SOLUTION: Jane and Fred live 4 miles apart. Jane travels at a rate of 2mi/h and Fred at a rate of 3 mi/h. How long will it be before they meet?

Question 13119: Jane and Fred live 4 miles apart. Jane travels at a rate of 2mi/h and Fred at a rate of 3 mi/h. How long will it be before they meet?
Answer by glabow(165) (Show Source): You can put this solution on YOUR website!
Assume you know where they meet. It is at point M. They take the same amount of time (x) to travel to point M.
She travels at 2 mi/hr. In x hours she will travel 2x mi. This is the distance M. So, 2x = M.
He travels at 3 mi/hr. In x hours he will travel 3x mi. This is the distance from where he is to M. But he is already at 4 mi. (That's how far apart they are when they start!) So the distance he travels to get to M is 4-M. [Can you see why?]
So, 3x = 4-M.
Now we solve for x.
2x = M
3x = 4 - M
3x = 4 - 2x [substituting 2x for M]
5x = 4
x = 4/5
Check: In 4/5 of an hour she will travel 8/5 mi.
In 4/5 of an hour he will travel 12/5 mi.
The total distance they traveled is 8/5 mi. + 12/5 mi. = 20/5 mi. = 4 mi.

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