Video Transcript: The Elimination Method

Systems of equations with elimination: King's cupcakes 

After you cross the troll's bridge and you save the prince or princess, you return them back to  their father, the King. And he's so excited that you returned their son or daughter to him that  he wants to throw a brunch in your honor. But he has a little bit of a conundrum in throwing the brunch. He wants to figure out how many cupcakes should he order? He doesn't want to waste  any, but he wants to make sure that everyone has enough to eat. And you say, well what's the  problem here? And he says, well I know adults eat a different number of cupcakes than children eat. And I know that in my kingdom, an adult will always eat the same amount and a child will  always eat the same amount. And so you say, King, well what information can you give me? I  might be able to help you out a little bit. You're feeling very confident after this troll situation.  And he says, well I know at the last party we had 500 adults and we had 200 children, and  combined they ate 2,900 cupcakes. And you say, OK, that's interesting. But I think I'll need a  little bit more information. Have you thrown parties before then? And the king says, of course I  have. I like to throw parties. Well what happened at the party before that? And he says, well  there we also had 500 adults and we had 300 children. And you say, well how many cupcakes  were eaten at that party? And he says, well we know it was 3,100 cupcakes. And so you get a  tingling feeling that a little bit of algebra might apply over here. And you say, well let me see.  What do we need to figure out? We need to figure out the number of cupcakes on average that  an adult will eat. So number of cupcakes for an adult. And we also need to figure out the  number of cupcakes for a child. So these are the two things that we need to figure out because  then we can know how many adults and children are coming to the next branch that are being  held in your honor and get the exact right number of cupcakes. So those are things you're  trying to figure out. And we don't know what those things are. Let's define some variables that  represent those things. Well let's do a for adults. Let's let a equal the number of cupcakes that,  on average, each adult eats. And let's do c for children. So c is the number of cupcakes for a  child on average. So given that information, let's see how we can represent what the King has  told us algebraically. So let's think about this orange information first. How could we represent  this algebraically? Well let's think about how many cupcakes the adults ate at that party. You  had 500 adults, and on average, each of them exactly a cupcakes. So the total number of  cupcakes that the adults ate were 500 times a. How many did the children eat? Well same logic.  You had 200 children and they each ate c cupcakes. So 200 times c is the total number of  cupcakes that the children ate. Well how much did they eat in total? Well it's the total number  that the adults ate plus the total number that the children ate which is 2,900 cupcakes. So let's  do that and apply that same logic to the blue party right over here, this blue information. How  can we represent this algebraically? Well once again, how many total cupcakes did the adults  eat? Well, you had 500 adults and they each ate a cupcakes, which is an unknown right now.  And then what about the children? Well you had 300 children and they each ate c cupcakes.  And so if you add up all the cupcakes that the adults ate plus all the cupcakes that the children  ate, you get to 3,100 cupcakes. So this is starting to look interesting. I have two equations. I  have a system of two equations with two unknowns. And you know from your experience with  the troll that you should be able to solve this. You could solve it graphically like you did in the  past, but now you feel that there could be another tool in your tool kit which is really just an  application of the algebra that you already know. So think a little bit about how you might do 

this. So let's rewrite this first equation right over here. So we have 500a plus 200c is equal to  2,900 cupcakes. Now, it would be good if we could get rid of this 500a somehow. Well you  might say, well let me just subtract 500a. So you might say, oh, I just want to subtract 500a. But if you subtracted 500a from the left hand side, you'd also have to subtract 500a from the right  hand side. And so the a wouldn't just disappear. It would just end up on the right hand side, and  you would still have one equation with two unknowns which isn't too helpful. But you see  something interesting. You're like, well this is a 500a here. What if I subtracted a 500a and this  300c? So if I subtracted the 500a and the 300c from the left hand side. And you're like, well why  is that useful? You're going to do the same thing on the right hand side and then you're going to have an a and a c on the right hand. And you just say, hold on, hold on one second here. Hold  on, I guess you're talking to yourself. Hold on one second. I'm subtracting the left hand side of  this equation, but this left hand side is the exact same thing as this right hand side. So here I  could subtract 500a and 300c, and I could do 500a and subtract 300c over here. But we know  that subtracting 500a and 300c, that's the exact same thing as subtracting 3,100. Let me make  it clear. This is 500a minus 300c is the exact same thing as subtracting 500a plus 300c. And we  know that 500a plus 300c is exactly 3,100. This is 3,100. This is what the second information  gave us. So instead of subtracting 500a and minus 300c, we can just subtract 3,100. So let me  do that. This is exciting. So let me clear that out. So let's clear that out. And so here, instead of  doing this, I can subtract the exact same value, which we know is 3,100, subtract 3,100. So  looking at it this way, it looks like we're subtracting this bottom equation from the top  equation, but we're really just subtracting the same thing from both sides. This is just very basic algebra here. But if we do that, let's see what happens. So on the left hand side, 500a minus  500a, those cancel out. 200c minus 300c, that gives us negative 100c. And on the right hand  side, 2,900 minus 3,100 is negative 200. Well now we have one equation with one unknown,  and we know how to solve this. We can divide both sides by negative 100. These cancel out.  And then over here, you end up with a positive 2. So c is equal to positive 2. So we've solved  one of the unknowns, the each child on average drinks two cups. So c is equal to 2. So how can  we figure out what a is? Well now we can take this information and go back into either one of  these and figure out what a has to be. So let's go back into the orange one right over here and  figure out what a has to be. So we had 500a plus 200c, but we know what c is, c is 2. So 200  times 2 is equal to 2,900. And now we just have to solve for a, one equation with one unknown.  So we have 500a, 200 times 2 is 400, plus 400 is equal to 2,900. We can subtract 400 from both  sides of this equation. Let me do that. Subtracting 400, and we are left with this cancels out.  And on the left hand side, we have 500a. This is very exciting. We're in the home stretch. On the right hand side, you have 2,500. 500a equals 2,500. We can divide both sides by 500, and we are left with 2,500 divided by 500 is just 5. So you have a is equal to 5 and you're done. You have  solved the King's conundrum. Each child on average drinks 2 cups of water-- sorry, not cups of  water. I don't know where I got that from. Each child 2 cupcakes and each adult will eat 5  cupcakes. a is equal to 5. And so based on how many adults and children are coming to the  brunch in your honor, you now know exactly the number of cupcakes that the King needs to  order. 

Elimination strategies

We're asked which of these strategies would eliminate a variable in the system of equations?  Choose all answers that apply. So this first one says add the equations. So pause this video.  Would adding the equations eliminate a variable in this system? All right, now let's do it  together. So if we add these equations, we have, on the left hand side, we have five x plus five  x, which is going to be 10 x, and then you have negative three y plus four y which is just a  positive one y or just plus y is equal to negative three plus six, which is just going to be equal to  positive three. We haven't eliminated any variables, so choice A, I could rule out. That did not  eliminate a variable. Let me cross it out and not check it. Subtract the bottom equation from  the top. Well when we subtract the bottom from the top, five x minus five x, that's going to be  zero x's, so I won't even write it down, and we've already seen we've eliminated an x so I'm  already feeling good about choice B, but then we can see negative three y minus four y is  negative seven y. Negative three minus six is going to be negative nine and so choice B does  successfully eliminate the x's. So I will select that. Choice C, multiply the top equation by two,  then add the equations. Pause the video. Does that eliminate a variable? Well we're gonna  multiply the top equation by two so it's going to become 10 x minus six y is equal to negative  six, and you could already see if you then add the equations, 10 x plus five x, you're gonna have  15 x, that's not gonna get eliminated. Negative six y plus four y is negative two y. That's not  going to be eliminated, so we can rule that out, as well. Let's do another example. One, they're  asking us the same question. Which of these strategies would eliminate a variable in the system of equations? The first choice says multiply the bottom equation by two, then add the  equations. Pause this video, does that work? All right, so if we multiply the bottom equation by  two, we are going to get, if we multiply it by two, we're gonna get two x minus two minus four  y, I should say. Two x, I'm just multiplying everything by two, minus four y is equal to 10. And  then if we were to add the equations. Four x plus two x is six x, so that doesn't get eliminated.  Positive four y plus negative four y is equal to zero y, so the y's actually do get eliminated when  you add four y to negative four y. So I like choice A and I'm gonna delete this so I have space to  work on the other choices, so I like one. What about choice B? Pause the video, does that work? Multiply the bottom equation by four, then subtract the bottom equation from the top  equation. All right, let's multiply the bottom equation by four. What do we get? We're going to  get four x minus eight y is equal to 20, yup. We multiplied it by four and then subtract the  bottom equation from the top. So we would subtract four x from four x. Well that's looking  good. That would eliminate the x's, so I'm feeling good about choice B. And then we could see if we subtract negative eight y from four y, well, subtracting a negative's the same thing as  adding a positive, so that would actually get us to 12 y if we're subtracting negative eight y  from four y. And then if we subtract 20 from negative two, we get to negative 22, but we see  that four x minus four x is going to eliminate our x's, so that does definitely eliminate a variable, so I like choice B. Now what about choice C? Multiply the top equation by 1/2, then add the  equations. Let's try that out, pause the video. All right, let's just multiply times 1/2, so the left  hand side times 1/2, we distribute the one half is one way to think about it. Four x times 1/2 is  going to be two x plus four y times 1/2 is two y is equal to negative two times 1/2 is equal to  negative one. Now and then they say add the equations. So two x plus x is going to be three x,  so that's not going to eliminate the x's. Two y plus negative two y, well that's going to be no y's.  So that actually will eliminate the y's, so I like this choice, as well. So actually, all three of these 

strategies would eliminate a variable in the system of equations. This is useful to see 'cause you  can see there's multiple ways to approach solving a system like this through elimination. Let's  do another example. Which of these strategies would eliminate a variable in the system of  equations? Same question again. So the first one, they suggest to subtract the bottom  equation from the top equation. Pause this video, does that work? Well if we subtract the  bottom from the top, so if you subtract a negative two x, that's the same thing as adding two x, 'cause you're adding two x to three x, that's five x. The x's don't get eliminated. Subtracting four y from negative three y's just gonna get us to negative seven y. The y's don't get eliminated, so I would rule this one out. Nothing's getting eliminated there. Multiply the top equation by three,  multiply the bottom equation by two, then add the equations. Pause the video, does that  work? All right, so if I multiply the top equation by three, I'm going to get nine x minus nine y is  equal to 21, and then if I multiply the bottom by two, so this is times two, I'm going to get two  times negative two is negative four x plus eight y is equal to 14, and then they say add the  equations. Well if I add nine x to negative four x, that doesn't eliminate the x's. That gets me to  positive five x, and if I add negative nine y to a positive eight y, that also doesn't eliminate the  y's. That gets me to a negative y, so choice B, I can also rule out. Once again deleting all of this  so I have space to try to figure out choice C. Multiply the top equation by two, multiply the  bottom equation by three, then add the equation. So they're telling us to do it the other way  around. Pause the video, does this work? All right, so we multiply the top equation by two and  we're gonna multiply the bottom equation by three. So the top equation times two is going to  be six x minus six y is equal to, is equal to 14. And then with this bottom equation, when you  multiply it by three, both sides, that's the only way to ensure that the equation is saying the  same thing is if you do the same thing to both sides. That's really the heart of algebra. So  negative two times three is negative six x, and I already like where this is going, 'cause when I  add these two, they're going to get eliminated, plus four y times three is gonna be plus 12 y is  going to be equal to 21. And then they say add the equations, well, you immediately see when  you add the x terms on the left hand side, they are going to cancel out. So I like choice C. 

Systems of equations with elimination: x-4y=-18 * -x=3y=11 

So we have a system of two linear equations here. This first equation, X minus four Y is equal to  negative 18, and the second equation, negative X plus three Y is equal to 11. Now what we're  gonna do is find an X and Y pair that satisfies both of these equations. That's what solving the  system actually means. As you might already have seen, there's a bunch of X and Y pairs that  satisfy this first equation. In fact, if you were to graph them, they would form a line, and there's  a bunch of other X and Y pairs that satisfy this other equation, the second equation, and if you  were to graph them, it would form a line. And so if you find the X and Y pair that satisfy both,  that would be the intersection of the lines, so let's do that. So actually, I'm just gonna rewrite  the first equation over here, so I'm gonna write X minus four Y is equal to negative 18. So, we've  already seen in algebra that as long as we do the same thing to both sides of the equation, we  can maintain our equality. So what if we were to add, and our goal here is to eliminate one of  the variables so we have one equation with one unknown, so what if we were to add this  negative X plus three Y to the left hand side here? So negative X plus three Y, well, that looks  pretty good because an X and a negative X are going to cancel out, and we are going to be left 

with negative four Y, plus three Y. Well, that's just going to be negative Y. So by adding the left  hand side of this bottom equation to the left hand side of the top equation, we were able to  cancel out the Xs. We had X, and we had a negative X. That was very nice for us. So what do we  do on the right hand side? We've already said that we have to add the same thing to both sides  of an equation. We might be tempted to just say, well, if I have to add the same thing to both  sides, well, maybe I have to add a negative X plus three Y to that side. But that's not going to  help us much. We're gonna have negative 18 minus X plus three Y. We would have introduced an X on the right hand side of the equation, but what if we could add something that's equivalent  to negative X plus three Y that does not introduce the X variable? Well, we know that the  number 11 is equivalent to negative X plus three Y. How do we know that? Well, that second  equation tells us that. So once again, all I'm doing is I'm adding the same thing to both sides of  that top equation. On the left, I'm expressing it as negative X plus three Y, but the second  equation tells us that negative X plus three Y is going to be equal to 11. It's introducing that  second constraint, and so let's add 11 to the right hand side, which is, once again, I know I keep  repeating it, it's the same thing as negative X plus three Y. So negative 18 plus 11 is negative  seven, and since we added the same thing to both sides, the equality still holds, and we get  negative Y is equal to negative seven, or divide both sides by negative one or multiply both  sides by negative one. So multiply both sides by negative one. We get Y is equal to seven, so we have the Y coordinate of the X, Y pair that satisfies both of these. Now how do we find the X?  Well, we can just substitute this Y equals seven to either one of these. When Y equals seven, we  should get the same X regardless of which equation we use. So let's use the top equation. So  we know that X minus four times, instead of writing Y, I'm gonna write four times seven 'cause  we're gonna figure out what is X when Y is seven? That is going to be equal to negative 18, and  so let's see, negative four or four times seven. That is 28. So let's see, I could just solve for X. I  could add 28 to both sides, so add 28 to both sides. On the left hand side, negative 28, positive  28, those cancel out. I'm just left with an X, and on the right hand side, I get negative 18 plus 28  is 10. So there you have it. I have the X, Y pair that satisfies both. X equals 10. Y equals seven. I  could write it here. So I could write it as coordinates. I could write it as 10 comma seven, and  notice, what I just did here, I encourage you. Substitute Y equals seven here, and you would  also get X equals 10. Either way, you would have come to X equals 10, and to visualize what is  going on here, let's visualize it really fast. Let me draw some coordinate axes. Whoops, I meant  to draw a straighter line than that. Alright, there you go, so let's say that is our Y axis, and that  is, whoops, that is our X axis, and then, let's see, the top equation is gonna look something like  this. It's gonna look something like this, and then that bottom equation is gonna look  something like, lemme draw a little bit nicer than that. It's going to look something like this,  something like that. Lemme draw that bottom one here so you see the point of intersection,  and so the point of intersection right over here, that is an X, Y pair that satisfies both of these  equations, and that we just saw, it happens when X is equal to 10, and Y is equal to seven. Once  again, this white line, that's all the X and Y pairs that satisfy the top equation. This orange line,  that's all the X and Y pairs that satisfy the orange equation, and where they intersect, that point is on both lines. It satisfies both equations, and once again, take X equals 10, Y equals seven.  Substitute it back into either one of these, and you will see that it holds.

Systems of equations with elimination: potato chips 

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. 

Systems of equations with elimination (and manipulation) 

Let's solve a few more systems of equations using elimination, but in these it won't be kind of a  one-step elimination. We're going to have to massage the equations a little bit in order to  prepare them for elimination. So let's say that we have an equation, 5x minus 10y is equal to 15. And we have another equation, 3x minus 2y is equal to 3. And I said we want to do this using  elimination. Once again, we could use substitution, we could graph both of these lines and  figure out where they intersect. But we're going to use elimination. But the first thing you  might say, hey, Sal, you know, with elimination, you were subtracting the left-hand side of one  equation from another, or adding the two, and then adding the two right-hand sides. And I  could do that, because it was essentially adding the same thing to both sides of the equation.  But here, it's not obvious that that would be of any help. If we added these two left-hand sides,  you would get 8x minus 12y. That wouldn't eliminate any variables. And on the right-hand side,  you would just be left with a number. And if you subtracted, that wouldn't eliminate any  variables. So how is elimination going to help here? And the answer is, we can multiply both of  these equations in such a way that maybe we can get one of these terms to cancel out with one of the others. And you could really pick which term you want to cancel out. Let's say we want to cancel out the y terms. So I'll just rewrite this 5x minus 10y here. 5x minus 10y is equal to 15.  Now, is there anything that I can multiply this green equation by so that this negative 2y term  becomes a term that will cancel out with the negative 10y? So I essentially want to make this  negative 2y into a positive 10y. Right? Because if this is a positive 10y, it'll cancel out when I add the left-hand sides of this equation. So what can I multiply this equation by? Well, if I multiply it  by negative 5, negative 5 times negative 2 right here would be positive 10. So let's do that. Let's multiply this equation times negative 5. So you multiply the left-hand side by negative 5, and  multiply the right-hand side by negative 5. And what do you get? Remember, we're not  fundamentally changing the equation. We're not changing the information in the equation.  We're doing the same thing to both sides of it. So the left-hand side of the equation becomes  negative 5 times 3x is negative 15x. And then negative 5 times negative 2y is plus 10y, is equal  to 3 times negative 5 is negative 15. And now, we're ready to do our elimination. If we add this  to the left-hand side of the yellow equation, and we add the negative 15 to the right-hand side  of the yellow equation, we are adding the same thing to both sides of the equation. Because  this is equal to that. So let's do that. So 5x minus 15y-- we have this little negative sign there,  we don't want to lose that-- that's negative 10x. The y's cancel out. Negative 10y plus 10y, that's 0y. That was the whole point behind multiplying this by negative 5. Is going to be equal to-- 15  minus 15 is 0. So negative 10x is equal to 0. Divide both sides by negative 10, and you get x is  equal to 0. And now we can substitute back into either of these equations to figure out what y  must be equal to. Let's substitute into the top equation. So we get 5 times 0, minus 10y, is equal to 15. Or negative 10y is equal to 15. Let me write that. Negative 10y is equal to 15. Divide both  sides by negative 10. And we are left with y is equal to 15/10, is negative 3/2. So if you were to  graph it, the point of intersection would be the point 0, negative 3/2. And you can verify that it  also satisfies this equation. The original equation over here was 3x minus 2y is equal to 3. 3  times 0, which is 0, minus 2 times negative 3/2 is, this is 0, this is positive 3. Right? These cancel 

out, these become positive. Plus positive 3 is equal to 3. So this does indeed satisfy both  equations. Let's do another one of these where we have to multiply, and to massage the  equations, and then we can eliminate one of the variables. Let's do another one. Let's say we  have 5x plus 7y is equal to 15. And we have 7-- let me do another color-- 7x minus 3y is equal to  5. Now once again, if you just added or subtracted both the left-hand sides, you're not going to  eliminate any variables. These aren't in any way kind of have the same coefficient or the  negative of their coefficient. So let's pick a variable to eliminate. Let's say we want to eliminate  the x's this time. And you could literally pick on one of the variables or another. It doesn't  matter. You can say let's eliminate the y's first. But I'm going to choose to eliminate the x's first.  And so what I need to do is massage one or both of these equations in a way that these guys  have the same coefficients, or their coefficients are the negatives of each other, so that when I  add the left-hand sides, they're going to eliminate each other. Now, there's nothing obvious-- I  can multiply this by a fraction to make it equal to negative 5. Or I can multiply this by a fraction  to make it equal to negative 7. But even a more fun thing to do is I can try to get both of them  to be their least common multiple. I could get both of these to 35. And the way I can do it is by  multiplying by each other. So I can multiply this top equation by 7. And I'm picking 7 so that this  becomes a 35. And I can multiply this bottom equation by negative 5. And the reason why I'm  doing that is so this becomes a negative 35. Remember, my point is I want to eliminate the x's.  So if I make this a 35, and if I make this a negative 35, then I'm going to be all set. I can add the  left-hand and the right-hand sides of the equations. So this top equation, when you multiply it  by 7, it becomes-- let me scroll up a little bit-- we multiply it by 7, it becomes 35x plus 49y is  equal to-- let's see, this is 70 plus 35 is equal to 105. Right? 15 and 70, plus 35, is equal to 105.  That's what the top equation becomes. This bottom equation becomes negative 5 times 7x, is  negative 35x, negative 5 times negative 3y is plus 15y. The negatives cancel out. And then 5--  this isn't a minus 5-- this is times negative 5. 5 times negative 5 is equal to negative 25. Now, we can start with this top equation and add the same thing to both sides, where that same thing is  negative 25, which is also equal to this expression. So let's add the left-hand sides and the right hand sides. Because we're really adding the same thing to both sides of the equation. So the  left-hand side, the x's cancel out. 35x minus 35x. That was the whole point. They cancel out, and on the y's, you get 49y plus 15y, that is 64y. 64y is equal to 105 minus 25 is equal to 80. Divide  both sides by 64, and you get y is equal to 80/64. And let's see, if you divide the numerator and  the denominator by 8-- actually you could probably do 16. 16 would be better. But let's do 8  first, just because we know our 8 times tables. So that becomes 10/8, and then you can divide  this by 2, and you get 5/4. If you divided just straight up by 16, you would've gone straight to  5/4. So y is equal to 5/4. Let's figure out what x is. So we can substitute either into one of these  equations, or into one of the original equations. Let's substitute into the second of the original  equations, where we had 7x minus 3y is equal to 5. That was the original version of the second  equation that we later transformed into this. So we get 7x minus 3 times y, times 5/4, is equal to 5. Or 7x minus 15/4 is equal to 5. Let's add 15/4-- Oh, sorry, I didn't do that right. This would be  7x minus 3 times 4-- Oh, sorry, that was right. What am I doing? 3 times is 15/4. Is equal to 5.  Let's add 15/4 to both sides. And what do we get? The left-hand side just becomes a 7x. These  guys cancel out. And that's going to be equal to 5, is the same thing as 20/4. 20/4 plus 15/4. Or  we get that-- let me scroll down a little bit-- 7x is equal to 35/4. We can multiply both sides by 

1/7, or we could divide both sides by 7, same thing. Let's multiply both sides by 1/7. The same  thing as dividing by 7. So these cancel out and you're left with x is equal to-- Here, if you divide  35 by 7, you get 5. You divide 7 by 7, you get 1. So x is equal to 5/4 as well. So the point of  intersection of this right here is both x and y are going to be equal to 5/4. So if you looked at it  as a graph, it'd be 5/4 comma 5/4. And let's verify that this satisfies the top equation. And if you  take 5 times 5/4, plus 7 times 5/4, what do you get? It should be equal to 15. So this is equal to  25/4, plus-- what is this? This is plus 35/4. Which is equal to 60/4, which is indeed equal to 15. So  it does definitely satisfy that top equation. And you could check out this bottom equation for  yourself, but it should, because we actually used this bottom equation to figure out that x is  equal to 5/4.