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Difficulty: HardLinear Inequalities and Absolute Value

If xx is a real number that satisfies both of the inequalities ∣3x+4∣≥11|3x + 4| \ge 11 and ∣x−1∣<6|x - 1| < 6, which of the following could be the value of xx? Select all that apply.

  1. A
    −5-5
  2. B
    −3-3
  3. 33Answer
  4. 55Answer
  5. E
    77

Answer

The values 3 and 5 satisfy both inequalities.
Solving the first inequality ∣3x+4∣≥11|3x + 4| \ge 11 yields x≤−5x \le -5 or x≥73x \ge \frac{7}{3}. Solving the second inequality ∣x−1∣<6|x - 1| < 6 yields −5<x<7-5 < x < 7. Intersecting these two regions, the interval x≤−5x \le -5 does not overlap with −5<x<7-5 < x < 7 because −5-5 is excluded from the second inequality. The overlap occurs only for 73≤x<7\frac{7}{3} \le x < 7. Among the choices, 3 and 5 fall within this valid interval.

Step-by-Step Solution

1
Solve the absolute value inequality ∣3x+4∣≥11|3x + 4| \ge 11.
x≤−5x \le -5 or x≥73x \ge \frac{7}{3}.
An absolute value inequality of the form ∣u∣≥k|u| \ge k (where k>0k > 0) splits into two separate inequalities: u≥ku \ge k or u≤−ku \le -k. Solving 3x+4≥113x + 4 \ge 11 gives 3x≥7  ⟹  x≥733x \ge 7 \implies x \ge \frac{7}{3}. Solving 3x+4≤−113x + 4 \le -11 gives 3x≤−15  ⟹  x≤−53x \le -15 \implies x \le -5.
2
Solve the absolute value inequality ∣x−1∣<6|x - 1| < 6.
−5<x<7-5 < x < 7.
An absolute value inequality of the form ∣u∣<k|u| < k is equivalent to the compound inequality −k<u<k-k < u < k. Thus, −6<x−1<6-6 < x - 1 < 6. Adding 1 to all parts yields −5<x<7-5 < x < 7.
3
Determine the intersection of the two solution sets.
73≤x<7\frac{7}{3} \le x < 7.
The portion x≤−5x \le -5 has no overlap with −5<x<7-5 < x < 7 because −5-5 is excluded by the strict inequality in the second condition. The portion x≥73x \ge \frac{7}{3} overlaps with −5<x<7-5 < x < 7 to yield the interval [73,7)[\frac{7}{3}, 7).
4
Evaluate which of the given options fall inside [73,7)[\frac{7}{3}, 7).
The numbers 3 and 5 belong to the interval, while −5-5, −3-3, and 77 do not.
Since 73≈2.33\frac{7}{3} \approx 2.33, the values 3 and 5 fall strictly between 2.33 and 7.

Key Concept

Solving systems of linear absolute value inequalities by finding the intersection of compound solution intervals
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Topics

Linear Inequalities and Absolute Value

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