Let the school be at point x
Mr. Beecher lives 8 miles away from school
Therefore, his starting point is x – 8
Every minute, he gets 1/6 of a mile closer to school
Therefore, at time, t(in minutes), Mr. Beecher’s distance
from school, Db = x – 8 + (1/6)t
Db = x – 8 + (1/6)t … (Equation I)
Mrs. Carter lives 20 miles away from school
Therefore, her starting point is x – 20
Every minute, she gets 1/2 of a mile closer to school
Therefore, at time, t(in minutes), Mrs. Carter’s distance
from school, Dc = x – 20 + (1/2)t
Dc = x – 20 + (1/2)t … (Equation II)
After how many minutes will Mr. Beecher and Mrs. Carter
first be the same distance away from school?
That’s when Db = Dc, given that t is same value in both
equations
Db = x – 8 + (1/6)t … (Equation I)
Dc = x – 20 + (1/2)t … (Equation II)
Db = Dc. Therefore,
x – 8 + (1/6)t = x – 20 + (1/2)t
Subtract x from both sides of the equation
x – x – 8 + (1/6)t = x – x – 20 + (1/2)t
– 8 + (1/6)t = – 20 + (1/2)t
(1/6)t – 8 = (1/2)t – 20
Add 20 to both sides of the equation
(1/6)t – 8 + 20 = (1/2)t – 20 + 20
(1/6)t + 16 = (1/2)t
(1/2)t = (1/6)t + 16
Subtract (1/6)t from both sides of the equation
(1/2)t – (1/6)t = (1/6)t – (1/6)t + 16
1/3t = 16
Multiply both sides of the equation by 3
1/3t x 3 + 16 x 3
t = 48 minutes
Therefore, after 48 minutes, Mr. Beecher and Mrs. Carter will
first be the same distance away from school.
