Question

Jonah and Jesse are taking a trip. Jonah left at 7 a.m. and travel at an average of 40 miles per hour. Jesse left at 10 a.m. and travel at average of 50 mph. At what time are they at the same place at the same time? Write a system of equations to represent this situation. Then use the substitution method with that system to determine at what time they will be in the same location. How many miles away from will they be at that time?
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Answer 1

Answer:

• y=40(t -7); y=50(t-10)
• t=22; 10 p.m.
• y=600 miles from home

Step-by-step explanation:

a) Let t represent the clock time (hours after midnight). The distance from home will be ...

  distance = speed × time

  jonah = 40(t -7)

  jesse = 50(t -10)

  jonah = jesse . . . . . . . same distance from home at the same time

(Note: you could let y = distance traveled and use y in both equations. That would give 2 equations in 2 unknowns, instead of 3 equations in 3 unknowns.)

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b) substituting the first two equations into the second, we get ...

  40(t -7) = 50(t -10)

  4t -28 = 5t -50 . . . . . . . divide by 10, eliminate parentheses

  22 = t

Jonah and Jesse will be at the same place at 10 p.m.

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c) We can find the distance by substituting into either equation.

  jonah = 40(22 -7) = 40(15) = 600

They will be 600 miles from home at that time.