# 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?

Jun 9, 2018

$x = \frac{5}{3}$ OR $x = 10$

#### Explanation:

We know that Rate$\times$Time = Distance
Therefore, Time = Distance$\div i \mathrm{de}$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$\div i \mathrm{de}$Rate can be determined.

$\frac{8}{x} + \frac{8}{x - 5} = \frac{12}{5}$

In the above equation, I added the time (Distance$\div i \mathrm{de}$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)

$8 \left(x - 5\right) \left(5\right) + 8 \left(x\right) \left(5\right) = 12 \left(x\right) \left(x - 5\right)$
$40 x - 200 + 40 x = 12 {x}^{2} - 60 x$
$10 x - 50 + 10 x = 3 {x}^{2} - 15 x$
$3 {x}^{2} - 35 x + 50 = 0$
$3 {x}^{2} - 30 x - 5 x + 50 = 0$
$3 x \left(x - 10\right) - 5 \left(x - 10\right) = 0$
$\left(3 x - 5\right) \left(x - 10\right) = 0$
$3 x - 5 = 0$ OR $x - 10 = 0$
$x = \frac{5}{3}$ OR $x = 10$