Aboat traveled upstream a distance of 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average of (v+3) miles per hour. if the trip upstream took half an hour longer than the trip downstream, how many hours did it take the boat to travel downstream?

Trip upstream took 90v−390v−3 hours and trip downstream took 90v+390v+3 hours. Also given that the difference in times was 1212 hours --> 90v−3−90v+3=1290v−3−90v+3=12;

90v−3−90v+3=1290v−3−90v+3=12 --> 90(v+3)−90(v−3)v2−9=1290(v+3)−90(v−3)v2−9=12 --> 90∗6v2−9=1290∗6v2−9=12 --> v2=90∗6∗2+9v2=90∗6∗2+9 --> v2=9∗(10∗6∗2+1)v2=9∗(10∗6∗2+1) --> v2=9∗121v2=9∗121 --> v=3∗11=33v=3∗11=33;

Trip downstream took 90v+3=9033+3=2.590v+3=9033+3=2.5 hours.

Yes, that is the correct formula. to solve this problem given the number of years, plug in the number of years for t and plug the whole thing into the

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