SOLUTION: Two people leave their homes at 11:30 am and between the two of them, they drive a total of 37.5 miles. one drives an average of 15mph faster then the other . they meet at noon fin

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Question 622266: Two people leave their homes at 11:30 am and between the two of them, they drive a total of 37.5 miles. one drives an average of 15mph faster then the other . they meet at noon find the average driving speed for each?
d=rt
d=37.5
t=30 min
let x =rate
x(15) + x * 30 min = 37.5
16x * 30 min = 37.5
I don't know if I am on the right track but I don't know where to go from here
Please help with a formula
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = the speed of the slower driver in mi/hr
+s+%2B+15+ = the speed of the faster driver
Let +d+ = distance driven by slower driver
+37.5+-+d+ = distance driven by faster driver
+.5+ hrs is the time for both drivers
-------------
Slower driver's equation:
(1) +d+=+s%2A%28.5%29+
Faster driver's equation:
(2) +37.5+-+d+=+%28+s+%2B+15+%29%2A%28.5%29+
---------------------------
Substitute (1) into (2)
(2) +37.5+-+d+=+%28+s+%2B+15+%29%2A%28.5%29+
(2) +37.5+-+.5s+=+%28+s+%2B+15+%29%2A%28.5%29+
(2) +37.5+-+.5s+=+.5s+%2B+7.5+
(2) +s+=+37.5+-+7.5+
(2) +s+=+30+
and
+s+%2B+15+=+45+
The speed of the slower driver is 30 mi/hr
The speed of the faster driver is 45 mi/hr
check:
(1) +d+=+s%2A%28.5%29+
(1) +d+=+.5%2A30+
(1) +d+=+15+ mi
and
(2) +37.5+-+15+=+%28+s+%2B+15+%29%2A%28.5%29+
(2) +22.5+=+.5%2A%28+s+%2B+15+%29+
(2) +45+=+s+%2B+15+
(2) +s+=+30+
OK