# Systems of linear equations uniform motion

>>All right, so these are

uniform motion problems with the current,

against the current, and what’s happening is you have

a boat, and it’s going upstream, which means the current

will be going against it. So you can kind of

think of the boat, and the current’s pushing

the opposite direction. Downstream the current’s

going with the boat. So if you’re in this situation,

the boat is slowing down. In this situation

the boat is speeding up because the current

is literally pushing it. All right, so if x is the

speed of the boat in calm water and y is the speed of

the stream in calm water, then the actual effective rate

of the boat would be x minus y, because the current is

subtracting speed from the boat. Going downstream the effect

of speed would be x plus y, because the current is

actually pushing the boat, and there’s your

setup for the begin. Now all we have to do is

go through the word problem and find out the distance

traveled in each situation and the time traveled

in each situation. So when Dwayne drives his

boat to work upstream, the trip takes 45 minutes. Now, with rates, we want miles

per hour, not miles per minute. So that 45 minutes would

have to be divided by a 60 to give us 0.75, and if we

plug it in a calculator, and there you go, .75. When he comes home

downstream, it takes 30 minutes. So again out of 60 and that’s

a clean division of 0.5. So now we have its hours. The distance to work

is 90 miles. So whether he comes

home or goes to work, the distance is 90 miles, and what we’ve just done

now is created our formula, and it’s distance

equals rate times time. So for the first

situation, which is upstream, we have an equation of this, and for the second

equation we have… All right, and then now we

have our linear equations, and before, we would

actually distribute this, but since these are nice, what we’re going to

do is divide them. So what’s 90 divided by .75? 120. So we get 120 equals x minus y, and we’ll do the same thing

here, 90 divided by .5. That’s 180, x plus y,

and now if you notice, the y’s are opposites. So we can just add them up,

and we get two x’s equals 300, which would allow us to divide. So we get 150 equals x, and

what x represents is the speed of the boat. It’s a very, very fast boat. So the boat, the boat’s speed, is 150 miles per hour. Then how do we find

the current speed? Just take any of the

equations and plug it in. So we have 120, or

we’ll do the 180. So 180 equals x plus

y. Plug in our x. Oops, that’s the y. So

that means the speed of the current is

30 miles per hour. That’s it. So that gives us an effective

rate of 180 miles per hour for the boat going with

the current upstream — I mean, downstream. Then going against the current

would be 120 miles per hour. All right, let’s try one more, and then I’ll give

you one to try. So a jet plane flying through

the wind went 2100 miles in four hours. Against the wind the plane could

fly only 1760 miles per hour. Find the rate of the

plane in calm air and the rate of the wind. So that’s where we have the x

and the y, and we always want to make x the rate

of the object. So in this case it’s

a plane in calm. I don’t even care about the air. All right, and the other one

is just the rate of the wind, and it’s the same situation. So if it goes against the wind,

you would be subtracting them, and if you go with the wind,

you would be adding them. Now we can fill in the rest. So a jet plane flying with

the wind, that’s this one, went 2100 miles in four hours. Against the wind the

plane could only fly 1760 in the same amount of time. So that gives us

our two equations. 1760 equals x minus y times 4, and 2100 equals x

plus y times 4. Then again you want

to divide them. So we take 1760 divided

by 4 and we get 440. Then the same thing. 2100 divided by 4 is 525, and you notice the y’s

are nice and clean. So we get 2x equals,

that’s 965 divided by 2. So x equals 482.5. And what does x stand

for, again? It’s the rate of the

plane in calm air. So rate of plane in calm

air is 482.5 miles per hour. Make sure there are miles here. Yup, miles per hour. To find the rest,

just plug it back in, so 525 equals 482.5 plus y. So

y would equal, just subtract it, 525 minus 482.5, 42.5. So rate of wind is 42.5 miles

per hour, and there you go. That’s our answer. So hopefully you kind of get

the idea of these systems of linear equations using

currents and wind speed. Let’s have you try

one right here. This is for a rowing team. Hit pause and give it a shot. All right, and so this is

what you should have so far. All right, let’s figure… So here’s what you should

have for your answer. All right; thank you.