SOLUTION: In three days Carter drove 2196 miles in 36 hours behind the wheel. The first day he averaged 64 mph, the second day 62 mph, and the third day 58 mph. If he drove 4 more hours on t

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: In three days Carter drove 2196 miles in 36 hours behind the wheel. The first day he averaged 64 mph, the second day 62 mph, and the third day 58 mph. If he drove 4 more hours on t      Log On


   

Question 577612: In three days Carter drove 2196 miles in 36 hours behind the wheel. The first day he averaged 64 mph, the second day 62 mph, and the third day 58 mph. If he drove 4 more hours on the third day than on the first day, then how many hours did he drive each day?
Here is what I have so far:

Let x= hours day 1
Let y= hours day 2
Let z= hours day 3
x+y+z=2196
day 1= 64x
day 2= 62y
day 3= 58z
z= x+4
64x+62y+58(x+4)=2196
64x+62y+58x+232=2196
and then
122x+62y=1964
and after that I am completely stuck...
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Let x= hours day 1
Let y= hours day 2
Let z= hours day 3
x+y+z=2196
day 1= 64x
day 2= 62y
day 3= 58z


x+y+z=36
y=36-x-z
64x+62y+58z= 2196
64x+62(36-x-z)+58z= 2196
64x+62*36-62x-62z+58z=2196
2x-4z=-2232+2196
2x-4z=-36
Now z= x+4
2x-4(x+4)=-36
2x-4x-16=-36
-2x=-20
/-2
x=10 hours first day
z=x+4 = 14 hours
y=12 hours