SOLUTION: During the first part of a trip, a canoeist travels 98 miles at a certain speed. The canoeist travels 10 miles on the second part of the trip at a speed 5 mph slower. The total tim

Algebra ->  Rational-functions -> SOLUTION: During the first part of a trip, a canoeist travels 98 miles at a certain speed. The canoeist travels 10 miles on the second part of the trip at a speed 5 mph slower. The total tim      Log On


   

Question 473643: During the first part of a trip, a canoeist travels 98 miles at a certain speed. The canoeist travels 10 miles on the second part of the trip at a speed 5 mph slower. The total time for the trip is 3 hours. What was the speed on the FIRST part of the trip? What was the speed on the SECOND part of the trip?
(type an integer or a decimal. Round to the nearest hundredth)
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
first part x mph
second part -5 mph

first part 98 miles
second part 10 miles

speed in first part x mph
speed second part x -5 mph
Total rowing time 3 hours
Time first part 98 / x
Time second part 10/(x-5)
Time first part + time second part = 3 hours
98/x+10 /(x-5) = 3
LCD =x*(x-5)
multiply the equation by the LCD
we get
98*(x-5)+10x = 3
98x-490 +10x=3x^2 + -15 x
123x-490=3x^2
3x^2-123x=490 = 0
3x^2-123x+490 =0
/3
x^2-41x+163.33 =0
Find the roots of the equation by quadratic formula
a= 1 b= -41 c= 163.33

b^2-4ac= 1681 - -653.33
b^2-4ac= 1027.67
sqrt%28%091027.67%09%29= 32.06
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=(41+32.06)/ 2
x1= 36.53
x2=(41-32.06)/ 2
x2= 4.47 4 4/9
Ignore 4.47
x=36.53 mph speed - first part
31.53 mph speed second part
m.ananth@hotmail.ca