bicyclist rides 30 miles against a wind and returns 30 miles with the wind. his average speed for the return trip is 5mph faster. how fast did the cyclist ride against the wind if the total time of the trip was 5 hours

Answer 1

the distance is the same.

v=riding velocity against wind
v+5=riding velocity with wind.

time there=(30/v)
time back=(30/(v+5))

so 5=30/v + 30/v+5
v(v+5)=6(v+5)+6v
v^2+5v-6v-6v=30
v^2-7v-30=0
(v-10)(v+3)=0
v=10