SOLUTION: One day, Jeff rides his bicycle out into the country 40 miles (see figure). On the way back, he takes a different route that is 2 miles longer. It takes him 6 hours longer to ride

Question 294627: One day, Jeff rides his bicycle out into the country 40 miles (see figure). On the way back, he takes a different route that is 2 miles longer. It takes him 6 hours longer to ride out than to return. If his rate on the way out into the country is 16 miles per hour slower than his rate back, find both rates.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
One day, Jeff rides his bicycle out into the country 40 miles (see figure).
On the way back, he takes a different route that is 2 miles longer.
It takes him 6 hours longer to ride out than to return.
If his rate on the way out into the country is 16 miles per hour slower than
his rate back, find both rates.
:
Let s = his speed into the country
then
(s+16) = his return speed
:
Write a time equation, Time = dist/speed
:
out time - return time = 6 hrs
- = 6
Multiply equation by s(s+16) to eliminate the denominators, results
40(s+16) - 42s = 6s(s+16)
:
40s + 640 - 42s = 6s^2 + 96s
;
640 - 2s = 6s^2 + 96s
Combine like terms on the right
0 = 6s^2 + 96s + 2s - 640
A quadratic equation
6s^2 + 98s - 640 = 0
simplify, divide by 2
3s^2 + 49s - 320 = 0
factors to
(3s + 64)(s - 5) = 0
the positive solution is what we want here
s = 5 mph his outbound rate
and
5 + 16 = 21 mph his return rate
:
:
Check solution by finding the times
40/5 = 8 hrs
42/21 = 2 hrs
------------
diff: 6 hrs

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