Examkin

Question

Difficulty: MediumLinear Inequalities and Absolute Value

If xx is a real number that satisfies both ∣5−2x∣≤11|5 - 2x| \le 11 and −3x+7<1-3x + 7 < 1, what is the least possible integer value of xx?

  1. A
    -3
  2. B
    2
  3. 3Answer
  4. D
    -8
  5. E
    8

Answer

The least possible integer value of xx is 3.
Solving ∣5−2x∣≤11|5 - 2x| \le 11 leads to −11≤5−2x≤11-11 \le 5 - 2x \le 11. Subtracting 5 gives −16≤−2x≤6-16 \le -2x \le 6, and dividing by −2-2 (flipping the inequalities) yields −3≤x≤8-3 \le x \le 8. Next, solving −3x+7<1-3x + 7 < 1 gives −3x<−6-3x < -6, which upon dividing by −3-3 (and flipping the inequality sign) gives x>2x > 2. Combining these two requirements yields 2<x≤82 < x \le 8. The integer values satisfying this inequality are 3, 4, 5, 6, 7, and 8. The least possible integer value among these is 3.

Step-by-Step Solution

1
Solve the absolute value inequality ∣5−2x∣≤11|5 - 2x| \le 11.
−11≤5−2x≤11  ⟹  −16≤−2x≤6  ⟹  −3≤x≤8-11 \le 5 - 2x \le 11 \implies -16 \le -2x \le 6 \implies -3 \le x \le 8.
An absolute value inequality ∣u∣≤k|u| \le k expands to −k≤u≤k-k \le u \le k. Dividing by −2-2 flips the inequality direction.
2
Solve the linear inequality −3x+7<1-3x + 7 < 1.
−3x<−6  ⟹  x>2-3x < -6 \implies x > 2.
Subtract 7 from both sides, then divide by −3-3, remembering to reverse the inequality sign.
3
Determine the intersection of both solution sets.
2<x≤82 < x \le 8.
xx must be strictly greater than 2 and less than or equal to 8.
4
Identify the smallest integer within the range 2<x≤82 < x \le 8.
3
Since x>2x > 2 is strict, 2 is excluded, making 3 the smallest integer in the range.

Key Concept

Solving systems of linear inequalities involving absolute values and correctly applying sign-flipping rules when multiplying or dividing by negative quantities.

Alternative Method

Test integer candidates directly: for x=2x = 2, −3(2)+7=1-3(2) + 7 = 1, which is not strictly less than 1. For x=3x = 3, −3(3)+7=−2<1-3(3) + 7 = -2 < 1 (valid) and ∣5−2(3)∣=∣−1∣=1≤11|5 - 2(3)| = |-1| = 1 \le 11 (valid), confirming 3 is the smallest integer solution.
Estimated Time:1m 30s
Rate this question

Topics

Linear Inequalities and Absolute Value

More questions from this topic