Solve a system of equations with three variables using substitution

Learn how to solve a system of equations with three variables with one solution using substitution 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 one equation and solve for one of its variables. Then, substitute the expression for that variable into each of the other two equations. Once you have done that, you will be left with a pair of equations with just 2 variables. 

Step 1

Choose 1. and solve for x.

x - 3y + 3z = -6

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

x + 3z = -6 + 3y

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

x = -6 + 3y - 3z

Step 2

Substitute -6 + 3y - 3z for x into 2. and 3.

Substitute -6 + 3y - 3z for x into 2.

2x + 3y - z = 15

2(-6 + 3y - 3z) + 3y - z = 15

-12 + 6y - 6z + 3y - z = 15

-12 + 9y - 7z = 15

4. 9y - 7z = 27

Substitute -6 + 3y - 3z for x into 3.

4x - 3y - z =  21

4(-6 + 3y - 3z) - 3y - z =  21

-24 + 12y - 12z - 3y - z = 21

-24 + 9y - 13z = 21

5. 9y - 13z = 45

Step 3

Write the two new equations as a system and solve for y and z

4. 9y - 7z = 27

5. 9y - 13z = 45

Multiply 5. by -1

-1(9y - 13z) = -1(45)

-9y + 13z = -45

4. 9y - 7z = 27

5. -9y + 13z = -45

Add the left sides of 4. and 5. and add the right sides of 4. and 5.

4. 9y - 7z = 27

5. -9y + 13z = -45
_________________
            6z    = -18

z = -3

Substitute the value of z in 4. to get y 

4. 9y - 7z = 27

9y - 7(-3) = 27

9y + 21 = 27

9y = 6

y = 6/9

y = 2/3

Step 4

Substitute the values for y and z into one of the original equations and solve for x. Let us replace in 1.

1. x - 3y + 3z = -6

x - 3(2/3) + 3(-3) = -6

x - 2 + -9 = -6

x + -11 = -6

x = -6 + 11

x = 5

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

Example #2:

1. 2x + y - z = 5

2. x + 4y + 2z = 16

3. 15x + 6y - 2z =  12

Step 1

Choose 1. and solve for z.

2x + y - z = 5

2x + y - z + z = 5 + z

2x + y - 5 = 5 - 5 + z

2x + y - 5 = z

Step 2

Substitute 2x + y - 5 for z into 2. and 3.

Substitute 2x + y - 5 for z into 2.

x + 4y + 2z = 16

x + 4y + 2(2x + y - 5) = 16

x + 4y + 4x + 2y - 10 = 16

5x + 6y = 16 + 10

4. 5x + 6y = 26

Substitute 2x + y - 5 for z into 3.

15x + 6y - 2z =  12

15x + 6y - 2(2x + y - 5) =  12

15x + 6y - 4x - 2y + 10 = 12

11x + 4y = 12 - 10

5. 11x + 4y = 2

Step 3

Write the two new equations as a system and solve for x and y

4. 5x + 6y = 26

5. 11x + 4y = 2

Multiply 4. by -2 and 5. by 3

4. -2(5x + 6y) = -2(26)

5. 3(11x + 4y) = 3(2)

4. -10x + -12y = -52

5. 33x + 12y = 6

Add the left sides of 4. and 5. and add the right sides of 4. and 5.

4. -10x + -12y = -52

5. 33x + 12y   = 6
_________________
   23x              = -46

x = -2

Substitute the value of x in 4. to get y 

4. 5x + 6y = 26

5(-2) + 6y = 26

-10 + 6y = 26

6y = 36

y = 6

Step 4

Substitute the values for x and y into one of the original equations and solve for z. Let us replace in 1.

1. 2x + y - z = 5

2(-2) + 6 - z = 5

-4 + 6 - z = 5

2 - z = 5

2 - 5 = z

z = -3

The solution of the system is (-2, 6, -3)

Recent Articles

  1. Write a Polynomial from Standard Form to Factored Form

    Oct 14, 21 05:41 AM

    Learn how to write a polynomial from standard form to factored form

    Read More

Enjoy this page? Please pay it forward. Here's how...

Would you prefer to share this page with others by linking to it?

  1. Click on the HTML link code below.
  2. Copy and paste it, adding a note of your own, into your blog, a Web page, forums, a blog comment, your Facebook account, or anywhere that someone would find this page valuable.