dan and peg paddled their canoe 40 mi in 5 hr with the current and 16mi in 8hr against the current. find the speed of the current and speed of canoe in still...

Question

i need to apply the distance rate and time formula using 2 variables in systems of linear equations. ( the last word was water, i couldnt fit the entire question in.)

Answer 1

Joseph,

The current moves in one direction.  If they don't paddle, the canoe will go with the current.  So the current either helps them along or pushes them back.

Let x = speed of the canoe and y = speed of the current.

You are correct, distance = rate * time is key

40 = 5(x +y)

16 = 8 (x - y)

Now we could multiply through with the 5 and the 8 in those equations but they would end up  much simpler is we divided equation 1 by 5 on both sides and equation 2 by 8 on both sides.

40/5 = 5/5(x+y)

8 = x + y

16/8 =8/8 ( x - y)

2 = x - y

 

Look at the two together

8 = x + y

2 = x  - y

You can add the two equations, and solve for x, then substitute the x value back into an equation and solve for y.  Check this by substituting the x,y pair into the original equations.

Make sense?  I hope so!

Mary

Answer 2

Distance = rate*time

 

Rate = distance/time

 

Let c = canoe speed in still water

Let w = rate of water

 

With current the speed is c+w:

c + w = 40/5 = 8 mph with the current

 

Against current the speed is c-w:

c - w = 16/8 = 2 mph against the current

 

c + w = 8

c - w = 2     add equations

----------------

2c = 10

c = 5 mph canoe speed in still water

 

c-w =2

w = 3 mph current speed