SOLUTION: during the first part of the trip a canoeist travels 40 miles at a certain speed the canoeist travels 20 miles on the seconf part of the trip at a speed 5 mph slower the total time

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Question 574461: during the first part of the trip a canoeist travels 40 miles at a certain speed the canoeist travels 20 miles on the seconf part of the trip at a speed 5 mph slower the total time for the trip is 3 hours what was the speed on each part of the trip?
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Forward 40 miles
Returning speed to DC 20 miles

Forward speed x mph
return speed x-5 mph
Total time =3hours
Time forward 40 / x
time return 20/(x-5)

add up the time =3 hours

40/x + 20/(x-5)=3
LCD = (x)*(x-5)
multiply the equation by the LCD
we get
40*(x-5 )+20x = 3
40x-200 +20x=3x^2-15x
75x-200 =3x^2
3x^2-75x+200= 0
3X^2-75 x+200 = 0
Find the roots of the equation by quadratic formula

a= 3 b= -75 c= 200

b^2-4ac= 5625 - 2400
b^2-4ac= 3225
sqrt%28%093225%09%29= 56.79 2.99
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=( 75 + 56.79 )/ 6
x1= 21.96
x2=( 75 -56.79 ) / 6
x2= 3.04
Forward speed = 22 mph
return speed = 17 mph