Question

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?

Answer 1

`x=5/3` OR `x=10`

Explanation:

We know that Rate`times`Time = Distance
Therefore, Time = Distance`divide`Rate
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, Distance`divide`Rate can be determined.

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

In the above equation, I added the time (Distance`divide`Rate) 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`