Lesson Objectives
  • Demonstrate an understanding of how to solve a linear system in two variables using elimination
  • Learn how to write the solution for a linear equation in three variables using an ordered triple
  • Learn how to solve a linear system in three variables
  • Learn how to identify a linear system in three variables with no solution
  • Learn how to identify a linear system in three variables with infinitely many solutions

How to Solve a Linear System in Three Variables


So far, we have learned how to solve a system of linear equations in two variables using graphing, substitution, and elimination. In this lesson, we will take the next step and learn how to solve a system of linear equations in three variables. At this point, we know the solution for a linear equation in two variables is displayed as an ordered pair (x,y). When we work with a linear equation in three variables, the solution is displayed as an ordered triple (x,y,z). Let's look at a quick example.
5x + 2y + 7z = 53
With a linear equation in two variables, we could pick a value for x and solve for y or vice versa. In the case of a linear equation in three variables, we need to pick the values for two of the variables and solve for the third. Let's let x be (-3), and y be (-1). What is the value of z that makes the equation true?
(-3,-1, ?)
Just plug in a (-3) for x, and a (-1) for y.
5(-3) + 2(-1) + 7z = 53
-15 - 2 + 7z = 53
-17 + 7z = 53
7z = 70
z = 10
(-3,-1,10)
Just like with a linear equation in two variables, there are infinitely many solutions for a linear equation in three variables. Although it is not practical, a system of equations in three variables can be solved using graphing. Since this requires three dimensional graphing, it is best to stick with algebraic methods.

Solving a Linear System in Three Variables

  • Write each equation in the format of: ax + by + cz = d
    • Move all variable terms to the left and the constant term to the right
  • Pick one of the variables to eliminate
    • Use elimination to eliminate one of the variables from any two equations
    • Use elimination again to eliminate the same variable from any other two equations
  • Solve the linear system in two variables
  • Find the final unknown
    • Plug in the values for the two known variables into any original equation, solve for the final unknown
  • Check
    • Plug in for x, y, and z in each original equation
Let's look at an example.
Example 1: Solve each linear system in three variables
3x + 3y - 5z = 19
-2x - 2y - 2z = -18
-x - 5y + 5z = -11
Step 1) Write each equation in the format of: ax + by + cz = d.
In our case, each equation is in the proper format. We will label our equations as 1,2, and 3.
1) 3x + 3y - 5z = 19
2) -2x - 2y - 2z = -18
3) -x - 5y + 5z = -11
Step 2) Pick one of the variables to eliminate.
In our case, it looks easiest to eliminate the variable z. In equation 1, we have (-5z) and in equation 3, we have (5z). Add the left sides and set this equal to the sum of the right sides. $$\require{cancel}+3x + 3y + \cancel{-5x} = 19$$ $$\underline{-1x - 5y + \cancel{5z} = -11}$$ $$\hspace{1em}2x-2y \hspace{2.5em} = 8$$ Let's label this as equation 4:
4) 2x - 2y = 8
Now we need to eliminate the same variable (z) from any other two equations. Let's eliminate (z) from equations 2 and 3. We can multiply equation 2 by (5) and multiply equation 3 by (2). This will give us -10z and 10z.
2) -10x - 10y - 10z = -90
3) -2x - 10y + 10z = -22
Add the left sides and set this equal to the sum of the right sides. $$-10x - 10y + \cancel{-10z} = -90 $$ $$\underline{-2x - 10y + \cancel{10z} = -22}$$ $$-12x-20y \hspace{2.5em} = -112$$ Let's label this as equation 5:
5) -12x - 20y = -112
Step 3) Solve the linear system in two variables.
We have created a linear system in two variables, equations 4 and 5.
4) 2x - 2y = 8
5) -12x - 20y = -112
To solve this system, we can multiply equation 4 by (6), this will give us 12x and -12x.
4) 12x - 12y = 48
5) -12x - 20y = -112
Add the left sides and set this equal to the sum of the right sides. $$\cancel{+12x} - 12y = 48 $$ $$\underline{\cancel{-12x} - 20y =-112}$$ $$\hspace{3em}-32y = -64$$ -32y = -64
y = 2
Now that we know that y is 2, let's plug back into either equation 4 or 5 and find the value of x.
Let's plug in a 2 for y in equation 4:
2x - 2y = 8
2x - 2(2) = 8
2x - 4 = 8
2x = 12
x = 6
Step 4) Find the final unknown.
We will plug in a 6 for x and a 2 for y in any of the three original equations. Let's use equation 3 since it is the simplest.
-x - 5y + 5z = -11
-(6) - 5(2) + 5z = -11
-6 - 10 + 5z = -11
-16 + 5z = -11
5z = 5
z = 1
This tells us our solution is the ordered triple: (6,2,1)
Step 5) Check.
1) 3x + 3y - 5z = 19
3(6) + 3(2) - 5(1) = 19
18 + 6 - 5 = 19
24 - 5 = 19
19 = 19
2) -2x - 2y - 2z = -18
-2(6) - 2(2) - 2(1) = -18
-12 - 4 - 2 = -18
-18 = -18
3) -x - 5y + 5z = -11
-(6) - 5(2) + 5(1) = -11
-6 - 10 + 5 = -11
-16 + 5 = -11
-11 = -11
Example 2: Solve each linear system in three variables
6x - 2y - 8z = 12
-5x + 6y + 4z = 25
9x + 6y - 8z = 43
Step 1) Write each equation in the format of: ax + by + cz = d.
In our case, each equation is in the proper format. We will label our equations as 1,2, and 3.
1) 6x - 2y - 8z = 12
2) -5x + 6y + 4z = 25
3) 9x + 6y - 8z = 43
Step 2) Pick one of the variables to eliminate.
In our case, it looks easiest to eliminate the variable y or z. In equation 2, we have (6y) and in equation 3, we have (6y). Let's multiply equation 3 by (-1). This will give us 6y and -6y.
3) -9x - 6y + 8z = -43
Add the left sides and set this equal to the sum of the right sides. $$-5x + \cancel{6y} + 4z = 25$$ $$\underline{-9x + \cancel{-6y} + 8z = -43}$$ $$-14x + 12z \hspace{2em} = -18$$ Let's label this as equation 4:
4) -14x + 12z = -18
Now we need to eliminate the same variable (y) from any other two equations. Let's eliminate (y) from equations 1 and 2. We can multiply equation 1 by (3). This will give us -6y and 6y.
1) 18x - 6y - 24z = 36
Add the left sides and set this equal to the sum of the right sides. $$18x + \cancel{-6y} - 24z = 36$$ $$\underline{-5x + \cancel{6y} + 4z = 25}$$ $$13x - 20z\hspace{2.5em} = 61$$ Let's label this as equation 5:
5) 13x - 20z = 61
Step 3) Solve the linear system in two variables.
We have created a linear system in two variables, equations 4 and 5.
4) -14x + 12z = -18
5) 13x - 20z = 61
To solve this system, we can multiply equation 4 by (5), and equation 5 by (3). This will give us 60z and -60z.
4) -70x + 60z = -90
5) 39x - 60z = 183
Add the left sides and set this equal to the sum of the right sides. $$-70x + \cancel{60z} = -90$$ $$\underline{39x + \cancel{-60z} = 183}$$ $$-31x \hspace{3em} = 93$$ -31x = 93
x = -3
Now that we know that x is -3, let's plug back into either equation 4 or 5 and find the value of z.
Let's plug in a -3 for x in equation 4:
-14x + 12z = -18
-14(-3) + 12z = -18
42 + 12z = -18
12z = -60
z = -5
Step 4) Find the final unknown.
We will plug in a -3 for x and a -5 for z in any of the three original equations. Let's use equation 1 since it is the simplest.
6x - 2y - 8z = 12
6(-3) - 2y - 8(-5) = 12
-18 - 2y + 40 = 12
22 - 2y = 12
-2y = -10
y = 5
This tells us our solution is the ordered triple: (-3,5,-5)
Step 5) Check.
1) 6x - 2y - 8z = 12
6(-3) - 2(5) - 8(-5) = 12
-18 - 10 + 40 = 12
12 = 12
2) -5x + 6y + 4z = 25
-5(-3) + 6(5) + 4(-5) = 25
15 + 30 - 20 = 25
25 = 25
3) 9x + 6y - 8z = 43
9(-3) + 6(5) - 8(-5) = 43
-27 + 30 + 40 = 43
43 = 43

Solving a System of Equations with Missing Terms

A common scenario involves a linear system in three variables where terms are missing. Let's look at an example.
Example 3: Solve each linear system in three variables
7y + 4z = -26
-6x + y = -24 - 3z
z = x + 4y + 23
Step 1) Write each equation in the format of: ax + by + cz = d.
We will also label our equations as 1, 2, and 3.
1) 7y + 4z = -26
2) -6x + y + 3z = -24
3) x + 4y - z = -23
Step 2) Pick one of the variables to eliminate.
Notice how equation 1 does not have the x-variable. This actually makes our process much easier. We just need to eliminate the x-variable from equation 2 and 3. This will produce a linear system in two variables.
We will multiply equation 3 by (6), this will give us -6x and 6x.
2) -6x + y + 3z = -24
3) 6x + 24y - 6z = -138
Add the left sides and set this equal to the sum of the right sides. $$-6x + y + 3z = -24$$ $$\underline{+6x + 24y - 6z = -138}$$ $$\hspace{3em}25y - 3z = -162$$ Let's label this as equation 4:
4) 25y - 3z = -162
Step 3) Solve the linear system in two variables.
We have created a linear system in two variables, equations 1 and 4.
1) 7y + 4z = -26
4) 25y - 3z = -162
Let's multiply equation 1 by (3) and equation 4 by (4). This will give us 12z and -12z.
1) 21y + 12z = -78
4) 100y - 12z = -648
Add the left sides and set this equal to the sum of the right sides. $$21y + \cancel{12z} = -78 $$ $$\underline{100y + \cancel{-12z} = -648}$$ $$\hspace{3em}121y = -726$$ 121y = -726
y = -6
Now that we know that y is -6, let's plug back into either equation 1 or 4 and find the value of z.
Let's plug in a -6 for y in equation 1:
7y + 4z = -26
7(-6) + 4z = -26
-42 + 4z = -26
4z = 16
z = 4
Step 4) Find the final unknown.
We will plug in a -6 for y and a 4 for z in either equation 2 or 3. Let's use equation 3 since it is the simplest.
x + 4y - z = -23
x + 4(-6) - 4 = -23
x - 24 - 4 = -23
x - 28 = -23
x = 5
This tells us our solution is the ordered triple: (5,-6,4)
Step 5) Check.
7y + 4z = -26
7(-6) + 4(4) = -26
-42 + 16 = -26
-26 = -26
-6x + y = -24 - 3z
-6(5) - 6 = -24 - 3(4)
-30 - 6 = -24 - 12
-36 = -36
z = x + 4y + 23
4 = 5 + 4(-6) + 23
4 = 5 - 24 + 23
4 = 4

Systems of Linear Equations in Three Variables with No Solution

When we work with linear systems in three variables, we may come across an "inconsistent system". This means the system does not have a solution. When this scenario occurs, the variables will drop out and we will be left with a false statement. Let's look at an example.
Example 4: Solve each linear system in three variables
x + y + 2z = -8
-7x + y - 4z = -31
-4x + 4y + 2z = -10
Step 1) Write each equation in the format of: ax + by + cz = d.
In this case, each equation is in the proper format. We will label our equations as 1, 2, and 3.
1) x + y + 2z = -8
2) -7x + y - 4z = -31
3) -4x + 4y + 2z = -10
Step 2) Pick one of the variables to eliminate.
In our case, it looks easiest to eliminate the variable z. In equation 1, we have (2z) and in equation 3, we have (2z). Let's multiply equation 3 by (-1). This will give us 2z and -2z.
3) 4x - 4y - 2z = 10
Add the left sides and set this equal to the sum of the right sides. $$x + y + 2z = -8$$ $$\underline{4x - 4y - 2z = 10}$$ $$5x - 3y \hspace{2em} = 2$$ Let's label this as equation 4:
4) 5x - 3y = 2
Now we need to eliminate the same variable (z) from any other two equations. Let's eliminate (z) from equations 2 and 3. We can multiply equation 3 by (2). This will give us -4z and 4z.
3) -8x + 8y + 4z = -20
Add the left sides and set this equal to the sum of the right sides. $$-7x + y - 4z = -31$$ $$\underline{-8x + 8y + 4z = -20}$$ $$-15x + 9y \hspace{1.5em} = -51$$ Let's label this as equation 5:
5) -15x + 9y = -51
Step 3) Solve the linear system in two variables.
We have created a linear system in two variables, equations 4 and 5.
4) 5x - 3y = 2
5) -15x + 9y = -51
To solve this system, let's multiply equation 4 by (3), this will give us 15x and -15x. Notice how we will also have -9y and 9y.
4) 15x - 9y = 6
5) -15x + 9y = -51
Add the left sides and set this equal to the sum of the right sides. $$\cancel{+15x} + \cancel{-9y} = 6$$ $$\underline{\cancel{-15x} + \cancel{9y} = -51}$$ $$\hspace{5em}0 = -45$$ 0 = -45 (false)
This tells us we have an inconsistent system. We can state our answer as "no solution".

Systems of Linear Equations in Three Variables with Infinitely Many Solutions

Lastly, a linear system in three variables may also give us an answer of "infinitely many solutions". When this scenario occurs, the variables will drop out and we will be left with a true statement. Let's look at an example.
Example 5: Solve each linear system in three variables
-4x = -y + 23
-4y + 4z = -2x - 22
-5x = 34 - 3y + 2z
Step 1) Write each equation in the format of: ax + by + cz = d.
We will also label our equations as 1, 2, and 3.
1) -4x + y = 23
2) 2x - 4y + 4z = -22
3) -5x + 3y - 2z = 34
Step 2) Pick one of the variables to eliminate.
Notice how equation 1 does not have the z-variable. This actually makes our process much easier. We just need to eliminate the z-variable from equations 2 and 3. This will produce a linear system in two variables. We will multiply equation 3 by (2), this will give us 4z and -4z.
3) -10x + 6y - 4z = 68
Add the left sides and set this equal to the sum of the right sides. $$+2x - 4y + \cancel{4z} = -22$$ $$\underline{-10x + 6y + \cancel{-4z} = 68}$$ $$-8x + 2y \hspace{3.5em} = 46$$ Let's label this as equation 4:
4) -8x + 2y = 46
Step 3) Solve the linear system in two variables.
We have created a linear system in two variables, equations 1 and 4.
1) -4x + y = 23
4) -8x + 2y = 46
Let's multiply equation 1 by (-2). This will give us 8x and -8x. Notice how we will also have -2y and 2y.
1) 8x - 2y = -46
4) -8x + 2y = 46
Add the left sides and set this equal to the sum of the right sides. $$\cancel{8x} + \cancel{-2y} = \cancel{-46}$$ $$\underline{\cancel{-8x} + \cancel{2y} = \cancel{46}}$$ $$0 = 0$$ 0 = 0 (true)
We can state our answer as "infinitely many solutions".