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Difficulty: Very hardAbsolute Value Equations and Inequalities

For all real values of xx where the expression is defined, which of the following inequality expressions represents the complete set of values that satisfy the inequality below?

∣x−2∣−3∣x+1∣−2≤0\frac{|x - 2| - 3}{|x + 1| - 2} \leq 0
  1. A
    −3≤x≤−1-3 \leq x \leq -1 or 1≤x≤51 \leq x \leq 5
  2. B
    −1≤x≤5-1 \leq x \leq 5
  3. −3<x≤−1-3 < x \leq -1 or 1<x≤51 < x \leq 5Answer
  4. D
    1<x≤51 < x \leq 5
  5. E
    x<−3x < -3 or −1≤x<1-1 \leq x < 1 or x≥5x \geq 5

Answer

−3<x≤−1-3 < x \leq -1 or 1<x≤51 < x \leq 5
The correct answer represents the union of the two valid cases where the numerator and denominator have opposite signs. Specifically, the numerator ∣x−2∣−3|x - 2| - 3 is non-positive when −1≤x≤5-1 \leq x \leq 5, and the denominator ∣x+1∣−2|x + 1| - 2 is positive when x<−3x < -3 or x>1x > 1. Their intersection is the interval 1<x≤51 < x \leq 5. Alternatively, the numerator is non-negative when x≤−1x \leq -1 or x≥5x \geq 5, and the denominator is negative when −3<x<1-3 < x < 1. Their intersection is the interval −3<x≤−1-3 < x \leq -1. Combining these yields the solution set −3<x≤−1-3 < x \leq -1 or 1<x≤51 < x \leq 5.

Step-by-Step Solution

1
Analyze the conditions required for a rational expression to be non-positive.
The fraction ND≤0\frac{N}{D} \leq 0 is satisfied when the numerator NN and denominator DD have opposite signs, while ensuring the denominator is not equal to zero (D≠0D \neq 0). Thus, we must evaluate two cases: Case 1 where N≤0N \leq 0 and D>0D > 0, and Case 2 where N≥0N \geq 0 and D<0D < 0.
This establishes the logical framework for solving the rational inequality.
2
Determine the critical intervals for the numerator, N=∣x−2∣−3N = |x - 2| - 3.
Solve ∣x−2∣−3≤0  ⟹  ∣x−2∣≤3  ⟹  −3≤x−2≤3  ⟹  −1≤x≤5|x - 2| - 3 \leq 0 \implies |x - 2| \leq 3 \implies -3 \leq x - 2 \leq 3 \implies -1 \leq x \leq 5. Thus, the numerator is non-positive on [−1,5][-1, 5] and non-negative on (−∞,−1]∪[5,∞)(-\infty, -1] \cup [5, \infty).
This identifies the values of xx where the numerator changes sign.
3
Determine the critical intervals for the denominator, D=∣x+1∣−2D = |x + 1| - 2, and identify restrictions.
Solve ∣x+1∣−2<0  ⟹  ∣x+1∣<2  ⟹  −2<x+1<2  ⟹  −3<x<1|x + 1| - 2 < 0 \implies |x + 1| < 2 \implies -2 < x + 1 < 2 \implies -3 < x < 1. Thus, the denominator is negative on (−3,1)(-3, 1) and positive on (−∞,−3)∪(1,∞)(-\infty, -3) \cup (1, \infty). The boundary points x=−3x = -3 and x=1x = 1 make the denominator zero and must be excluded from the domain.
This identifies the values of xx where the denominator changes sign or makes the expression undefined.
4
Evaluate Case 1 and Case 2, then combine the solutions.
For Case 1 (N≤0N \leq 0 and D>0D > 0), find the intersection of [−1,5][-1, 5] and (−∞,−3)∪(1,∞)(-\infty, -3) \cup (1, \infty), which yields 1<x≤51 < x \leq 5. For Case 2 (N≥0N \geq 0 and D<0D < 0), find the intersection of (−∞,−1]∪[5,∞)(-\infty, -1] \cup [5, \infty) and (−3,1)(-3, 1), which yields −3<x≤−1-3 < x \leq -1. Combining these intervals gives the final solution set: −3<x≤−1-3 < x \leq -1 or 1<x≤51 < x \leq 5.
This determines the overall solution set that satisfies the starting inequality.

Key Concept

Solving rational inequalities containing absolute value expressions by finding critical intervals.
Estimated Time:3m 0s
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Topics

Absolute Value Equations and Inequalities

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