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X - 5 = 2(15 - x)
X - 5 = 30 - 2x
3x = 35
X = 35/3
X = 11 2/3

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Bunuel
If x is the number on the number line between 5 and 15 that is twice as far from 5 as from 15, then x is


A. \(5 \frac{2}{3}\)

B. \(10\)

C. \(11 \frac{2}{3}\)

D. \(12 \frac{1}{2}\)

E. \(13 \frac{1}{3}\)
Since x is BETWEEN 5 and 15, we know that:
15 - x = the distance from x to 15
x - 5 = the distance from x to 5

GIVEN: x is twice as far from 5 as x is from 15
In other words, the distance from x to 5 is TWICE the distance from x to 15.
So, we can write: x - 5 = 2(15 - x)
Expand: x - 5 = 30 - 2x
Rearrange values to get: 3x = 35
Solve: x = 35/3 = 11 2/3

Answer: C

Cheers,
Brent
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Set up a Ratio: If the Distance from 5 is TWICE the Distance from 15


Distance from 5 : Distance from 15 : TOTAL = 2g : 1g : 3g


Actual Value of the TOTAL Distance between 5 and 15 = 15 - 5 = 10

3g = 10

g = 3 + 1/3***


2g = 2 (3 + 1/3) = 6 + 2/3

X is therefore + (6 and 2/3) AWAY from 5 ----

5 + (6 and 2/3) = 11 and 2/3

-C-
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