SOLUTION: You travel 575 miles in 8 hours. Part of the trip is made at 60mph and part of the trip is made at 80mph. Find the time traveled at each rate. This is the question. I have that d

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: You travel 575 miles in 8 hours. Part of the trip is made at 60mph and part of the trip is made at 80mph. Find the time traveled at each rate. This is the question. I have that d      Log On


   

Question 96328This question is from textbook
: You travel 575 miles in 8 hours. Part of the trip is made at 60mph and part of the trip is made at 80mph. Find the time traveled at each rate. This is the question. I have that distance =rate * time and 575=r*8. I also have that r=(60T+80t)/2 and T+t=8 but when i try to solve it I get a negative time. Thanks for trying to help. This question is from textbook

Found 2 solutions by checkley71, Earlsdon:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
575=60T+80(8-T)
575=60T+640-80T
575=-20T+640
-20T=575-640
-20T=-65
T=-65/-20
T=3.25 HOURS DRIVING @ 60MPH.
8-3.25=4.75 HOURS DRIVING @ 80MPH.
PROOF
575=60*3.25+80*4.75
575=195+380
575=575


Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
You are taking the right approach! Let's see how it goes:
d%5B1%5D+=+r%5B1%5Dt%5B1%5D and...
d%5B2%5D+=+r%5B2%5Dt%5B2%5D We are given that:
r%5B1%5D+=+60
r%5B2%5D+=+80
d%5B1%5D%2Bd%5B2%5D+=+575 or d%5B1%5D+=+575-d%5B2%5D
and we have to find:
t%5B1%5D and t%5B2%5D but we also know that:
t%5B1%5D%2Bt%5B2%5D+=+8 or t%5B1%5D+=+8-t%5B2%5D
Let's put in some values.
d%5B1%5D+=+60t%5B1%5D or d%5B1%5D+=+60%288-t%5B2%5D%29
d%5B2%5D+=+80t%5B2%5D
Now, since d%5B1%5D+=+d%5B1%5D we can write:
575-d%5B2%5D+=+60%288-t%5B2%5D%29 Solve this for d%5B2%5D
d%5B2%5D+=+575-60%288-t%5B2%5D%29
d%5B2%5D+=+575-480%2B60t%5B2%5D
d%5B2%5D+=+95%2B60t%5B2%5D substitute d%5B2%5D+=+80t%5B2%5D
80t%5B2%5D+=+95%2B60t%5B2%5D Solve this for t%5B2%5D
20t%5B2%5D+=+95
t%5B2%5D+=+95%2F20
t%5B2%5D+=+4.75hours.
t%5B1%5D+=+8-t%5B2%5D
t%5B1%5D+=+8-4.75
t%5B1%5D+=+3.25hours.
Check:
r%5B1%5Dt%5B1%5D%2Br%5B2%5Dt%5B2%5D+=+575 The total trip distance.
60%283.25%29%2B80%284.75%29+=+195%2B380 = 575