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Systems of equations with elimination: potato chips

Sal solves another system of equations using elimination. Created by Sal Khan.

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Video transcript

Everyone in the kingdom is very impressed with your ability to help with the party planning, everyone except for this gentlemen right over here. This is Arbegla. And he is the king's top adviser, and also chief party planner. And he seems somewhat threatened by your ability to solve these otherwise unsolvable problems, or at least from his point of view, because he keeps over-ordering or under-ordering things like cupcakes. And he says, king, that cupcake problem was easy. Ask them about the potato chip issue, because we could never get the potato chips right. And so the king says, Arbegla, that's a good idea. We need to get the potato chips right. So he comes to you and says, how do we figure out, on average, how many potato chips we need to order? And to do that, we have to figure out how much, on average, does each man eat and how much each woman eats. You say, well, what about the children? The king says, in our kingdom, we forbid potato chips for children. You say, oh, well, that's all and good. Tell me what happened at the previous parties. And so the king says, you might remember, at the last party, in fact, the last two parties, we had 500 adults. At the last party, 200 of them were men, and 300 of them were women. And in total, they ate 1,200 bags of potato chips. And you say, what about the party before that? He says, that one, we had a bigger skew towards women. We only had 100 men, and we have 400 women. And that time, we actually had fewer bags consumed-- 1,100 bags of potato chips. So you say, OK, king and Arbegla, this seems like a fairly straightforward thing. Let me define some variables to represent our unknowns. So you go ahead and you say, well, let's let m equal the number of bags eaten by each man. And you could think of it on average, or maybe all the men in that kingdom are completely identical. Or maybe it's the average number of bags eaten by each man. And let's let w equal the number of bags eaten by each woman. And so with these definitions of our variables, let's think about how we can represent this first piece of information, this piece of information in green. Well, let's think about the total number of bags that the men ate. You had 200 men. Let me scroll over a little bit. You had 200 men, and they each ate m bags, m bags per man. So the men at this first party collectively ate 200 times m bags. If m is 10 bags per man, then this would be 2,000. If m was 5 bags per man, then this would be 5,000. We don't know what m is, but 200 times m is the total eaten by the men. Same logic-- total eaten by the women is 300 women times the number of bags eaten by each woman. And so if you add the total to eaten by the men and the women, you get the 1,200 bags. So this is information, written algebraically, given these variable definitions. Now, let's do the same thing with the second part of the information that they gave us right over here. Let's think about how we can represent this algebraically. Well, similar logic-- what was a total that the men ate at that party? It was 100 men times m bags per man. And we're assuming that m is the same across parties, that men, on average, always eat the same number of bags. And how many did the women eat at that second party? Well, you had 400 women. And on average, they ate w bags per woman. So this is 400 times w is the total number that the women ate. You add those two together, you get the total number that all the adults ate. So this is going to be 1,100 bags. So it looks pretty similar now. You have a system of two equations with two unknowns. And so you try your best to solve it. But when you solve it, you see something interesting. Last time, it was very convenient. You had a, I think it was a 500 here, for 500 adults, and you had another 500. And so it seemed like it was pretty easy to cancel out one of the variables. Here it seems a little bit more difficult. What's multiplying by the m's, it's different here. The coefficient on the w is different over here. You say, well, maybe I can change one of these equations so it makes it a little bit easier to cancel out with the other equation. So what if, for example, I were to take this blue equation right over here and multiply it by negative 2? And you might say, well, Sal, why are we multiplying it by negative 2? Well, if were to multiply it by negative 2, this 100m would become a negative 200m. And if it was a negative 200m, then that would cancel out with a positive 200m when we add the two. So let's see what happens. So let's multiply this blue equation by negative 2. We're going to multiply by negative 2. Let me scroll over to the left a little bit. So what happens? Remember, when we multiply an equation, we can't just do one side of the equation. We have to do the entire equation in order for the equality to hold true. So negative 2 times 100m is negative 200m. Negative 2 times 400w-- there's a positive right over there. So it becomes negative 800w. And then negative 2-- now, we did the left hand side, but we also have to do the right hand side. Negative 2 times 1,100 is negative 2,200. So just to be clear, this equation that I just wrote here essentially has the same information we just manipulated. We just changed this equation, multiplied both sides by negative 2. But it's kind of the same constraint. But what makes this interesting is, now, we can rewrite this green equation. Let me do it over here, this first one. 200m plus 300w is equal to 1,200. And the whole reason why I multiplied by negative 2 is, so that if I were to add these two things, I might be able to get rid of that variable over there. And so let's do that. Let's add the left hand sides, and let's add the right hand sides. And you could literally view it as, we're starting with this blue equation. We're adding this quantity, the left hand side of the yellow equation to the left hand side of the blue. And then 1,200 is the exact same thing that we're adding to the right hand side. We know that this is equal to this. So we can add this to the left hand side and this to the right hand side. So let's see what happens. So the good thing is, the whole reason we multiplied it by negative 2, so that these two characters cancel out. You add those two together. You just get 0m or just 0. You have negative 800w plus 300w. Well, that's negative 500w. And then on the right hand side, you have negative 2,200 plus 1,200. So that's negative 1,000. And now this is pretty straightforward-- one equation, one unknown, a fairly straightforward equation. We divide both sides by the coefficient of w, multiplying w. So divide by negative 500 on the left, divide by negative 500 on the right. And we are left with w is equal to 2. On average, women ate two bags of potato chips at these parties. We're assuming that's constant across the parties. So let's think about how you would then figure out how many bags, on average, each man ate. Well, to do that, we just go back to either one of these equations. In the last set of videos, I went to the first equation. I'll show that the second equation should also work. Either one should work. So let's substitute back into the second equation. And you could either pick this version of it or this one. But I'll pick the original one. So you have 100 times m, which we're trying to figure out, plus 400 times-- well, we now know that w is equal to 2-- 400 times 2 is equal to 1,100. So you have 100m plus 800 is equal to 1,100. And now, to solve for m, we could subtract 800 from both sides. And we are left with 100m is equal to 300. And now, divide both sides by 100. And we are left with m, which is, on average, the number of bags of chips each man eats is equal to 3. So you have solved Arbegla's problem, what he thought was a difficult problem, using the magical, mystical powers of algebra. You were able to tell the king in his party planning process that, on average, the men will eat three bags of potato chips each. And on average, the women will eat two bags of potato chips each.