SOLUTION: at 900am two people leave their homes that are 15 miles apart walking toward each other. if one person walks at a rate 4mph faster than the other and they meet after 1.5 hours, how
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Question 497557: at 900am two people leave their homes that are 15 miles apart walking toward each other. if one person walks at a rate 4mph faster than the other and they meet after 1.5 hours, how fast was each person walking? Answer by htmentor(1343) (Show Source):
You can put this solution on YOUR website! Let d = the distance traveled by the slower person
Then 15 - d = the distance traveled by the faster person
Let s = the slower person's speed
Then s + 4 = the faster person's speed
Since speed = distance/time, and the time is 1.5 h, the slower person's speed is:
s = d/1.5 [1]
And the faster person's speed is:
s + 4 = (15 - d)/1.5 [2]
Solve for s in [1] and substitute into [2]:
s + 4 = (15 - 1.5s)/1.5
s + 4 = 10 - s
2s = 6
s = 3
Therefore, their speeds are 3 mph and 7 mph
