I really need help

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. If the round trip

takes two hours and 24

minutes—that

​is,

125

hours—what

is your average rate on the return​ trip?

Answer 1

8 miles back and forth, total 16 miles

5 mph faster downhill than uphill

total time back&forth =2 hours and 24 minutes = 124 minutes = 2.4 hours

d=rt

distance = rate of speed times time

8 = rt = (r-5)(2.4-t) where r-5<r slower speed uphill and 2.4-t>t longer time uphill

expand

8= 2.4r +5t -rt - 12

8+12 = 2.4r +5t -rt

20 = 2.4r +5t -rt

8 = rt add to eliminate rt

28= 2.4r +5t

2.4r +5t -28 =0

rt = 8

t =8/r

2.4r +5(8/r)- 28 =0

2.4r +40/r -28 =0

.6r+10/r -7 =0

3r^2 +50 - 35r =0

(3r-5)(r-10) =0

r =10 mph (ignore the other factor as it leads to a negative speed for uphill)

10-5 = 5mph uphill

time downhill = t = 8/r = 8/10 = .8 hours = 48 minutes

time uphill = 2.4-.8 = 1.6 hours = 1 hour and 36 minutes

total time back & forth = 2 hours and 24 minutes

check the answer

downhill d=rt = 8 miles = 10 x .8 = 8 downhill

uphill d =(r-5)(2.4-.8) = 5 x 1.6 = 8 uphill