SOLUTION: Can someone please help me with a word problem? Here is the word problem that I am trying to solve: "A tugboat goes 180 miles upstream in 10 minutes. The return trip downstream t

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Can someone please help me with a word problem? Here is the word problem that I am trying to solve: "A tugboat goes 180 miles upstream in 10 minutes. The return trip downstream t      Log On


   



Question 630771: Can someone please help me with a word problem?
Here is the word problem that I am trying to solve:
"A tugboat goes 180 miles upstream in 10 minutes. The return trip downstream takes 5 minutes. Find the speed of the tugboat without a current and the speed of the current."
I assigned the x variable to the tugboat and the y variable to the current. It makes sense to me that if I divide 180 by 10, the speed of the tugboat without the current would be 18 mph. It also makes sense to me that if the tugboat made the return trip downstream in half the time, the current needs to be moving at the same speed as the tugboat to increase the speed of the boat by 2. If the boat made the trip in half the time, it had to be moving twice as fast. We are working on linear equations in two variables, and I am having difficulty writing two equations to solve this problem. If I knew the two equations, I would have no trouble solving them. Thanks!

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
There isn't a boat in the world let alone a tugboat that can go 180 miles in 10 minutes.
There isn't a car in the world that can go that fast.
You will be hard pressed to find a plane that can go that fast. Most commercial planes go about 500 mph.
10 minutes not 10 hours. Yet you have mph miles per hour.
10 minutes is 1/6 hour
r*t=d
r=d/t
r-c=180/1/6 =1080
r+c=180/1/12=2160
add them
2r=3240
r=1620
1620-1080=c=540
1620+c=2160
c=2160-1620=540