Hi, and welcome. My name is Anthony Varela. And today, I'm going to be talking about distance, rate, and time with a system of equations. So we're going to review this relationship between distance, rate, and time.
We're going to be talking about combined rates, so when rates work together or against each other. And we're going to be using all of this to setup and solve for a system of equations. So first, let's review distance, rate, and time.
We can actually draw this from something we see almost every day, speed limit sign. So 45 miles per hour, I'm going to write that as 45 miles over one hour. And taking a look at what these represent, well, miles per hour is a rate. Miles is distance and hour is a unit of time.
So here, we know that rate equals distance over time. Now, if I multiply it both sides of this equation by time, this would then tell me that distance equals rate times time. And we're going to be using this in our example in just a moment, so I'd like to develop a system of equations that represent swimming out half a mile and then turning around and swimming back.
Now when you're swimming out, you might be going with the current, so there are actually two rates at work here. One rate is the rate at which you're swimming, and you have the rate of the current. And because the current helps you move along, we can combine those rates through addition, two rates working together.
Now, however, on your return trip, the rate of the current is actually working against you. So we still have the rate at which you're swimming, r sub one, but the rates of the current is opposing. So we're going to be subtracting r2 rather than adding it when you were swimming with the current. So we're going to use then this idea of combined rates swimming with the current and swimming against the current to set up a system of equations and solve some problems.
So the first thing that I want to do, though, is define some variables. So I'm going to define x as the rate at which I'm swimming that could also easily be r sub 1, but I like to use x. And y is going to be the rate of the current. That could easily be r sub 2, but I like to use x and y.
So now that I've defined my variables, what I'm going to do is give off some information about me swimming with the current, swimming against the current, and then we're going to find out the rate at which I'm swimming. And we're going to calculate then the rate of the current, so the distance in both cases is 0.5 miles. Now I was able to assume that half a mile with the currents in 10 minutes, but against the current, it took me 25 minutes to get back.
So now how can I talk about the rate? Well, the rate with the current is the two rates working with each other, so that would be x plus y. And coming back, this is the two rates working against each other. So this would be x minus y.
Now before I develop my equations, I'm just used to thinking about rate as being miles per hour. I have miles and minutes. So I'm just going to quickly convert my minutes into hours by dividing by 60, so 10 minutes is the same as 0.16 repeating hours. And 25 minutes is the same as 0.416 repeating hours just dividing those two numbers by 60 to convert from minutes into hours.
So now, I'm going to develop these equations using distance equals rate times time. One equation is going to be with the current, and one equation is going to be against the current. So in developing my equation with the current, distance is 0.5 miles, rate is x plus y, and time is 0.16 repeating. So that's one equation.
This is just distance equals rate times time. Now I'm going to develop distance equals rate times time, again, but against the current. So my distance is still 0.5. My rate is x minus y, and my time is 0.416 repeating. So now, I have these two equations. X and y mean the same thing and both of them, the rate at which I am swimming and the rate of the current.
So here is our system of equations. How are we going to solve for this system of equations? Well, one of the first things I'd like to do is just rearrange these equations a little bit. It's not exactly how I'm used to seeing linear equations. So I'm just going to distribute them, this 0.16 repeating and this 0.416 repeating. So distributing that, and then I'm just going to write the distance here on the other side.
Because I'm familiar with seeing an x term and a y term equaling something. So I've just distributed there, so it's the same system of equations. I've just rewritten it in a different way, so we're going to solve by substitution. And how we solved by substitution is we're going to write an equivalent expression from one of our variables, and that's going to give us a single variable equation.
And once we have a single variable equation, then we can solve for that variable. And once we have a value then, we can substitute that to solve for our other variable. So that's the general step to solving by substitution, so we're going to go through all of these steps.
So I'm going to take one of my equations. It doesn't matter which one, and I'm going to rewrite this as either x equals or y equals. Once again, the choice is up to you, so I'm going to subtract 0.16 repeating x from both sides of my equation. And then I'm going to divide by that coefficient, so what I have is y equals three minus x. And how did I get this equation of 0.5 divided by this coefficient is three? And because, here, these coefficients are the same, that's just one. And they're combined by subtraction.
So now, I have y equals three minus x. What can I do with that? Well, I can take my other equation. And instead of writing y, I'm going to write what y equals. That is three minus x, so just making that substitution.
Now notice this equation only has one variable, so I know how to solve one variable equations. So first, I'm going to distribute then this negative 0.416 repeating into the three and the minus x. So this doesn't change.
This would be then three times 0.416 repeating. Make sure that I carry my negative, and then I have a negative times a negative. So it's a positive, and then I just have my 0.416 repeating as the coefficient in front of x.
I can combine some like terms, because this is an x term and this is an x term. So let's go ahead and combine those. So I have 0.83 repeating in front of x, and I still have my minus 1.25. Well, I'm going to move then that 1.25 to the other side of the equation by adding it to both sides and dividing by that coefficient.
So I know now that x equals 2.1. Well, this isn't the solution to my system. All I've have done so far is solved for one of my variables. Now this makes the rest of this pretty easy, because I already know what one variable is. I can solve for the other.
Now what it could do is I can come back to any of these equations here, plug-in 2.14 x, and solve for y. But what I really want to do is use this equation right here, just because it's the simplest, right? Not a lot of messy decimal numbers, like we see everywhere else. So I'm going to be using x equals 2.1 and y equals three minus x to solve for my other variable just by substituting then 2.1 in for x.
So y equals three minus 2.1, so we know then that y equals 0.9. So what does this all mean? Well, my rate is the variable x, and this is measured in miles per hour. So I'm swimming at a rate of 2.1 miles per hour, and then y is the rate of the current also measured in miles per hour. So the water is flowing at a rate of 0.9 miles per hour.
So another way to interpret this solution then is that when x is 2.1 and when y is 0.9, we can plug those into our original equations that make up our system. And we'll get true statements. They will satisfy both equations in our system. So let's review distance, rate, and time with systems of equations.
We talked about two rates can work with each other or against each. Swimming with the current, those rates can be added together. Swimming against the current, you would take your rate and subtract the rate of the current. Because it's opposing you. We use the equation distance equals rate times time to develop equations that make up our system.
So distance would equal the rates either working with you times time, or your distance would be the rate working against you times time that made up our system. And we solved our system by substitution, and that meant rewriting one of the equations to get an equivalent expression for a variable. That will then gets you a single variable equation, which we know how to solve for, and then we can substitute that value to solve for the other variable in our system. So Thanks for watching this tutorial on distance, rate, and time with systems of equations. Hope to see you next time.