SOLUTION: two cars started from the same place at the same time and traveled in opposite directions. at the end of 6 hours, they were 420 miles apart. if the average rate of the slow car exc

Question 166009: two cars started from the same place at the same time and traveled in opposite directions. at the end of 6 hours, they were 420 miles apart. if the average rate of the slow car exceeded 1/2 of the average rate of the fast car by 13 mph, find the rate of each car.
Found 2 solutions by elima, Mathtut:
Answer by elima(1433) (Show Source): You can put this solution on YOUR website!
fast car = r
slow car= 1/2r+13
d=rt
420=6(r + 1/2r + 13)
420 = 6r + 3r + 78
342=9r
38=r
=================
fast car rate = 38
slow car rate = 1/2(38)(6)
slow car = 192
:)
Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
lets call s the rate of travel of the slow car and f the rate of travel of the fast car.
we know that s=1/2f+13 and we know that the rate multiplied by the time which is 6s+6f=420
so plugging the value for s in the first equation into the second we get 6(1/2f+13)+6f=420
3f+78+6f=420 distribute
9f=342 combined like terms
f=38 so the fast car travels at 38 mph
s=1/2(38)+13=32 so the slow car travels and 32 mph.
the fast car has to be faster than the slow car otherwise the slow car is the fast car........wow
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