Questions on Algebra: Expressions involving variables, substitution answered by real tutors!

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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
: 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?
Answer by orca(336) 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/9.
The time walking is (8-x)/5
As the time jogging + the time walking = 1 hour, we have
x/9+(8-x)/5=1
Solving for x, we have
45(x/9+(8-x)/5)=1*45 (multiply both sides by 45)\
5x +9(8-x)=45
5x+72-9x=45
-4x=-27
x = 27/4
so the student jogs 27/4 miles.