Posted by **Stella** on Sunday, April 22, 2012 at 3:36pm.

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**, Tuesday, April 24, 2012 at 7:46pm
V1 = X mi/h on 1st part of trip.

t1 = d1/V1 = 57 / X hrs.

t2 = d2/V2 = 13 / (x-5) hrs.

t1 + t2 = 5 hrs.

57/x + 13/(x-5) = 5.

Multiply both sides by x(x-5).

57(x-5) + 13x = 5x(x-5).

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.

x-5 = 15,3-5 = 10.3 mi/h = Speed on 2nd

part of trip.

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