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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 x7|x - 7| and the distance between xx and 5-5 is x+5|x + 5|. Equating x7=3x+5|x - 7| = 3|x + 5| leads to two equations: x7=3(x+5)x - 7 = 3(x + 5) giving x=11x = -11, and x7=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 x7|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 x7=3x+5|x - 7| = 3|x + 5|.
Distance between two points aa and bb on a real number line is expressed as ab|a - b|.
2
Solve Case 1 where x7x - 7 and x+5x + 5 have the same sign
x7=3(x+5)    x7=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, x7=3x+5|x - 7| = 3|x + 5| simplifies directly to x7=3(x+5)x - 7 = 3(x + 5).
3
Solve Case 2 where x7x - 7 and x+5x + 5 have opposite signs
x7=3(x+5)    x7=3x15    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, x7=3x+5|x - 7| = 3|x + 5| simplifies to x7=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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