Algebra
posted by Stella .
During the first part of a trip, a canoeist travels 57 miles at a certain speed. The canoeist travels 13 miles on the second trip at a speed 5 mph slower. The total time for the trip is 5 hrs. what was the speed on each part of the trip?
The speed on the first part is?
The speed on the second part is? Someone was kind enough to help, but it did not work into the equation. Help? Please..

Algebra 
Henry
V1 = X mi/h on 1st part of trip.
t1 = d1/V1 = 57 / X hrs.
t2 = d2/V2 = 13 / (x5) hrs.
t1 + t2 = 5 hrs.
57/x + 13/(x5) = 5.
Multiply both sides by x(x5).
57(x5) + 13x = 5x(x5).
57x  285 + 13x = 5x^2  25x.
5x^2 + 57x +13x +25x = 285.
5x^2 + 95x = 285.
5x^2 + 95x  285 = 0.
Divide both sides by 5:
x^2  19x + 57 = 0.
Use Quadratic Formula.
X = 15.3 mi/h = Speed on 1st part of trip.
x5 = 15,35 = 10.3 mi/h = Speed on 2nd
part of trip.
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