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Distance, Rate, and Time in a System of Equations

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Today we're going to talk about solving problems involving distance, rate, and time with systems of equations. Remember the relationship with distance, rate, and time can be thought of with an equation, d is equal to r times t, where d is the distance that you've traveled, r is the rate, or the speed, that you're traveling at, and t is the amount of time that you've been traveling. So we're going to see how we can use this relationship to solve problems with systems of equations involving distance, rate, and time.

So let's do an example. Suppose I have a motorboat that takes a 24 mile trip against the current, or upstream. And it takes the boat three hours. The return trip, with the current, takes only two hours. We want to find both the speed, or the rate, of the boat, and the speed, or the rate, of the current. So I'm going to start by defining two variables. I'm going to let r1 be the speed, or the rate, of the boat. And I'm going to let r2 be the speed, or the rate, of the current.

Now when we have rates working together, we can define the combined rate with a sum of the two rates. So when the two currents are working together, when the boat is traveling downstream, then the combined rate could be represented as r1 plus r2. And when we have rates working against each other, such as when the boat is traveling upstream, we can define the overall rate as a difference. In this case, it would r1 minus r2. Because the two rates are working against each other, one of the rates is opposing. And in this case, the r2, or the rate of the current, is the opposing rate.

So let's see how we can use these combination of rates to write a system of equations for our situation and solve to find the rate of both the boat and the current. All right. So to write my equations, I'm going to think about my relationship between distance, rate, and time, being that distance is equal to the rate times the time. So let's write our first equation for the boat traveling upstream to represent the distance, rate, and time.

So I know that when I'm traveling upstream and traveling downstream, my distance is going to be 24 miles. And that's going to be equal to my rate times the time. Now when I'm traveling upstream, then my rates are opposing each other. So that's going to be represented by a difference of my two rates, r1 and r2. And I know that on that trip, against the current, that took me 3 hours. So my time is 3.

So let's think about our second equation. The return trip, when I was traveling downstream, also was 24 miles. So my distance is 24. My rate is going to be the sum of my two rates, because I'm going with the current. So that's going to be represented by r1 plus r2. And this time, it only took me two hours, which makes sense. It should take a less time if I'm traveling with the current.

So I'm going to simplify these two equations, and then we'll decide which method we'll use to solve. So I'm going to distribute my 3 to both variables in my top equation. That's going to give me 24 is equal to 3 r1 minus 3 r2. And my second equation distributing, I'm going to have 24 is equal to 2 r1 plus 2 r2.

So I've got my two equations simplified. And now to determine which method I want to use, I could either, to solve algebraically, use the addition method or the elimination method. Or I could use the substitution method. I'm going to go ahead and use the addition, or the elimination method, which means I need the coefficients in front of one of my variables to be the same, so that when I add them, they cancel out.

So to do that, I'm going to multiply my top equation by 2. And I'm going to multiply my bottom equation by 3, so that I'll have sixes in front of actually both my r1 and r2 variables. So multiplying this top equation by 2, and the terms in my bottom equation by 3, my two new equations are going to become-- so let's see, 24 times 2 will give me 48, is equal to 6 r1 minus 6 r2.

And my second equation, multiplying all the terms inside by 3, is going to give me 72 is equal to 6 r1 plus 6 r2. So now, let's see how we can go ahead and solve this system of equations for both r1 and r2.

OK. So using the elimination method, or the addition method, as it's sometimes called, I'm going to add these two equations together. And in doing that, I can see that my r2 variable is going to cancel out or be eliminated. Because a negative 6 r2 plus a positive 6 r2 will be 0. So starting here, 48 plus 72 is going to give me 120. 6 r1 plus another 6 r1 is going to give me 12 r1. And as I said, this last two terms will add together to be 0.

So simplifying this, I just have 120 is equal to 12 r1. I can divide both sides by 12, which gives me a value for r1 of 10. So that means that the rate, or the speed, of my boat is 10, or 10 miles per hour. So then, we can-- since I have my value for r1, I can substitute it into either of my equations to determine a value for r2, which again is the rate, or the speed, of my current.

So I'm going to go ahead and pick my first equation. 48 is equal to 6 times r1, which I know is equal to 10, minus 6 r2. Multiplying here, 6 times 10 gives me 60. I'm going to subtract by 60 on both sides. This is going to give me negative 12 is equal to negative 6 r2. Dividing both sides by negative 6, I find that r2 is equal to 2. Or the speed or rate of my current is equal to 2, or 2 miles per hour.

So let's go over our key points from today. A system of equations can be used to represent two rates working with and against each other. When rates work together, the combined rate is the sum of the individual rates. And when rates work against each other, the opposing rate is negative. And you can use the graphing, substitution, or addition method to solve the system of equations.

So I hope that these key points and examples helped you understand a little bit more about solving problems with distance, rate, and time and systems of equations. Keep using your notes, and keep on practicing, and soon you'll be a pro. Thanks for watching.