SOLUTION: during the first part of a trip, a bicyclist travels 56 miles at a certain speed. the return trip is made at a speed that is 6 mph slower. total time for the round trip is 16 hr. f

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: during the first part of a trip, a bicyclist travels 56 miles at a certain speed. the return trip is made at a speed that is 6 mph slower. total time for the round trip is 16 hr. f      Log On

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Question 867889: during the first part of a trip, a bicyclist travels 56 miles at a certain speed. the return trip is made at a speed that is 6 mph slower. total time for the round trip is 16 hr. find the bicyclists average speed on each part of the trip.
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
to go there speed x mph
Return speed x -6 mph
Total time 16 hours
First Part time 56 / x
Second Part time 56 / ( x -6 )

Time first part + time second part = 16 hours

56 / x + 56 /(x -6 ) = 16
LCD =x* (x -6 )
multiply the equation by the LCD
we get
56 * (x -6) +56x=16x^2-96x
56x+56x-336=16x^2-96x
16x^2-208x+336=0
/16
x^2-13x+21=0Find the roots of the equation by quadratic formula

a= 1 b= -13 c= 21

b^2-4ac= 169 - 84
b^2-4ac= 85 sqrt%28%0985%09%29= 9.22
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=( 13 + 9.22 )/ 2
x1= 11.11
x2=( 13 - 9.22 )/ 2
x2= 1.89
Ignore negative value
x = 11.11 mph
To there speed = 11.11 mph
return speed subtract 6 mph