SOLUTION: Bill leaves his house for Makayla's house, riding his bicycle at 8 miles per hour. At the same time, Mikayla leaves her house, heading for bill's house walking at 3 miles per hour.

Question 349512: Bill leaves his house for Makayla's house, riding his bicycle at 8 miles per hour. At the same time, Mikayla leaves her house, heading for bill's house walking at 3 miles per hour. Considering they live 8.25 miles apart please write a linear system that shows at what time they would meet.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Bill leaves his house for Makayla's house, riding his bicycle at 8 miles per
hour. At the same time, Mikayla leaves her house, heading for bill's house
walking at 3 miles per hour.
Considering they live 8.25 miles apart please write a linear system that shows at what time they would meet.
:
Let t = time B bikes and time M walks
:
When they meet their total distance will = 8.25 miles
Write a distance equation; dist = speed * time
:
B's dist + M's dist = 8.25 mi
8t + 3t = 8.25
11t = 8.25
t =
t = .75 hrs or .75(60) = 45 minutes for them to meet
;
:
Check by finding the actual dist each traveled
8(.75) = 6 mi
3(.75) = 2.25
-------------
total: 8.25 mi; confirms our solution of t = .75
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