SOLUTION: suppose you drive 100 mile home for fall break. your return trip took 30 minutes less because you drive 10 mph faster. What was your rate on the return trip?

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Question 737804: suppose you drive 100 mile home for fall break. your return trip took 30 minutes less because you drive 10 mph faster. What was your rate on the return trip?
Answer by Ed Parker(21) About Me  (Show Source):
You can put this solution on YOUR website!
suppose you drive 100 mile home for fall break. your return trip took 30 minutes less because you drive 10 mph faster. What was your rate on the return trip?
             distance     rate      time=distance/rate
going home     100         x            100/x 
returning      100        x+10          100/(x+10)

             
           %28matrix%282%2C1%2C%0D%0A%0D%0Atime%2C+returning%29%29%22%22=%22%22%28matrix%282%2C1%2C%0D%0A%0D%0Atime%2C+going%29%29%22%22-%22%221%2F2hour

                  100%2F%28x%2B10%29%22%22=%22%22100%2Fx%22%22-%22%221%2F2

Solve that and get x=40 mph going home and therefore the returning rate
is x+10 or 40+10 = 50 mph.

Edwin