Solving 3 Variable Systems of Equations by Elimination


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In this tutorial we’ll explain solving 3 variable systems of equations by elimination. In other tutorials we’ve covered solving 2 variable systems of equations by elimination, and solving systems of equations by substitution.

For example, what is the value of x, y and z for 3x + 2y – z = 8, x – 3y + 2z = 7 and x – 2y + z = 4

Let’s number our three equations:

1) 3x + 2y – z = 8
2) x – 3y + 2z = 7
3) x – 2y + z = 4

The first step is to get rid of one of the variables by adding or subtracting two equations.
Since x has the same coefficient in equations 2 and 3, let’s subtract them:

(x – 3y + 2z = 7) – (x – 2y + z = 4)
x – 3y + 2z – x + 2y – z = 7 – 4
-3y + 2y + 2z – z = 3
-y + z = 3

We’ll call this equation 4:

4) -y + z = 3

This equation does not involve variable x, but we still can’t solve it because it has 2 unknowns, so we need 2 unique equations. The way to do this is to find another equation that does not involve variable x through a different combination of equations. We already used equations 2 and 3 to find equation 4, now we’ll use equations 1 and 3.

We can multiply both sides of equation 3 by 3 to get the same coefficients for x, resulting in 3x – 6y + 3z = 12
Now we subtract this from equation 1:

(3x + 2y – z = 8) – (3x – 6y + 3z = 12)
3x + 2y – z – 3x + 6y – 3z = 8 – 12
2y + 6y – z – 3z = -4
8y – 4z = -4
We can actually simplify this by dividing everything by 4 to get
2y – z = -1

We’ll call this equation 5:

5) 2y – z = -1

Now we have 2 equations that only involve y and z, so we can use the elimination method to solve. Since equation 4 has a +z term and equation 5 has a -z term, let’s add them together:

(-y + z = 3) + (2y – z = -1)
-y + z + 2y – z = 3 – 1
-y + 2y = 2
y = 2

Now we can plug the value of y into equation 4 or 5 to solve for z. Let’s plug it into equation 4:

-y + z = 3
-2 + z = 3
z = 5

If you were to plug y into equation 5 you also get z = 5. Now that we have the values of y and z, we can simply plug them into equation 1 or 2 to find x. Let’s use equation 2:

x – 3y + 2z = 7
x – 3(2) + 2(5) = 7
x – 6 + 10 = 7
x + 4 = 7
x = 3

If you want to check that this is correct we can plug our answers into the original equations:

1) 3x + 2y – z = 8
2) x – 3y + 2z = 7
3) x – 2y + z = 4

1) 3(3) + 2(2) – (5) = 8
2) (3) – 3(2) + 2(5) = 7
3) (3) – 2(2) + (5) = 4

Here’s a practice problem to try on your own:
What is the value of x, y and z for 5x – 3y – 2z = -42, 2x + 3y – 7z = -37, and 4x – y + z = -3
The answer is x = -1, y = 7, z = 8

And that’s it for solving 3 variable systems of equations using elimination!


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