SOLUTION: mr winkle jogs 4 miles and then turns and jogs back to his starting point.the first part of his jog he was going mostly uphill so his speed was 2 miles per hr slower than his speed

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Question 168706: mr winkle jogs 4 miles and then turns and jogs back to his starting point.the first part of his jog he was going mostly uphill so his speed was 2 miles per hr slower than his speed returning.if the total time he spent was 1 2/3 hour find his rate going and his rate returning
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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mr winkle jogs 4 miles and then turns and jogs back to his starting point.the first part of his jog he was going mostly uphill so his speed was 2 miles per hr slower than his speed returning.if the total time he spent was 1 2/3 hour find his rate going and his rate returning
:
Let s = speed returning
then
(s-2) = speed up the hill
:
Write a time equation: Time = dist/speed
;
Time return + time out = 1 & 2/3 hrs
4%2Fs + 4%2F%28%28s-2%29%29 = 12%2F3
:
4%2Fs + 4%2F%28%28s-2%29%29 = 5%2F3
Multiply equation by 3s(s-2) to get rid of the denominators, results
4*3(s-2) + 4*3s = 5s(s-2)
:
12s - 24 + 12s = 5s^2 - 10s
:
24s - 24 = 5s^2 - 10s
:
0 = 5s^2 - 10s - 24s + 24
:
A quadratic equation, solve for s
5s^2 - 34s + 24 = 0
Factors to:
(5s - 4)(s - 6) = 0
:
5s = 4
s = 4%2F5; this solution makes no sense.
and
s = 6 mph is his return speed, then obviously 4 mph is his up-hill speed
:
:
Check solution by find the times
4/6 + 4/4 = 12%2F3 hrs