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M8Adv Video Lesson 5-3 Solve Systems by Elimination

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Video Example - Systems by Elimination - Multiplication

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Video Example - Systems by Elimination - Multiplication

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Video Example - Solve a system of equations using elimination(addition)

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Video Example - Solve a system of equations using elimination(addition)

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Graph a Pair of Linear Equations

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Find the solution to a pair of linear equations by looking at their graphs | LearnZillion

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Systems by Elimination

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Solve a system of equations by elimination | LearnZillion

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Let's explore a few more methods for solving systems of equations. Let's say I have the equation, 3x plus 4y is equal to 2.5. And I have another equation, 5x minus 4y is equal to 25.5. And we want to find an x and y value that satisfies both of these equations. If you think of it graphically, this would be the intersection of the lines that represent the solution sets to both of these equations. So how can we proceed? We saw in substitution, we like to eliminate one of the variables. We did it through substitution last time. But is there anything we can add or subtract-- let's focus on this yellow, on this top equation right here-- is there anything that we can add or subtract to both sides of this equation? Remember, any time you deal with an equation you have to add or subtract the same thing to both sides. But is there anything that we could add or subtract to both sides of this equation that might eliminate one of the variables? And then we would have one equation in one variable, and we can solve for it. And it's probably not obvious, even though it's sitting right in front of your face. Well, what if we just added this equation to that equation? What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you're probably saying, Sal, hold on, how can you just add two equations like that? And remember, when you're doing any equation, if I have any equation of the form-- well, really, any equation-- Ax plus By is equal to C, if I want to do something to this equation, I just have to add the same thing to both sides of the equation. So I could, for example, I could add D to both sides of the equation. Because D is equal to D, so I won't be changing the equation. You would get Ax plus By, plus D is equal to C plus D. And we've seen that multiple, multiple times. Anything you do to one side of the equation, you have to do to the other side. But you're saying, hey, Sal, wait, on the left-hand side, you're adding 5x minus 4y to the equation. On the right-hand side, you're adding 25.5 to the equation. Aren't you adding two different things to both sides of the equation? And my answer would be no. We know that 5x minus 4y is 25.5. This quantity and this quantity are the same. They're both 25.5. This second equation is telling me that explicitly. So I can add this to the left-hand side. I'm essentially adding 25.5 to it. And I could add 25.5 to the right-hand side. So let's do that. If we were to add the left-hand side, 3x plus 5x is 8x. And then what is 4y minus 4y? And this was the whole point. When I looked at these two equations, I said, oh, I have a 4y, I have a negative 4y. If you just add these two together, they are going to cancel out. They're going to be plus 0y. Or that whole term is just going to go away. And that's going to be equal to 2.5 plus 25.5 is 28. So you divide both sides. So you get 8x is equal to 28. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. That's equal to 7 over 2. That's our x value. Now we want to solve for our y value. And we could substitute this back into either of these two equations. Let's use the top one. You could do it with the bottom one as well. So we know that 3 times x, 3 times 7 over 2-- I'm just substituting the x value we figured out into this top equation-- 3 times 7 over 2, plus 4y is equal to 2.5. Let me just write that as 5/2. We're going to stay in the fraction world. So this is going to be 21 over 2 plus 4y is equal to 5/2. Subtract 21 over 2 from both sides. So minus 21 over 2, minus 21 over 2. The left-hand side-- you're just left with a 4y, because these two guys cancel out-- is equal to-- this is 5 minus 21 over 2. That's negative 16 over 2. So that's negative 16 over 2, which is the same thing-- well, I'll write it out as negative 16 over 2. Or we could write that-- let's continue up here-- 4y-- I'm just continuing this train of thought up here-- 4y is equal to negative 8. Divide both sides by 4, and you get y is equal to negative 2. So the solution to this equation is x is equal to 7/2, y is equal to negative 2. This would be the coordinate of their intersection. And you could try it out on both of these equations right here. So let's verify that it also satisfies this bottom equation. 5 times 7/2 is 35 over 2 minus 4 times negative 2, so minus negative 8. That's equivalent to-- let's see, this is 17.5 plus 8. And that indeed does equal 25.5. So this satisfies both equations. Now let's see if we can use our newly found skills to tackle a word problem, our newly found skills in elimination. So here it says, Nadia and Peter visit the candy store. Nadia buys 3 candy bars and 4 Fruit Roll-Ups for $2.84. Peter also buys 3 candy bars, but can only afford 1 additional Fruit Roll-Up. His purchase costs $1.79. What is the cost of each candy bar and each Fruit Roll-Up? So let's define some variables. Let's just use x and y. Let's let x equal cost of candy bar-- I was going to do a c and a f for Fruit Roll-Up, but I'll just stick with x and y-- cost of candy bar. And let y equal the cost of a Fruit Roll-Up. All right. So what does this first statement tell us? Nadia buys 3 candy bars, so the cost of 3 candy bars is going to be 3x. And 4 Fruit Roll-Ups. Plus 4 times y, the cost of a Fruit Roll-Up. This is how much Nadia spends. 3 candy bars, 4 Fruit Roll-Ups. And it's going to cost $2.84. That's what this first statement tells us. It translates into that equation. The second statement. Peter also buys 3 candy bars, but could only afford 1 additional Fruit Roll-Up. So plus 1 additional Fruit Roll-Up. His purchase cost is equal to $1.79. What is the cost of each candy bar and each Fruit Roll-Up? And we're going to solve this using elimination. You could solve this using any of the techniques we've seen so far-- substitution, elimination, even graphing, although it's kind of hard to eyeball things with the graphing. So how can we do this? Remember, with elimination, you're going to add-- let's focus on this top equation right here. Is there something we could add to both sides of this equation that'll help us eliminate one of the variables? Or let me put it this way, is there something we could add or subtract to both sides of this equation that will help us eliminate one of the variables? Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? And remember, by doing that, I would be subtracting the same thing from both sides of the equation. This is $1.79. How do I know? Because it says this is equal to $1.79. So if we did that we would be subtracting the same thing from both sides of the equation. So let's subtract 3x plus y from the left-hand side of the equation. And let me just do this over on the right. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So let's subtract it. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. I'm just taking the second equation. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. And what do we get? When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out. 3x minus 3x is 0x. I won't even write it down. You get 4x minus-- sorry, 4y minus y. That is 3y. And that is going to be equal to $2.84 minus $1.79. What is that? That's $1.05. So 3y is equal to $1.05. Divide both sides by 3. y is equal to-- what's $1.05 divided by 3? So 3 goes into $1.05. It goes into 1 zero times. 0 times 3 is 0. 1 minus 0 is 1. Bring down a 0. 3 goes into 10 three times. 3 times 3 is 9. Subtract. 10 minus 9 is 1. Bring down the 5. 3 goes into 15 five times. 5 times 3 is 15. Subtract. We have no remainder. So y is equal to $0.35. So the cost of a Fruit Roll-Up is $0.35. Now we can substitute back into either of these equations to figure out the cost of a candy bar. So let's use this bottom equation right here. Which was originally, if you remember before I multiplied it by negative 1, it was 3x plus y is equal to $1.79. So that means that 3x plus the cost of a Fruit Roll-Up, 0.35 is equal to $1.79. If we subtract 0.35 from both sides, what do we get? The left-hand side-- you're just left with the 3x; these cancel out-- is equal to-- let's see, this is $1.79 minus $0.35. That's $1.44. And 3 goes into $1.44, I think it goes-- well, 3 goes into $1.44, it goes into 1 zero times. 1 times 3 is 0. Bring down the 1. Subtract. Bring down the 4. 3 goes into 14 four times. 4 times 3 is 12. I'm making this messy. 14 minus 12 is 2. Bring down the 4. 3 goes into 24 eight times. 8 times 3 is 24. No remainder. So x is equal to 0.48. So there you have it. We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.

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Solving systems of equations by elimination (video) | Khan Academy

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Let's explore a few more methods for solving systems of equations. Let's say I have the equation, 3x plus 4y is equal to 2.5. And I have another equation, 5x minus 4y is equal to 25.5. And we want to find an x and y value that satisfies both of these equations.

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Systems of equations with elimination (and manipulation)

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In some cases, we need to slightly manipulate a system of equations before we can solve it using the elimination method. So how it's done in this video.

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- [Instructor] 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.

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Systems of equations with elimination: x-4y=-18 & -x+3y=11 (video) | Khan Academy

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Instructor] 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.

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Systems of equations with elimination: King's cupcakes

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Sal uses simple elimination to figure out how many cupcakes are eaten by children and adults.

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In this module we have learned what a system of equations is and 3 different methods in which we can solve them. We can solve systems graphically, we can solve them using the substitution method, or we can solve them using the elimination method.

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How Do You Solve a System of Equations Using the Elimination by Addition Method? | Virtual Nerd

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Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. In this non-linear system, users are free to take whatever path through the material best serves their needs. These unique features make Virtual Nerd a viable alternative to private tutoring.