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Zorluk: OrtaReal Numbers, Number Line, and Absolute Value

On the real number line, point PP represents the real number xx, point QQ represents 77, and point RR represents −5-5. If the distance between PP and QQ is equal to 33 times the distance between PP and RR, what is the sum of all possible values of xx?

  1. −13-13Cevap
  2. B
    −11-11
  3. C
    −7-7
  4. D
    −2-2
  5. E
    1313

Cevap

The sum of all possible values of xx is −13-13.
The distance between xx and 77 is ∣x−7∣|x - 7| and the distance between xx and −5-5 is ∣x+5∣|x + 5|. Equating ∣x−7∣=3∣x+5∣|x - 7| = 3|x + 5| leads to two equations: x−7=3(x+5)x - 7 = 3(x + 5) giving x=−11x = -11, and x−7=−3(x+5)x - 7 = -3(x + 5) giving x=−2x = -2. Adding both solutions yields (−11)+(−2)=−13(-11) + (-2) = -13.

Adım Adım Çözüm

1
Set up the distance equation using absolute value notation
The distance between P(x)P(x) and Q(7)Q(7) is ∣x−7∣|x - 7|, and the distance between P(x)P(x) and R(−5)R(-5) is ∣x−(−5)∣=∣x+5∣|x - (-5)| = |x + 5|. The problem specifies that ∣x−7∣=3∣x+5∣|x - 7| = 3|x + 5|.
Distance between two points aa and bb on a real number line is expressed as ∣a−b∣|a - b|.
2
Solve Case 1 where x−7x - 7 and x+5x + 5 have the same sign
x−7=3(x+5)  ⟹  x−7=3x+15  ⟹  −22=2x  ⟹  x=−11x - 7 = 3(x + 5) \implies x - 7 = 3x + 15 \implies -22 = 2x \implies x = -11.
When both absolute value expressions have identical signs, ∣x−7∣=3∣x+5∣|x - 7| = 3|x + 5| simplifies directly to x−7=3(x+5)x - 7 = 3(x + 5).
3
Solve Case 2 where x−7x - 7 and x+5x + 5 have opposite signs
x−7=−3(x+5)  ⟹  x−7=−3x−15  ⟹  4x=−8  ⟹  x=−2x - 7 = -3(x + 5) \implies x - 7 = -3x - 15 \implies 4x = -8 \implies x = -2.
When the absolute value expressions have opposite signs, ∣x−7∣=3∣x+5∣|x - 7| = 3|x + 5| simplifies to x−7=−3(x+5)x - 7 = -3(x + 5).
4
Calculate the sum of all possible values of xx
(−11)+(−2)=−13(-11) + (-2) = -13.
Summing the two solutions gives the final required value.

Anahtar Kavram

Distance on a number line and absolute value equations
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