Question

Difficulty: MediumReal Numbers, Number Line, and Absolute Value

On the real number line, the distance between xx and 33 is equal to twice the distance between xx and −9-9. What is the sum of all possible real values of xx?

  1. −26-26Answer
  2. B
    −21-21
  3. C
    −14-14
  4. D
    −5-5
  5. E
    2222

Answer

The sum of all possible real values of xx is −26-26.
The distance between xx and aa on the real number line is expressed as ∣x−a∣|x - a|. Thus, the condition translates to ∣x−3∣=2∣x−(−9)∣|x - 3| = 2|x - (-9)|, which simplifies to ∣x−3∣=2∣x+9∣|x - 3| = 2|x + 9|. Setting up the two algebraic cases gives x−3=2(x+9)x - 3 = 2(x + 9), leading to x=−21x = -21, and x−3=−2(x+9)x - 3 = -2(x + 9), leading to x=−5x = -5. Adding these two solutions yields (−21)+(−5)=−26(-21) + (-5) = -26.

Step-by-Step Solution

1
Translate the geometric statement on the number line into an absolute value equation.
The distance between xx and 33 is ∣x−3∣|x - 3|, and the distance between xx and −9-9 is ∣x−(−9)∣=∣x+9∣|x - (-9)| = |x + 9|. Therefore, ∣x−3∣=2∣x+9∣|x - 3| = 2|x + 9|.
Distance between two points aa and bb on the number line is given by ∣a−b∣|a - b|.
2
Solve Case 1 where the expressions inside the absolute values have the same sign.
x−3=2(x+9)  ⟹  x−3=2x+18  ⟹  x=−21x - 3 = 2(x + 9) \implies x - 3 = 2x + 18 \implies x = -21.
Removing absolute values with identical signs gives a linear equation in xx.
3
Solve Case 2 where the expressions inside the absolute values have opposite signs.
x−3=−2(x+9)  ⟹  x−3=−2x−18  ⟹  3x=−15  ⟹  x=−5x - 3 = -2(x + 9) \implies x - 3 = -2x - 18 \implies 3x = -15 \implies x = -5.
Removing absolute values with opposite signs accounts for the second possible geometric location.
4
Sum the valid real solutions.
(−21)+(−5)=−26(-21) + (-5) = -26.
The question asks for the sum of all possible real values of xx.

Key Concept

Absolute Value as Distance on the Real Number Line
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