Question 1044407
Jessica drove to her cabin on the lake and back. it took one hour longer to go there then it did to come back. The average speed on the trip was 50 km/h. The average speed on the way back was 75 km/h. how many hours to the trip there take?
<pre>This is done based on the assumption that the speed to the cabin was 50 km/h
Let time taken to get to the cabin be T
Then time taken on return trip = T – 1
We then get the following DISTANCE equation: 50T = 75(T – 1)
50T = 75T – 75
50T – 75T = - 75 
– 25T = - 75
T, or time taken to get to the cabin = {{{(- 75)/(- 25)}}}, or {{{highlight_green(matrix(1,2, 3, hours))}}}