# Solve a system of equations with three variables using substitution

Learn how to solve a system of equations with three variables using substitution with a couple of good examples. The examples mentioned here have only one solution. 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. This is basically what you do when you try to solve a system of equations with three variables using substitution.

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)

## A word problem about how to solve a system of equations with three variables using substitution

Jose, Paul and Dan went fishing and all three of them caught a fish. Dan's fish weighed four times as much as Paul's fish. Jose's fish weighed half as much as Dan's fish. Altogether the three fish weighed 21 pounds. How much did each fish weigh?

Let x be Jose's weight
Let y be Paul's weight
Let z be Dan's weight

x + y + z = 21 (Equation #1)
z = 4 × y, so y = z/4 (Equation #2)

x = (1/2)z = z/2 (Equation #3)

Replace (Equation #2) and (Equation #3) in (Equation #1)
z/2 + z/4 + z = 21
2z/4 + z/4 + 4z/4 = 21
7z/4 = 21
(4/7) × 7z/4 = 21 * 4/7
z = 84/7 = 12
x = z/2 = 12/2 = 6
y = z/4 = 12/4 = 3

100 Tough Algebra Word Problems.

If you can solve these problems with no help, you must be a genius!