Question

Difficulty: HardInequalities and Absolute Value Equations

How many integer values of xx satisfy the inequality ∣∣2x−5∣−7∣≤4||2x - 5| - 7| \leq 4?

Answer: 10

Answer

The total number of integer values of xx satisfying the inequality is 10.
To solve ∣∣2x−5∣−7∣≤4||2x - 5| - 7| \leq 4, rewrite the inequality without the outer absolute value as −4≤∣2x−5∣−7≤4-4 \leq |2x - 5| - 7 \leq 4. Adding 7 across all parts yields 3≤∣2x−5∣≤113 \leq |2x - 5| \leq 11. The inequality ∣2x−5∣≤11|2x - 5| \leq 11 simplifies to −3≤x≤8-3 \leq x \leq 8. The inequality ∣2x−5∣≥3|2x - 5| \geq 3 simplifies to x≤1x \leq 1 or x≥4x \geq 4. Intersecting these two regions gives the set of real numbers x∈[−3,1]∪[4,8]x \in [-3, 1] \cup [4, 8]. The integer solutions within [−3,1][-3, 1] are −3,−2,−1,0,1-3, -2, -1, 0, 1 (5 integers), and within [4,8][4, 8] are 4,5,6,7,84, 5, 6, 7, 8 (5 integers). The total number of valid integer solutions is 5+5=105 + 5 = 10.

Step-by-Step Solution

1
Unfold the outer absolute value expression.
−4≤∣2x−5∣−7≤4-4 \leq |2x - 5| - 7 \leq 4
An inequality of the form ∣U∣≤C|U| \leq C with C>0C > 0 is equivalent to −C≤U≤C-C \leq U \leq C.
2
Isolate the inner absolute value expression by adding 7 throughout.
3≤∣2x−5∣≤113 \leq |2x - 5| \leq 11
Adding a constant to all parts preserves the direction of the inequality.
3
Solve the upper bound ∣2x−5∣≤11|2x - 5| \leq 11.
−3≤x≤8-3 \leq x \leq 8
−11≤2x−5≤11-11 \leq 2x - 5 \leq 11 adds 5 to give −6≤2x≤16-6 \leq 2x \leq 16, which divides by 2 to yield −3≤x≤8-3 \leq x \leq 8.
4
Solve the lower bound ∣2x−5∣≥3|2x - 5| \geq 3.
x≤1x \leq 1 or x≥4x \geq 4
An inequality ∣U∣≥C|U| \geq C splits into U≥CU \geq C (2x−5≥3  ⟹  x≥42x - 5 \geq 3 \implies x \geq 4) or U≤−CU \leq -C (2x−5≤−3  ⟹  x≤12x - 5 \leq -3 \implies x \leq 1).
5
Find the intersection of the upper and lower bound conditions and count the integers.
10 integer solutions: {−3,−2,−1,0,1,4,5,6,7,8}\{-3, -2, -1, 0, 1, 4, 5, 6, 7, 8\}.
Combining −3≤x≤8-3 \leq x \leq 8 with (x≤1x \leq 1 or x≥4x \geq 4) produces two disjoint intervals [−3,1][-3, 1] and [4,8][4, 8], containing 5 integers each.

Key Concept

Solving nested absolute value inequalities using double inequalities and boundary region intersections.
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