SOLUTION: Jose and Eduardo are going to run a race. Jose is willing to give Eduardo a 3 mile head start. Jose runs at a rate of 7.5 miles per hour. Eduardo's rate is 1.5 miles per hour SLOWE
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-> SOLUTION: Jose and Eduardo are going to run a race. Jose is willing to give Eduardo a 3 mile head start. Jose runs at a rate of 7.5 miles per hour. Eduardo's rate is 1.5 miles per hour SLOWE
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Question 1073991: Jose and Eduardo are going to run a race. Jose is willing to give Eduardo a 3 mile head start. Jose runs at a rate of 7.5 miles per hour. Eduardo's rate is 1.5 miles per hour SLOWER that Jose. Found 2 solutions by josgarithmetic, ikleyn:Answer by josgarithmetic(39620) (Show Source):
You can put this solution on YOUR website! Eduardo's rate is 6 miles per hour. If he had the 3 mile head start, then for this distance he had run , half an hour.
d is the distance they both go when Jose catches up with Eduardo.
t is how long it takes for Jose to reach Eduardo.
SPEED TIME DISTANCE
JOSE 7.5 t d
EDUARDO 6 t+1/2 d
If solve for time only, t for how long to catchup,
You can put this solution on YOUR website! .
Jose and Eduardo are going to run a race. Jose is willing to give Eduardo a 3 mile head start.
Jose runs at a rate of 7.5 miles per hour. Eduardo's rate is 1.5 miles per hour SLOWER that Jose.
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Your equation is this:
7.5*t = 3 + 6*t
The solution is
7.5t - 6t = 3,
1.5t = 3 ---> t = = 2 hours.
Answer. Jose catches up Eduardo in 2 hours.
Writing by "josgarithmetic" (including his equation) is WRONG, NONSENSICAL and does not correspond to the condition..
It would be good (much better) if you put a question in your post.