SOLUTION: Nancy jogs and walks to school every day. She does 4km/h walking and 8km/h jogging. From home to school it is 6km and it is an 1 hour trip. Using substitution how far does Nancy jo

Algebra ->  Expressions-with-variables -> SOLUTION: Nancy jogs and walks to school every day. She does 4km/h walking and 8km/h jogging. From home to school it is 6km and it is an 1 hour trip. Using substitution how far does Nancy jo      Log On


   

Question 70545: Nancy jogs and walks to school every day. She does 4km/h walking and 8km/h jogging. From home to school it is 6km and it is an 1 hour trip. Using substitution how far does Nancy jog?
Answer by funmath(2933) About Me  (Show Source):
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Nancy jogs and walks to school every day. She does 4km/h walking and 8km/h jogging. From home to school it is 6km and it is an 1 hour trip. Using substitution how far does Nancy jog?
Let the time jogged be: x
Let the time walked be: y
Then the total time is: x+y=1
Distance=rate * time
The distance jogged is: 8x
The distance walked is: 4y
Then the total distance is: 8x+4y=6
So the system your solving is:
x+y=1
8x+4y=6
Solve the 1st equation for y and substitute it into the second equation and solve for x.
x+y=1
x-x+y=1-x
y=1-x Substitute that in for y in the second eqaution:
8x+4(1-x)=6
8x+4-4x=6
4x+4=6
4x+4-4=6-4
4x=2
4x/4=2/4
x=1/2
The distance jogged is 8x, let x=1/2, and the distance jogged is: 8(1/2)=4km
Happy Calculating!!!!