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Difficulty: HardLinear Inequalities and Absolute Value

How many integer values of xx satisfy the inequality ∣3−2x∣+∣x+4∣≤9|3 - 2x| + |x + 4| \le 9?

  1. A
    3
  2. B
    4
  3. 5Answer
  4. D
    6
  5. E
    7

Answer

5 integer values (specifically -2, -1, 0, 1, and 2)
Solving the absolute value inequality using piecewise intervals yields the continuous solution set [−2,83][-2, \frac{8}{3}]. The integers contained in this range are −2,−1,0,1,-2, -1, 0, 1, and 22, giving a total of 5 integer values.

Step-by-Step Solution

1
Identify the critical points of the absolute value expressions.
The critical points are x=32x = \frac{3}{2} and x=−4x = -4. These split the real number line into three intervals: x<−4x < -4, −4≤x≤32-4 \le x \le \frac{3}{2}, and x>32x > \frac{3}{2}.
Absolute value expressions change definition at their zeroes.
2
Analyze Case 1: x<−4x < -4.
Here ∣3−2x∣=3−2x|3 - 2x| = 3 - 2x and ∣x+4∣=−(x+4)|x + 4| = -(x + 4). The inequality becomes (3−2x)−(x+4)≤9  ⟹  −3x−1≤9  ⟹  −3x≤10  ⟹  x≥−103≈−3.33(3 - 2x) - (x + 4) \le 9 \implies -3x - 1 \le 9 \implies -3x \le 10 \implies x \ge -\frac{10}{3} \approx -3.33. Since there is no overlap between x<−4x < -4 and x≥−3.33x \ge -3.33, no solutions exist in this interval.
Evaluating expressions according to the sign of terms when x<−4x < -4.
3
Analyze Case 2: −4≤x≤32-4 \le x \le \frac{3}{2}.
Here ∣3−2x∣=3−2x|3 - 2x| = 3 - 2x and ∣x+4∣=x+4|x + 4| = x + 4. The inequality becomes (3−2x)+(x+4)≤9  ⟹  7−x≤9  ⟹  −x≤2  ⟹  x≥−2(3 - 2x) + (x + 4) \le 9 \implies 7 - x \le 9 \implies -x \le 2 \implies x \ge -2. Combining with the case interval gives [−2,32][-2, \frac{3}{2}].
Determining valid values of xx within the middle interval.
4
Analyze Case 3: x>32x > \frac{3}{2}.
Here ∣3−2x∣=2x−3|3 - 2x| = 2x - 3 and ∣x+4∣=x+4|x + 4| = x + 4. The inequality becomes (2x−3)+(x+4)≤9  ⟹  3x+1≤9  ⟹  3x≤8  ⟹  x≤83≈2.67(2x - 3) + (x + 4) \le 9 \implies 3x + 1 \le 9 \implies 3x \le 8 \implies x \le \frac{8}{3} \approx 2.67. Combining with the case interval gives (32,83](\frac{3}{2}, \frac{8}{3}].
Determining valid values of xx within the upper interval.
5
Combine solution intervals and count integer solutions.
The total solution set is [−2,83][-2, \frac{8}{3}]. The integer values within this interval are −2,−1,0,1,-2, -1, 0, 1, and 22. Total count = 5.
Identifying all integer values within the bounded set [−2,2.67][-2, 2.67].

Key Concept

Solving absolute value inequalities with multiple absolute value terms using critical points and case analysis.
Estimated Time:2m 0s
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Topics

Linear Inequalities and Absolute Value

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