Asked by angela

A student walks and jogs to college each day. She averages 5km/h walking and 9km/h jogging. The distance from home to college is 8km and she makes the trip in 1 hour. How far does the student jog.

To set up the equation I got:

5(x)+9(8-x)=8

5x+72-9x=8
-4x=-64
x=16

but this gives me a negative number for the walking.

WHAT IS THE CORRECT WAY TO SET UP AND SOLVE THE EQUATION?
Answered by Damon
You x is TIME, not distance
5(x) + 9 (1-x) = 8
5 x + 9 -9 x = 8
- 4x = - 1
x = 1/4 or 15 minutes walking
3/4 hour jogging
3/4 * 9 = 27/4 = 6.75 km jogging
Answered by mark
OOOH! I see, thanks for the help!
Answered by surer
i need help with my homewok please help out and is hard that why i am asking help.
Answered by drwls
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