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Difficulty: Very hardLinear Inequalities and Absolute Value

If xx is a real number that satisfies the compound absolute value inequality ∣∣2x−5∣−7∣≤4||2x - 5| - 7| \le 4, which of the following values could be the value of xx? Select all such values.

  1. −2-2Answer
  2. 00Answer
  3. C
    33
  4. 66Answer
  5. E
    99

Answer

The real numbers −2-2, 00, and 66 satisfy the inequality.
The solution set to the compound inequality ∣∣2x−5∣−7∣≤4||2x - 5| - 7| \le 4 is the union of two intervals: [−3,1][-3, 1] and [4,8][4, 8]. Among the options provided, −2-2 and 00 fall into the interval [−3,1][-3, 1], and 66 falls into the interval [4,8][4, 8]. Therefore, these three values satisfy the original inequality.

Step-by-Step Solution

1
Unfold the outer absolute value inequality.
−4≤∣2x−5∣−7≤4-4 \le |2x - 5| - 7 \le 4
For any expression UU and constant c≥0c \ge 0, ∣U∣≤c|U| \le c is equivalent to −c≤U≤c-c \le U \le c.
2
Isolate the inner absolute value term by adding 77 across all parts.
3≤∣2x−5∣≤113 \le |2x - 5| \le 11
Adding a constant preserves inequality directions.
3
Break the compound inequality into two separate absolute value conditions.
Condition 1: ∣2x−5∣≤11|2x - 5| \le 11; Condition 2: ∣2x−5∣≥3|2x - 5| \ge 3
Both conditions must hold simultaneously for xx.
4
Solve Condition 1 (∣2x−5∣≤11|2x - 5| \le 11).
−11≤2x−5≤11  ⟹  −6≤2x≤16  ⟹  −3≤x≤8-11 \le 2x - 5 \le 11 \implies -6 \le 2x \le 16 \implies -3 \le x \le 8
Expanding the bounded absolute value inequality and solving for xx.
5
Solve Condition 2 (∣2x−5∣≥3|2x - 5| \ge 3).
2x−5≥32x - 5 \ge 3 or 2x−5≤−3  ⟹  2x≥82x - 5 \le -3 \implies 2x \ge 8 or 2x≤2  ⟹  x≥42x \le 2 \implies x \ge 4 or x≤1x \le 1
For ∣U∣≥c|U| \ge c, U≥cU \ge c or U≤−cU \le -c.
6
Find the intersection of the two solution sets.
x∈[−3,1]∪[4,8]x \in [-3, 1] \cup [4, 8]
Intersecting [−3,8][-3, 8] with (−∞,1]∪[4,∞)(-\infty, 1] \cup [4, \infty) yields [−3,1]∪[4,8][-3, 1] \cup [4, 8].
7
Test the provided choices against the solution set [−3,1]∪[4,8][-3, 1] \cup [4, 8].
−2∈[−3,1]-2 \in [-3, 1] (valid), 0∈[−3,1]0 \in [-3, 1] (valid), 3∉[−3,1]∪[4,8]3 \notin [-3, 1] \cup [4, 8] (invalid), 6∈[4,8]6 \in [4, 8] (valid), 9∉[−3,1]∪[4,8]9 \notin [-3, 1] \cup [4, 8] (invalid).
Values lying within the solution intervals satisfy the original inequality.

Key Concept

Solving nested absolute value inequalities using multi-step compound interval intersections.
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Topics

Linear Inequalities and Absolute Value

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