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Question

Difficulty: MediumLinear Inequalities and Absolute Value

If xx is a real number that satisfies both ∣4−x∣<3|4 - x| < 3 and 2x−15−3≤1\frac{2x - 15}{-3} \le 1, which of the following inequalities expresses all possible values of xx?

  1. 6≤x<76 \le x < 7Answer
  2. B
    1<x≤61 < x \le 6
  3. C
    1<x<71 < x < 7
  4. D
    x≥6x \ge 6
  5. E
    6<x<76 < x < 7

Answer

The inequality expressing all possible values of xx is 6≤x<76 \le x < 7.
The correct inequality is determined by finding the intersection of both given inequalities. The absolute value inequality ∣4−x∣<3|4 - x| < 3 simplifies to 1<x<71 < x < 7. The linear inequality 2x−15−3≤1\frac{2x - 15}{-3} \le 1 simplifies to x≥6x \ge 6 after reversing the inequality sign when multiplying by −3-3. Combining 1<x<71 < x < 7 and x≥6x \ge 6 yields the compound inequality stating xx is greater than or equal to 66 and strictly less than 77.

Step-by-Step Solution

1
Solve the absolute value inequality ∣4−x∣<3|4 - x| < 3.
−3<4−x<3  ⟹  −7<−x<−1  ⟹  1<x<7-3 < 4 - x < 3 \implies -7 < -x < -1 \implies 1 < x < 7.
An absolute value inequality ∣u∣<k|u| < k translates to −k<u<k-k < u < k. Multiplying or dividing by −1-1 flips the inequality signs.
2
Solve the linear inequality 2x−15−3≤1\frac{2x - 15}{-3} \le 1.
2x−15≥−3  ⟹  2x≥12  ⟹  x≥62x - 15 \ge -3 \implies 2x \ge 12 \implies x \ge 6.
Multiplying both sides of an inequality by a negative number reverses the direction of the inequality sign.
3
Find the intersection of the two solution sets.
1<x<71 < x < 7 and x≥6x \ge 6 combine to give 6≤x<76 \le x < 7.
Values of xx must satisfy both inequalities simultaneously.

Key Concept

Solving systems of absolute value inequalities and linear inequalities, including sign reversal rules for negative multipliers.
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Topics

Linear Inequalities and Absolute Value

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