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Let's explore a few more methods for

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solving systems of equations.

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Let's say I have the equation, 3x plus 4y is equal to 2.5.

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And I have another equation, 5x minus 4y is equal to 25.5.

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And we want to find an x and y value that satisfies both of

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these equations.

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If you think of it graphically, this would be the

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intersection of the lines that represent the solution sets to

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both of these equations.

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So how can we proceed?

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We saw in substitution, we like to

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eliminate one of the variables.

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We did it through substitution last time.

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But is there anything we can add or subtract-- let's focus

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on this yellow, on this top equation right here-- is there

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anything that we can add or subtract to both

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sides of this equation?

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Remember, any time you deal with an equation you have to

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add or subtract the same thing to both sides.

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But is there anything that we could add or subtract to both

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sides of this equation that might

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eliminate one of the variables?

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And then we would have one equation in one variable, and

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we can solve for it.

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And it's probably not obvious, even though it's sitting right

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in front of your face.

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Well, what if we just added this

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equation to that equation?

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What I mean by that is, what if we were to add 5x minus 4y

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to the left-hand side, and add 25.5 to the right-hand side?

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So if I were to literally add this to the left-hand side,

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and add that to the right-hand side.

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And you're probably saying, Sal, hold on, how can you just

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add two equations like that?

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And remember, when you're doing any equation, if I have

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any equation of the form-- well, really, any equation--

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Ax plus By is equal to C, if I want to do something to this

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equation, I just have to add the same thing to both sides

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of the equation.

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So I could, for example, I could add D to both sides of

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the equation.

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Because D is equal to D, so I won't be

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changing the equation.

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You would get Ax plus By, plus D is equal to C plus D.

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And we've seen that multiple, multiple times.

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Anything you do to one side of the equation, you have to do

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to the other side.

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But you're saying, hey, Sal, wait, on the left-hand side,

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you're adding 5x minus 4y to the equation.

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On the right-hand side, you're adding 25.5 to the equation.

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Aren't you adding two different things to both sides

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of the equation?

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And my answer would be no.

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We know that 5x minus 4y is 25.5.

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This quantity and this quantity are the same.

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They're both 25.5.

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This second equation is telling me that explicitly.

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So I can add this to the left-hand side.

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I'm essentially adding 25.5 to it.

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And I could add 25.5 to the right-hand side.

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So let's do that.

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If we were to add the left-hand side,

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3x plus 5x is 8x.

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And then what is 4y minus 4y?

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And this was the whole point.

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When I looked at these two equations, I said, oh, I have

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a 4y, I have a negative 4y.

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If you just add these two together, they are going to

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cancel out.

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They're going to be plus 0y.

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Or that whole term is just going to go away.

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And that's going to be equal to 2.5 plus 25.5 is 28.

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So you divide both sides.

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So you get 8x is equal to 28.

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And you divide both sides by 8, and we get x is equal to 28

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over 8, or you divide the numerator and the

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denominator by 4.

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That's equal to 7 over 2.

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That's our x value.

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Now we want to solve for our y value.

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And we could substitute this back into either

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of these two equations.

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Let's use the top one.

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You could do it with the bottom one as well.

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So we know that 3 times x, 3 times 7 over 2-- I'm just

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substituting the x value we figured out into this top

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equation-- 3 times 7 over 2, plus 4y is equal to 2.5.

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Let me just write that as 5/2.

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We're going to stay in the fraction world.

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So this is going to be 21 over 2 plus 4y is equal to 5/2.

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Subtract 21 over 2 from both sides.

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So minus 21 over 2, minus 21 over 2.

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The left-hand side-- you're just left with a 4y, because

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these two guys cancel out-- is equal to-- this is 5

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minus 21 over 2.

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That's negative 16 over 2.

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So that's negative 16 over 2, which is the same thing--

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well, I'll write it out as negative 16 over 2.

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Or we could write that-- let's continue up here-- 4y-- I'm

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just continuing this train of thought up here-- 4y is equal

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to negative 8.

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Divide both sides by 4, and you get y is

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equal to negative 2.

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So the solution to this equation is x is equal to 7/2,

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y is equal to negative 2.

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This would be the coordinate of their intersection.

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And you could try it out on both of these

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equations right here.

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So let's verify that it also

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satisfies this bottom equation.

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5 times 7/2 is 35 over 2 minus 4 times negative 2, so minus

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negative 8.

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That's equivalent to-- let's see, this is 17.5 plus 8.

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And that indeed does equal 25.5.

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So this satisfies both equations.

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Now let's see if we can use our newly found skills to

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tackle a word problem, our newly found skills in

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elimination.

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So here it says, Nadia and Peter visit the candy store.

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Nadia buys 3 candy bars and 4 Fruit Roll-Ups for $2.84.

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Peter also buys 3 candy bars, but can only afford 1

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additional Fruit Roll-Up.

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His purchase costs $1.79.

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What is the cost of each candy bar and each Fruit Roll-Up?

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So let's define some variables.

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Let's just use x and y.

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Let's let x equal cost of candy bar-- I was going to do

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a c and a f for Fruit Roll-Up, but I'll just stick with x and

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y-- cost of candy bar.

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And let y equal the cost of a Fruit Roll-Up.

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All right.

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So what does this first statement tell us?

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Nadia buys 3 candy bars, so the cost of 3 candy bars is

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going to be 3x.

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And 4 Fruit Roll-Ups.

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Plus 4 times y, the cost of a Fruit Roll-Up.

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This is how much Nadia spends.

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3 candy bars, 4 Fruit Roll-Ups.

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And it's going to cost $2.84.

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That's what this first statement tells us.

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It translates into that equation.

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The second statement.

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Peter also buys 3 candy bars, but could only afford 1

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additional Fruit Roll-Up.

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So plus 1 additional Fruit Roll-Up.

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His purchase cost is equal to $1.79.

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What is the cost of each candy bar and each Fruit Roll-Up?

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And we're going to solve this using elimination.

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You could solve this using any of the techniques we've seen

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so far-- substitution, elimination, even graphing,

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although it's kind of hard to eyeball

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things with the graphing.

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So how can we do this?

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Remember, with elimination, you're going to add-- let's

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focus on this top equation right here.

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Is there something we could add to both sides of this

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equation that'll help us

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eliminate one of the variables?

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Or let me put it this way, is there something we could add

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or subtract to both sides of this equation that will help

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us eliminate one of the variables?

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Well, like in the problem we did a little bit earlier in

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the video, what if we were to subtract this equation, or

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what if we were to subtract 3x plus y from 3x plus 4y on the

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left-hand side, and subtract $1.79 from

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the right-hand side?

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And remember, by doing that, I would be subtracting the same

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thing from both sides of the equation.

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This is $1.79.

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How do I know?

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Because it says this is equal to $1.79.

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So if we did that we would be subtracting the same thing

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from both sides of the equation.

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So let's subtract 3x plus y from the

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left-hand side of the equation.

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And let me just do this over on the right.

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If I subtract 3x plus y, that is the same thing as negative

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3x minus y, if you just distribute the negative sign.

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So let's subtract it.

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So you get negative 3x minus y-- maybe I should make it

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very clear this is not a plus sign; you could imagine I'm

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multiplying the second equation by negative 1-- is

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equal to negative $1.79.

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I'm just taking the second equation.

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You could imagine I'm multiplying it by negative 1,

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and now I'm going to add the left-hand side to the

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left-hand side of this equation, and the right-hand

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side to the right-hand side of that equation.

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And what do we get?

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When you add 3x plus 4y, minus 3x, minus y,

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the 3x's cancel out.

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3x minus 3x is 0x.

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I won't even write it down.

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You get 4x minus-- sorry, 4y minus y.

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That is 3y.

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And that is going to be equal to $2.84 minus $1.79.

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What is that?

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That's $1.05.

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So 3y is equal to $1.05.

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Divide both sides by 3.

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y is equal to-- what's $1.05 divided by 3?

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So 3 goes into $1.05.

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It goes into 1 zero times.

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0 times 3 is 0.

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1 minus 0 is 1.

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Bring down a 0.

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3 goes into 10 three times.

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3 times 3 is 9.

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Subtract.

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10 minus 9 is 1.

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Bring down the 5.

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3 goes into 15 five times.

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5 times 3 is 15.

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Subtract.

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We have no remainder.

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So y is equal to $0.35.

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So the cost of a Fruit Roll-Up is $0.35.

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Now we can substitute back into either of these equations

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to figure out the cost of a candy bar.

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So let's use this bottom equation right here.

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Which was originally, if you remember before I multiplied

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it by negative 1, it was 3x plus y is equal to $1.79.

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So that means that 3x plus the cost of a Fruit Roll-Up, 0.35

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is equal to $1.79.

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If we subtract 0.35 from both sides, what do we get?

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The left-hand side-- you're just left with the 3x; these

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cancel out-- is equal to-- let's see, this

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is $1.79 minus $0.35.

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That's $1.44.

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And 3 goes into $1.44, I think it goes-- well, 3 goes into

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$1.44, it goes into 1 zero times.

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1 times 3 is 0.

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Bring down the 1.

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Subtract.

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Bring down the 4.

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3 goes into 14 four times.

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4 times 3 is 12.

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I'm making this messy.

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14 minus 12 is 2.

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Bring down the 4.

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3 goes into 24 eight times.

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8 times 3 is 24.

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No remainder.

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So x is equal to 0.48.

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So there you have it.

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We figured out, using elimination, that the cost of

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a candy bar is equal to $0.48, and that the cost of a Fruit

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Roll-Up is equal to $0.35.

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