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Difficulty: HardLinear Inequalities and Absolute Value

If xx is an integer that satisfies both ∣3x−12∣≤15|3x - 12| \le 15 and ∣2x+4∣≥6|2x + 4| \ge 6, what is the sum of the maximum possible value of xx and the minimum possible value of xx?

Answer: 10

Answer

The correct answer is 10.
Solving the first inequality ∣3x−12∣≤15|3x - 12| \le 15 yields −1≤x≤9-1 \le x \le 9. Solving the second inequality ∣2x+4∣≥6|2x + 4| \ge 6 yields x≥1x \ge 1 or x≤−5x \le -5. The values of xx that satisfy both constraints are those in the interval [1,9][1, 9]. Within this interval, the minimum integer value is 11 and the maximum integer value is 99. Adding these values together gives 1+9=101 + 9 = 10.

Step-by-Step Solution

1
Solve the bounded absolute value inequality ∣3x−12∣≤15|3x - 12| \le 15.
-15 \le 3x - 12 \le 15 \implies -3 \le 3x \le 27 \implies -1 \le x \le 9.
An inequality of the form ∣A∣≤B|A| \le B (where B≥0B \ge 0) unwraps to the compound inequality −B≤A≤B-B \le A \le B.
2
Solve the boundary absolute value inequality ∣2x+4∣≥6|2x + 4| \ge 6.
2x + 4 \ge 6 \implies 2x \ge 2 \implies x \ge 1 ,or, or 2x + 4 \le -6 \implies 2x \le -10 \implies x \le -5$.
An inequality of the form ∣A∣≥B|A| \ge B (where B>0B > 0) splits into two separate cases: A≥BA \ge B or A≤−BA \le -B.
3
Find the intersection of the solution sets from Step 1 and Step 2.
The intersection of [−1,9][-1, 9] and (−∞,−5]∪[1,∞)(-\infty, -5] \cup [1, \infty) is [1,9][1, 9].
To satisfy both inequalities simultaneously, xx must belong to both set [−1,9][-1, 9] and set (−∞,−5]∪[1,∞)(-\infty, -5] \cup [1, \infty).
4
Identify the extreme integer values of xx and compute their sum.
\text{Minimum integer } x = 1, \quad \text{Maximum integer } x = 9; \quad 1 + 9 = 10.
The range of valid integer values for xx is all integers from 1 to 9 inclusive.

Key Concept

Linear Inequalities and Absolute Value
Estimated Time:2m 0s
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Topics

Linear Inequalities and Absolute Value

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