algebra 2
posted by esther .
During the first part of a trip a canoeist travels 93 miles at a certain speed. The conoiest travels 5 miles on the second trip at 5mph slower. The total time for the trip is 2 hrs. What was the speed on each part of the trip?
Thank you for all your help.

speed in first trip  x mph
time for 1st trip = 93/x hours
speed in 2nd trip  x5 mph
time for 2nd trip = 5/(x5)
93/x + 5/(x5) = 2
93(x5) + 5x = 2x(x5)
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 4.75, 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/(V5).
Adding, 93/V + 5/(V5) = 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.
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