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Systems of equations with elimination: TV & DVD

Sal solves a word problem about the weights of TVs and DVDs by creating a system of equations and solving it. Created by Sal Khan and Monterey Institute for Technology and Education.

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Video transcript

An electronics warehouse ships televisions and DVD players in certain combinations to retailers throughout the country. They tell us that the weight of 3 televisions and 5 DVD players is 62.5 pounds, and the weight of 3 televisions and 2 DVD players-- so they're giving us different combinations-- is 52 pounds. Create a system of equations that represents this situation. Then solve it to find out how much each television and DVD player weighs. Well, the two pieces of information they gave us in each of these statements can be converted into an equation. The first one is is that the weight of 3 televisions and 5 DVD players is 62.5 pounds. Then they told us that the weight of 3 televisions and 2 DVD players is 52 pounds. So we can translate these directly into equations. If we let t to be the weight of a television, and d to be the weight of a DVD player, this first statement up here says that 3 times the weight of a television, or 3 televisions, plus 5 times the weight of a DVD player, is going to be equal to 62.5 pounds. That's exactly what this first statement is telling us. The second statement, the weight of 3 televisions and 2 DVD players, so if I have 3 televisions and 2 DVD players, so the weight of 3 televisions plus the weight of 2 DVD players, they're telling us that that is 52 pounds. And so now we've set up the system of equations. We've done the first part, to create a system that represents the situation. Now we need to solve it. Now, one thing that's especially tempting when you have two systems, and both of them have something where, you know, you have a 3t here and you have a 3t here, what we can do is we can multiply one of the systems by some factor, so that if we were to add this equation to that equation, we would get one of the terms to cancel out. And that's what we're going to do right here. And you can do this, you can do this business of adding equations to each other, because remember, when we learned this at the beginning of algebra, anything you do to one side of an equation, if I add 5 to one side of an equation, I have to add 5 to another side of the equation. So if I add this business to this side of the equation, if I add this blue stuff to the left side of the equation, I can add this 52 to the right-hand side, because this is saying that 52 is the same thing as this thing over here. This thing is also 52. So if we're adding this to the left-hand side, we're actually adding 52 to it. We're just writing it a different way. Now, before we do that, what I want to do is multiply the second, blue equation by negative 1. And I want to multiply it by negative 1. So negative 3t plus-- I could write negative 2d is equal to negative 52. So I haven't changed the information in this equation. I just multiplied everything by negative 1. The reason why I did that is because now if I add these two equations, these 3t terms are going to cancel out. So let's do that. Let's add these two equations. And remember, all we're doing is we're adding the same thing to both sides of this top equation. We're adding essentially negative 52 now, now that we've multiplied everything by a negative 1. This negative 3t plus negative 2d is the same thing as negative 52. So let's add this left-hand side over here. The 3t and the negative 3t will cancel out. That was the whole point. 5d plus negative 2d is 3d. So you have a 3d is equal to 62.5 plus negative 52, or 62.5 minus 52 is 10.5. And now we can divide both sides of this equation by 3. And you get d is equal to 10.5 divided by 3. So let's figure out what that is. 3 goes into 10.5-- it goes into 10 three times. 3 times 3 is 9. Subtract. Get 1. Bring down the 5. Of course, you have your decimal point right here. 3 goes into 15 five times. 5 times 3 is 15. You've got to subtract, and you get a 0. So it goes exactly 3.5 times. So the weight of a DVD player-- that's what d represents-- is 3.5 pounds. Now we can substitute back into one of these equations up here to figure out the weight of a television. We can just use that top equation. So you get 3t plus 5 times the weight of a DVD player, which we just figured out is 3.5. Remember, we're just looking for values that satisfy both of these equations. So 5 times 3.5-- needs to be equal to 62.5. So you get 3t plus-- what is this going to be? This is going to be 15 plus 2.5, right? 5 times 0.5 is 2.5, 5 times 3 is 15. So it's 17.5, is equal to 62.5. Now we can subtract 17.5 from both sides of this equation. And what do we get? The left-hand side is just going to be 3t. This cancels out, that was the whole point of it. 3t is going to be equal to-- let's see. The 0.5 minus 0.5, that cancels out. So this is the same thing as 62 minus 17. 62 minus 7 would be 55. And so we're going to subtract another 10. So it's going to be 45. So this is going to be equal to 45. Now you can divide both sides of this equation by 3. And we get t is equal to 15. So we've solved our system. The weight of a DVD player is 3.5 pounds, and the weight of a television is 15 pounds. And we're done.