SOLUTION: Bill leaves for Mikes house traveling 8 miles per hour. At the same time Mike leaves his house heading for Bills house travelin 3 miles per hour. The distance between the two hous
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Question 624612: Bill leaves for Mikes house traveling 8 miles per hour. At the same time Mike leaves his house heading for Bills house travelin 3 miles per hour. The distance between the two houses in 8.25 miles. At what point will they meet? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Bill leaves for Mikes house traveling 8 miles per hour.
At the same time Mike leaves his house heading for Bills house traveling 3 miles per hour.
The distance between the two houses in 8.25 miles.
At what point will they meet?
:
Let t = time for them to meet
:
Write a distance equation; dist = speed * time
:
Mikes dist + Bills dist = 8.25
8t + 3t = 8.25
11t = 8.25
t = 8.25/11
t = .75 hrs traveling time when they meet
:
Find how far is the meeting point from Mikes house?
.75 * 8 = 6 mi
:
Check this by finding the meeting point from Bill's house
.75 * 3 = 2.25 mi
The sum of the distances should be 8.25 mi
