SOLUTION: alberto made a trip to his friend's house and back. the trip there took three hours and the trip back took four hours. he averaged 9.8 mph faster on the trip there than on the retu
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Question 220073: alberto made a trip to his friend's house and back. the trip there took three hours and the trip back took four hours. he averaged 9.8 mph faster on the trip there than on the return trip. find alberto's average speed on the outbound trip.
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let r=Alberto's average rate of speed on the outbound trip
Then (r-9.8)=Alberto's average rate of speed on the inbound trip
Distance travelled on the outbound trip=(r)*3
Distance travelled on the inbound trip=(r-9.8)*4
Now we know that the above two distances are equal, so our equation to solve is:
3r=4(r-9.8) get rid of parens
3r=4r-39.2 subtract 4r from each side
3r-4r=4r-4r-39.2 collect like terms
-r=-39.2 mph or
r=39.2 mph---Alberto's rate of speed on the outbound trip
r-9.4=39.2-9.8= 29.4 mph---Alberto's average speed on the inbound trip
CK
Distance travelled on outbound trip=39.2*3=117.6 mi
Distance travelled on the inbound trip=29.4*4=117.6 mi
Hope this helps---ptaylor
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