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# Systems of equations with elimination: x+2y=6 & 4x-2y=14

CCSS Math: 8.EE.C.8b

## Video transcript

Use elimination to solve for x and y. And they gave us two equations here-- x plus 2y is equal to 6 and 4x minus 2y is equal to 14. So to solve by elimination, what we do is we're going to add these two equations together so that one of the two variables essentially gets eliminated, gets canceled out. And what we could do right here, we see we have a plus 2y here, and we see we have a negative 2y right over here. So clearly if we added these two together, the y's would cancel out, and so that's exactly what we're going to do. We're going to add the left side of this equation to the left side of this equation, and the right-hand side of that equation to the right-hand side of the bottom equation. And just to make it clear that this should make sense is we're just using both of these constraints, whenever you learn about any type of equation. So if I have x plus 2y is equal to 6, you learned early on in algebra that you can manipulate this equation in any way as long as whatever you do to the left-hand side of the equation, you do the right-hand side. If this is equal to that, the only way that the equality will still hold is whatever you do to this-- whatever you add to this or multiply it by-- you also do to the right-hand side. So when we're adding these two equations, that's exactly what we're doing. We could say, hey, let's add 14 to both sides of this equation. So you could add 14 on this side, you could add 14 on that side. That wouldn't be anything new, but the second equation right here tells us that 4x minus 2y is the same thing as 14. So instead of adding 14 on the left-hand side, I could add 4x minus 2y. When we're adding these two equations, we're really just adding the same thing. You could view it as we're starting with this equation, and then we're adding the same thing to both sides. On the right-hand side it looks like we're adding 14 to the 6. On the left-hand side it looks like we're adding 4x minus 2y to whatever is on the left-hand side. But the second constraint tells us that 14 and 4x minus 2y are the same thing. So we're adding the same thing to both sides. So with that said, let's just do it. So the left-hand side, if we add it up, we have x plus 4x is 5x, and then the 2y cancels out with the negative 2y. And then on the right-hand side we have 6 plus 14. 6 plus 14 Is 20. So we're left with one equation with one unknown. 5x is equal to 20. We can divide both sides by 5, and we are left with x is equal to 4. Now we can go back and substitute in x equals 4 into either of these equations to solve for y. So let's use this top one. So we have 4 plus 2y is equal to 6. We can subtract 4 from both sides. So then we get 2y is equal to 2. Divide both sides by 2. We get y is equal to 1. So the solution, the x's and y's that satisfy both of these equations, are x is equal to 4, and y is equal to 1. So this is the solution for this system, or this coordinate would be the point of intersection of these two lines. And we can verify it. Let's verify that when we put x is equal to 4, and y is equal to 1, in this first equation it satisfies it. So we have 4 plus 2 times 1. That's 4 plus 2. That does, indeed, equal 6. And then the second equation right over here, you have 4 times 4 minus 2 times 1. This is equal to 16 minus 2, which does indeed equal 14. So it definitely does satisfy both of these equations. So we're done. X is equal to 4, and y is equal to 1.