Solve a system of equations with three variables using elimination

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Learn how to solve a system of equations with three variables with one solution using elimination method. We will number the equations in order to make the procedure easy to follow.

Example #1:

1. x - 3y + 3z = -6

2. 2x + 3y - z = 15

3. 4x - 3y - z =  21

The goal is to pick a pair of equations and eliminate a variable by adding the two equations. And then pick another pair of equations and eliminate the exact same variable. Once you have done that, you will be left with a pair of equations with just 2 variables. 

Step 1

Notice that you can easily eliminate y because the y terms are already additive inverse.

Therefore, pair 1. with 2. and 2. with 3. so you can easily get rid of the y terms.

Add the left sides of 1. and 2. and add the right sides of 1. and 2.

You get 3x + 2z = 9 and we call this new equation 4.

1. x - 3y + 3z = -6

2. 2x + 3y - z = 15
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4. 3x   +    2z = 9 

Add the left sides of 2. and 3. and add the right sides of 2. and 3.

You get 6x - 2z = 36 and we call this new equation 5.

2. 2x + 3y - z = 15

3. 4x - 3y - z =  21
_________________
5. 6x    -    2z = 36

Step 2

Write the two new equations 4. and 5. as a system. 

4. 3x   +    2z = 9 

5. 6x    -    2z = 36

Solve for x and z. 

Since the z terms are already additive inverses, you can just add the left sides of 4. and 5. and add the right sides of 4. and 5. 

4. 3x   +    2z = 9 

5. 6x    -    2z = 36
_________________
    9x             = 45

x = 5

Use either equation 4. or 5. to find z

4. 3x + 2z = 9

3(5) + 2z = 9

15 + 2z = 9

2z = 9 - 15

2z = -6

z = -3

Step 3

Use either equation 1. 2. or 3. to find y.

2. 2x + 3y - z = 15

2(5) + 3y - -3 = 15

10 + 3y + 3 = 15

13 + 3y = 15

3y = 15 - 13

3y = 2

y = 2/3

The solution of the system is (5, 2/3, -3)


Example #2:

1. 6x - y + 2z = 8

2. 2x + 3y - z = -9

3. 4x + 2y + 5z =  1

Step 1

Multiply equation 2. by 2.

2. 2(2x + 3y - z) = 2(-9)

2. 4x + 6y - 2z = -18

Looking at 1. and 2. you can easily eliminate z because the z terms are now additive inverse.

1. 6x - y + 2z = 8

2. 4x + 6y - 2z = -18

Add the left sides of 1. and 2. and add the right sides of 1. and 2. 

You get 10x + 5y = -10 and we call this new equation 4.

1. 6x - y + 2z = 8

2. 4x + 6y - 2z = -18
__________________
4. 10x + 5y = -10

Multiply equation 2. by 5.

2. 5(2x + 3y - z) = 5(-9)

2. 10x + 15y - 5z = -45

Looking at 3. and 2. you can easily eliminate z because the z terms are now additive inverse.

2. 10x + 15y - 5z = -45

3. 4x + 2y +  5z =  1

Add the left sides of 2. and 3. and add the right sides of 2. and 3. 

You get 14x + 17y = -44 and we call this new equation 5.

2. 10x + 15y - 5z  = -45

3. 4x + 2y + 5z  =  1
____________________
5. 14x + 17y     = -44

Step 2

Write the two new equations 4. and 5. as a system. 

4. 10x + 5y = -10

5. 14x + 17y     = -44

Solve for x and y. 

Multiply 4. by -14 and 5. by 10

4. -14(10x + 5y) = -14(-10)

5. 10(14x + 17y)     = 10(-44)

4. -140x + -70y = 140

5. 140x + 170y     = -440

Since the x terms are now additive inverses, you can just add the left sides of 4. and 5. and add the right sides of 4. and 5. 

4. -140x + -70y = 140

5. 140x + 170y   = -440
______________________
               100y   = -300

y = -3

Use either equation 4. or 5. to find x

4. 10x + 5y = -10

10x + 5(-3) = -10

10x + -15 = -10

10x = 5

x = 5/10

x = 1/2

Use either equation 1. 2. or 3. to find z.

1. 6x - y + 2z = 8

6(1/2) - -3 + 2z = 8

3 + 3 + 2z = 8

6 + 2z = 8

2z = 2

z = 1

The solution of the system is (1/2, -3, 1)