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In another tutorial we covered the substitution method for solving systems of equations, in this tutorial we’ll go over solving systems of equations by elimination.

The elimination method works by first eliminating one of the variables that exists in the given equations. The way to accomplish this is to multiply or divide the given equations by a constant, and try to one get one of the variables in all equations to have the same coefficient.

For example, if the given equations are 3x + 4y = 5 and x – 2y = 1

Our goal is for one of the variables to have the same coefficient in both equations (doens’t matter which one).

We can multiply both sides of the second equation by 2 to write x – 2y = 1 as 2x – 4y = 2, so the coefficient for variable y will match with the coefficient for y in the first equation.

The next step is to add or subtract the equations to eliminate one of the variables.

We now have the equations:

3x + 4y = 5

2x – 4y = 2

The first equation has a +4y term and the second equation has a -4y term, so if we add the 2 equations together, they will cancel out. The way to add the equations is to simply group the left side terms and right side terms of both equations together.

3x + 4y + 2x – 4y = 5 + 2

3x + 2x = 5 + 2

5x = 7

x = 7/5

Now we can plug this value of x into either of the equations to solve for y!

Let’s look at another example, the same one as the substitution method tutorial:

What is the value of x and y for 3x – 5y = -7, and x + 3y = 13

Let’s number our two equations:

1) 2x – 5y = -7

2) x + 3y = 13

We can get the same coefficient on both sides by multiplying equation 2 by 2, we’ll call this equation 3:

3) 2x + 6y = 26

Now if we subtract equation 1 from equation 3:

(2x – 5y = -7) – (2x + 6y = 26)

2x – 5y – 2x – 6y = -7 – 26

-5y – 6y = -33

-11y = -33

y = 3

Now we can plug this value of y into equation 1:

2x – 5y = -7

2x – 5(3) = -7

2x = 8

x = 4

Just to check, we can also plug the answer for y into equation 2

x + 3y = 13

x + 3(3) = 13

x + 9 = 13

x = 13 – 9

x = 4

It’s great that we get the same value, as expected if we did everything correctly.

We can go through another example, also solved in the substitution tutorial:

What is the value of x and y for 3x + 2y = 11, and 7x – 4y = 10?

Let’s number our equations again:

1) 3x + 2y = 11

2) 7x – 4y = 10

We need one of the variable coefficients to match, and we could do either one, but just looking at the equations, the y coefficients are a lot easier, because 4 is simply 2 doubled (3 and 7 are a lot more annoying to match). We’ll multiply both sides of equation 1 by 2 to get equation 3:

3) 6x + 4y = 22

Now since equation 2 has a -4y term and equation 3 has a +4y term, we simply add them together:

(7x – 4y = 10) + (6x + 4y = 22)

7x – 4y + 6x + 4y = 10 + 22

7x + 6x = 32

13x = 32

x = 32/13

Now we can plug this value of x into equation 1 or 2, let’s do equation 1:

3x + 2y = 11

3(32/13) + 2y = 11

96/13 + 2y = 11

2y = 143/13 – 96/13

2y = 47/13

y = 47/26

We can check if we found the correct values by plugging them into both equations:

1) 3x + 2y = 11

2) 7x – 4y = 10

1) 3(^{33}/_{13}) + 2(^{47}/_{26}) = 11

2) 7(^{33}/_{13}) – 4(^{47}/_{26}) = 10

If you find that this doesn’t hold true, it means we made an error somewhere.

The best way to learn is by practicing, so try out these examples on your own:

What is the value of x and y for 2x + 4y = 22 and 7x – 3y = 26?

Answer: x = 5, y = 3

What is the value of x and y for ^{1}/_{2}x – 6y = -1 and 8x + 3y = 17?

Answer: x = ^{1}/_{2}, y = ^{1}/_{3}

If you can solve these, you’re all set for solving systems of equations by elimination!

If you prefer to watch a video version of this tutorial:

You can also check out more great math tutorials.