Question 638441
"Matt traveled to his friends house and back. The trip there took 4 hours, the trip back took 6 hours. He averaged 20 mph faster on the trip there then on the trip back. Find his average speed on the out going trip.
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let x=speed of outgoing trip
x-20=speed of trip back
distance=travel time*speed (same for outgoing and trip back))
..
4x=6(x-20)
4x=6x-120
2x=120
x=60
speed of outgoing trip=60 mph