sum is 2 − 2x, and 2 − 2x ≤ 6 gives x ≥ −2. Combining the three regions gives −2 ≤ x ≤ 4, whose integers are −2, −1, 0, 1, 2, 3, 4, a total of 7.
Method 2 (test the boundaries): at x = −2 the sum is 5 + 1 = 6 and at x = 4 the sum is 1 + 5 = 6, both meeting ≤ 6, while x = −3 and x = 5 give 8 > 6. So the included integers run from −2 through 4, which is 7 values.
Choice 5 keeps only the integers in the constant-minimum stretch −1 through 3 and ignores the tails that the bound 6 still allows. Choice 6 finds the correct interval −2 ≤ x ≤ 4 but counts only one endpoint. Choice 8 solves the right tail as x ≤ 4 but the left tail as x ≥ −3, mis-placing the lower boundary by one. Choice 9 extends both tails one unit too far, counting −3 through 5 as if each boundary inequality were loose.