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Difficulty: Very hardReal Numbers, Number Line, and Absolute Value

On the real number line, the distance between two real numbers xx and yy is dd. If the midpoint of xx and yy is 77 and ∣x−3∣=2∣y−3∣|x - 3| = 2|y - 3|, what is the maximum possible value of dd?

  1. A
    12
  2. B
    16
  3. 24Answer
  4. D
    28
  5. E
    32

Answer

The maximum possible value of dd is 24.
The midpoint condition dictates that x+y=14x + y = 14, so y=14−xy = 14 - x and the distance between them is d=∣x−y∣=2∣x−7∣d = |x - y| = 2|x - 7|. Substituting y=14−xy = 14 - x into ∣x−3∣=2∣y−3∣|x - 3| = 2|y - 3| gives ∣x−3∣=2∣11−x∣|x - 3| = 2|11 - x|. Solving the positive case x−3=2(x−11)x - 3 = 2(x - 11) gives x=19x = 19 and y=−5y = -5, resulting in distance d=∣19−(−5)∣=24d = |19 - (-5)| = 24. Solving the negative case x−3=−2(x−11)x - 3 = -2(x - 11) gives x=25/3x = 25/3 and y=17/3y = 17/3, resulting in distance d=8/3d = 8/3. Thus, 24 is the maximum possible value of dd.

Step-by-Step Solution

1
Express yy in terms of xx using the midpoint formula
Since the midpoint of xx and yy is 77, x+y2=7\frac{x + y}{2} = 7, which gives y=14−xy = 14 - x. The distance d=∣x−y∣=∣x−(14−x)∣=∣2x−14∣=2∣x−7∣d = |x - y| = |x - (14 - x)| = |2x - 14| = 2|x - 7|.
Relating yy to xx reduces the problem to a single variable.
2
Substitute y=14−xy = 14 - x into the absolute value equation
∣x−3∣=2∣(14−x)−3∣  ⟹  ∣x−3∣=2∣11−x∣=2∣x−11∣|x - 3| = 2|(14 - x) - 3| \implies |x - 3| = 2|11 - x| = 2|x - 11|.
Setting up the single-variable absolute value equation allows finding all possible values for xx.
3
Solve the absolute value equation for all possible cases
Case 1: x−3=2(x−11)  ⟹  x−3=2x−22  ⟹  x=19x - 3 = 2(x - 11) \implies x - 3 = 2x - 22 \implies x = 19. Then y=14−19=−5y = 14 - 19 = -5.
Case 2: x−3=−2(x−11)  ⟹  x−3=−2x+22  ⟹  3x=25  ⟹  x=253x - 3 = -2(x - 11) \implies x - 3 = -2x + 22 \implies 3x = 25 \implies x = \frac{25}{3}. Then y=14−253=173y = 14 - \frac{25}{3} = \frac{17}{3}.
Absolute value equations ∣A∣=B|A| = B split into A=BA = B and A=−BA = -B.
4
Calculate the distance dd for each case and select the maximum
For Case 1 (x=19,y=−5x = 19, y = -5): d=∣19−(−5)∣=24d = |19 - (-5)| = 24.
For Case 2 (x=25/3,y=17/3x = 25/3, y = 17/3): d=∣25/3−17/3∣=8/3d = |25/3 - 17/3| = 8/3.
The maximum possible value of dd is 2424.
Comparing the distance values determined in each case identifies the maximum distance.

Key Concept

Absolute Value as Distance and Multi-Case Equations on the Real Number Line
Estimated Time:2m 30s
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Topics

Real Numbers, Number Line, and Absolute Value

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