Posted by **angela** on Thursday, February 26, 2009 at 7:54am.

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?

- math algebra -
**Damon**, Thursday, February 26, 2009 at 8:02am
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

- math algebra -
**mark**, Thursday, February 26, 2009 at 8:06am
OOOH! I see, thanks for the help!

- math algebra -
**surer**, Thursday, February 26, 2009 at 10:26am
i need help with my homewok please help out and is hard that why i am asking help.

- math algebra (surer) -
**drwls**, Thursday, February 26, 2009 at 5:57pm
Post your question separately, not as part of this question. Click on the link at the top labeled "Post a new question"

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