SOLUTION: Walking and Jogging: A student walks and jogs to college each day. She averages 5km/h walking and 9km/h jogging. The distance from the home to college is 8km, and she makes the

Algebra ->  Expressions-with-variables -> SOLUTION: Walking and Jogging: A student walks and jogs to college each day. She averages 5km/h walking and 9km/h jogging. The distance from the home to college is 8km, and she makes the       Log On


   

Question 154051This question is from textbook
: Walking and Jogging: A student walks and jogs to college each day. She averages 5km/h walking and 9km/h jogging. The distance from the home to college is 8km, and she makes the trip in 1 hour. How far does the student jog? This question is from textbook

Answer by orca(409) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be the distance the student jogs, the the distance the student walks is 8-x.
The time jogging is x%2F9.
The time walking is %288-x%29%2F5
As the time jogging + the time walking = 1 hour, we have
x%2F9%2B%288-x%29%2F5=1
Solving for x, we have
45%28x%2F9%2B%288-x%29%2F5%29=1%2A45 (multiply both sides by 45)\
5x+%2B9%288-x%29=45
5x%2B72-9x=45
-4x=-27
x+=+27%2F4
so the student jogs 27%2F4 miles.