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

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: A student walks and jogs to college each day. The student averages 5 km/h walking and 9 km/h jogging. The distance from home to college is 8 km, and the student makes the trip in       Log On


   

Question 67070This question is from textbook Algebra 1
: A student walks and jogs to college each day. The student averages 5 km/h walking and 9 km/h jogging. The distance from home to college is 8 km, and the student makes the trip in 1 hour. How far does the student jog?
THANK YOU!!! This question is from textbook Algebra 1

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A student walks and jogs to college each day. The student averages 5 km/h walking and 9 km/h jogging. The distance from home to college is 8 km, and the student makes the trip in 1 hour. How far does the student jog?
:
Let x = the distance jogged
Then (8-x) = distance walked
:
Write a time equation: Time = dist/speed
:
Jog time + walk time = 1 hr
x%2F9 + %28%288-x%29%29%2F5%29 = 1
:
Multiply thru by 45 to get rid of the denominators, resulting in:
5x + 9(8-x) = 45
:
5x + 72 - 9x = 45
-4x = 45 - 72
-4x = -27
x = -27/-4
x = + 6.75 km jogged
:
:
Check: 8 - 6.75 = 1.25 km walked
6.75/9 + 1.25/5 =
.75 + .25 = 1
:
How about this, did it make sense to you?