Question 404644
jerri jogs 1 hour each day over the same 9 mile course.
 five miles of the course is downhill, whereas the other 4 miles is on level ground.
 jerri figures that she runs 2 miles per hour faster downhill than she runs on level ground.
 find the rate at which jerri runs on level ground. 
:
Let s = the speed on level ground
then
(s+2) = the speed down hill
:
Write a time equation: Time = dist/speed
:
Level ground time + downhill time = 1 hr
{{{4/s}}} + {{{5/((s+2))}}} = 1
:
Multiply by s(s+2) to clear the denominators, results:
4(s+2) + 5s = s(s+2)
4s + 8 + 5s = s^2 + 2s
9s + 8 = s^2 + 2s
Arrange as a quadratic equation on the right
0 = s^2 + 2s - 9s - 8
:
s^2 - 7s - 8 = 0
Factors to
(s-8)(s+1) = 0
The positive solution is all we want here
s = 8 mph on level ground
:
:
Check solution by finding the actual times (downhill speed = 10 mph)
4/8 + 5/10 =
1/2 + 1/2 = 1 hr, confirms our solution of s = 8