speed in first trip ---- x mph
time for 1st trip = 93/x hours
speed in 2nd trip ---- x-5 mph
time for 2nd trip = 5/(x-5)
93/x + 5/(x-5) = 2
93(x-5) + 5x = 2x(x-5)
93x - 465 + 5x = 2x^2 - 10x
2x^2 - 108x + 465 = 0
x = 49.3 or x = 4.7
since the speed in the second trip is reduced by 5, then the new speed would have to be 4.7-5, which is negative. So we'll reject the second answer
check: if speed = 49.3
time = 93/49.3 + 5/44.3 = 1.999 (not bad)
This answer is wrong, If I can just get the time of the first trip, I can take it from there. I don't need to know the second trip, just the first trip. I appreciate your help. Thank you
The answer I gave you is correct for the way the question is given.
The answer was "x=49.3" or the speed on the first part was 49.3 mph and on the second part it was 44.3 mph.
I then checked the answer and got 2 hours as the total time.
The question itself is totally flawed.
I have done enough canoeing in my life time to know that you cannot do a 98 mile canoe trip in 2 hours.
Reiny is totally correct Ester.
You cannot get the correct speed by considering only the first leg of the trip.
The time of the first leg is T1 = 93/V while the time of the second leg is T2 = 5/(V-5).
Adding, 93/V + 5/(V-5) = 2
Multiplying out and simplifying yields 2V^2 - 108V + 465 = 0 which produces a positive V = 49.282 mph, the speed during the 93 miles trip, the speed during the second leg being 44.282 mph.