You ride your bike to campus a distance of 8 miles and return home on the same route. Going to​ campus, you ride mostly downhill and average 5 miles per hour faster than on your return trip home. Continued in details?

If the round trip takes two hours and 24 minutes—that ​is, 12/5 hours—what is your average rate on the return​ trip?

1 Answer
Jun 9, 2018

x=5/3 OR x=10

Explanation:

We know that RatetimesTime = Distance
Therefore, Time = DistancedivideRate
We can also create two equations to solve for the rate: one for to campus and one for coming back home.

TO FIND THE AVERAGE RATES
Let x = your average rate on the return trip.
If we define x as above, we know that x-5 must be your average rate on the way to campus (going home is 5mph faster)

TO CREATE AN EQUATION
We know that both trips were 8 miles. Therefore, DistancedivideRate can be determined.

8/x+8/(x-5)=12/5

In the above equation, I added the time (DistancedivideRate) of both trips to equal the given total time.

TO SOLVE THE EQUATION
Multiply the whole equation through by the LCM (the product of all the denominators in this case)

8(x-5)(5)+8(x)(5)=12(x)(x-5)
40x-200+40x=12x^2-60x
10x-50+10x=3x^2-15x
3x^2-35x+50=0
3x^2-30x-5x+50=0
3x(x-10)-5(x-10)=0
(3x-5)(x-10)=0
3x-5=0 OR x-10=0
x=5/3 OR x=10