Problem Solving using Linear Systems

Question

Iggy travels in his boat 60 km upstream in 5 hours. The boat travels the same distance downstream in 3 hours. What is the rate of Iggy’s boat? What is the rate of the current of the stream?

Answer 1

Answer 2

Hi...

For a problem like this, think of the current working against the boat as it travels upstream and helping the boat as it travels downstream. That's why it takes longer to go the 60km upstream than it does going downstream.

You need TWO variables, so take;

X as the rate of the boat

Y as the rate of the current

Distance = Rate x Time

Upstream (against the current), the distance traveled = 60 = 5 (x - y)

Downstream (with the current), the distance traveled = 60 = 3 (x + y)

Eq 1 = 5(x - y) = 60

Eq 2 = 3(x + y) = 60

You want to be able to ADD the two equations and eliminate ONE variable so as to solve for the other.

Best thing to do is multiply Eq 1 and Eq by 5 and 3, respectively. You should always try to do what would make the math easy to cancel out one of the variables.

Eq 1 = ¹/₅ •[5(x - y)] = 60•¹/₅ = x - y = 12

Eq 2 = ¹/₃•[3(x + y)] = 60•¹/₃ = x + y = 20

When you add the TWO equations, the y terms cancel out and you're left with;

2x = 32

x = 16 km/hr - this is the rate of the boat

Then substituting into Eq 2. y = 4km/hr - this is the rate of the current

NOTE: In keeping the existing equations you will end up with 5x - 5y = 60 and 3x + 3y = 60, you will need to multiply Eq 1 and Eq 2 by 5 and 3, respectively so that you can cancel out the y term. You will have:

3•(5x - 5y = 60)•3 = 15x - 15y = 180

5•(3x + 3y = 60)•5 = 15x + 15y = 300

The 15y terms cancel out and you will be left with 30x = 480, with x = 16 and subsequently y = 4

Hope this helps :)