Steps for solving systems of equations by elimination

Another way of solving a linear system is to use the elimination method. In the elimination method you either add or subtract the equations to get an equation in one variable.

When the coefficients of one variable are opposites you add the equations to eliminate a variable and when the coefficients of one variable are equal you subtract the equations to eliminate a variable.

Example

\begin{cases} 3y+2x=6\\ 5y-2x=10 \end{cases}

We can eliminate the \(x\)-variable by addition of the two equations.

\begin{cases} 3y+2x=6 \\ \underline{+\: 5y-2x=10} \end{cases} 

$$=8y\: \: \: \: \; \; \; \; =16$$

$$\begin{matrix} \: \: \: y\: \: \: \: \: \; \; \; \; \; =2 \end{matrix}$$

The value of \(y\) can now be substituted into either of the original equations to find the value of \(x\)

$$3y+2x=6$$

$$3\cdot {\color{green} 2}+2x=6$$

$$6+2x=6$$

$$x=0$$

The solution of the linear system is \((0, 2)\).

To avoid errors make sure that all like terms and equal signs are in the same columns before beginning the elimination.

If you don't have equations where you can eliminate a variable by addition or subtraction you directly you can begin by multiplying one or both of the equations with a constant to obtain an equivalent linear system where you can eliminate one of the variables by addition or subtraction.

Example

\begin{cases} 3x+y=9\\ 5x+4y=22 \end{cases}

Begin by multiplying the first equation by \(-4\) so that the coefficients of \(y\) are opposites

\begin{cases} \color{green}{-4} \cdot (3x + y) = \color{green}{-4} \cdot 9\\ 5x + 4y = 22 \end{cases}

$$\Rightarrow$$

\begin{cases}-12x-4y=-36 \\ \underline{+5x+4y=22 }\end{cases}

$$=-7x\: \: \: \: \: \: \: \: \: \: =-14$$

$$\begin{matrix} \: \:\; \:\: x\: \: \: \: \: \: \: \: \: \: \:=2 \end{matrix}$$

Substitute \(x\) in either of the original equations to get the value of \(y\)

$$3x+y=9$$

$$3\cdot {\color{green} 2}+y=9$$

$$6+y=9$$

$$y=3$$

The solution of the linear system is \((2, 3)\)

Video lesson

Solve the following linear system using the elimination method

\begin{cases} 2y - 4x = 2 \\ y = -x + 4 \end{cases}

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Video transcript

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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