Question 118448
A SAILBOAT TRAVELS 20 MILES DOWNSTREAM IN 3 HOURS.IT RETURNS IN 4 HOURS. FIND THE SPEED OF THE SAILBOAT IN STILL WATER AND THE RATE OF THE CURRENT.
I CAN FIGURE OUT THAT THE RATE OF THE BOAT GOING DOWN STREAM WOULD TRAVEL 6.23 MPH AND COMING BACK IT WOULD TRAVEL 5 MPH.
:
One way is to use two unknowns and use the elimination method
:
Let x = speed of the boat in still water
Let y = speed of the current
then:
(x+y) = speed downstream
(x-y) = speed upstream
:
Write two distance equations: Distance = speed * time
Down stream: 3(x+y) = 20
Upstream:    4(x-y) = 20
:
3x + 3y = 20
4x - 4y = 20
:
Multiply the 1st equation by 4 and the 2nd equation by 3
12x + 12y = 80
12x - 12y = 60
---------------adding eliminates y, find x
24x + 0y = 140
x = 140/24
x = 5.83 mph speed of the boat in still water
:
Find y
3(5.833) + 3y = 20
17.5 + 3y = 20
3y = 20 - 17.5
3y = 2.5
y = 2.5/3
y = .833 mph is the speed of the current
:
:
Check solutions in the 2nd equation
4(5.833) - 4(.833) =
 23.33 - 3.33 = 20 confirms our solution