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Difficulty: HardInequalities and Absolute Value Equations

How many integer values of xx satisfy the inequality ∣x−2∣−3∣x+1∣−4≤0\frac{|x - 2| - 3}{|x + 1| - 4} \leq 0?

Answer: 6

Answer

The correct answer is 6.
Analyzing the signs of the numerator ∣x−2∣−3|x - 2| - 3 and denominator ∣x+1∣−4|x + 1| - 4 shows that the quotient is non-positive when the numerator and denominator have opposite signs or when the numerator is zero. This yields two intervals: (−5,−1](-5, -1] and (3,5](3, 5]. The integers contained in these intervals are −4,−3,−2,−1,4,5-4, -3, -2, -1, 4, 5, which totals 6 integer values.

Step-by-Step Solution

1
Analyze the sign of the numerator ∣x−2∣−3|x - 2| - 3
The numerator is zero at x=−1x = -1 and x=5x = 5, negative for −1<x<5-1 < x < 5, and positive for x<−1x < -1 or x>5x > 5.
Solving ∣x−2∣=3|x - 2| = 3 yields x−2=3  ⟹  x=5x - 2 = 3 \implies x = 5 and x−2=−3  ⟹  x=−1x - 2 = -3 \implies x = -1.
2
Analyze the sign of the denominator ∣x+1∣−4|x + 1| - 4
The denominator is zero at x=−5x = -5 and x=3x = 3, negative for −5<x<3-5 < x < 3, and positive for x<−5x < -5 or x>3x > 3. Exclude x=−5x = -5 and x=3x = 3.
Solving ∣x+1∣=4|x + 1| = 4 yields x+1=4  ⟹  x=3x + 1 = 4 \implies x = 3 and x+1=−4  ⟹  x=−5x + 1 = -4 \implies x = -5. Denominators cannot be zero.
3
Find intervals where numerator and denominator have opposite signs or numerator is zero
The solution set is the union of (−5,−1](-5, -1] and (3,5](3, 5].
A fraction ND≤0\frac{N}{D} \leq 0 requires N≥0,D<0N \ge 0, D < 0 or N≤0,D>0N \le 0, D > 0.
4
Count the total number of integer solutions
The valid integers are −4,−3,−2,−1,4,5-4, -3, -2, -1, 4, 5, giving a total count of 6.
Listing integers in (−5,−1](-5, -1] yields −4,−3,−2,−1-4, -3, -2, -1, and in (3,5](3, 5] yields 4,54, 5.

Key Concept

Solving Rational Inequalities with Absolute Values
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Topics

Inequalities and Absolute Value Equations

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