3-6
Solving Systems of Linear
Equations in Three
Variables
Objective: CA 2.0: Students solve
systems of linear equations and
inequalities in three variables by
substitution, with graphs.
The linear combination method (3variable systems):
Step 1:
Use the linear combination method to
rewrite the linear system in three
variables as a linear system in two
variables.
Step 2:
Solve the new linear system for both of
its variables.
Step 3
Substitute the value found in step 2 into
one of the original equations and solves
for the remaining variable.
If you obtain a false equation, such as:
• 0 = 1, then the system has no
solutions.
• 0 = 0, then the system has
infinitely many solutions.
Example 1: Solve the system using
the Linear Combination Method.
3x  2 y  4 z  11
2 x  y  3z  4
5x  3 y  5 z  1
Equation 1
Equation 2
Equation 3
Step 1: Eliminate one of the variables
in two of the original equations.
3x + 2y + 4z = 11
Equation 1
2(2x – y + 3z = 4)
Equation 2
4x – 2y + 6z = 8
7x + 0y + 10z = 19
7x + 10z = 19
new Equation 1
Step 1: Eliminate one of the
variables in two of the original
equations.
-3(2x – y + 3z = 4)
Equation 2
5x – 3y + 5z = -1
Equation 3
-6x + 3y – 9z = -12
-x – 4z = -18
new Equation 2
Step 2: Solve the new system of
equations
7x + 10z = 19
-x – 4z = -13
7x + 10z = 19
7(-x – 4z) = -13(7)
-7x – 28x = -91
7x + 10z = 19
+ -7x - 28z = -91
-18z = -72
z=4
Substitute z = 4 into new
equation 1 or 2 and solve for x
-x – 4(4) = -13
-x -16 = -13
-x = 3
x = -3
Step 3: Substitute x = -3 and z = 4 into
any of the original equations and solve
for y.
2x – y + 3z = 4
2(-3) – y + 3(4) = 4
-6 – y + 12 = 4
-y + 6 = 4
-y = -2
y=2
The solution is x = -3, y = 2, z = 4
or
(-3, 2, 4)
Example 2: Solve the system of
Equations
x+y+z=2
3x + 3y + 3z = 14
x – 2y + z = 4
Step 1: Eliminate one of the variables
in two of the original equations.
-3(x + y + z = 2)
-3x – 3y -3z = -6
Add -3 times Equation 1 to Equation 2
-3x- 3y - 3z = -6
3x + 3y + 3z = 14
new Eq.1
Eq. 2
0 =8
Since 0 ≠ 8 the result is a false
equation. This system has NO
SOLUTIONS.
Example 3: Solve the system of
equations.
x+y+z=2
x+y–z=2
2x + 2y + z = 4
Step 1: Use the linear combination
method to rewrite the linear system in
three variables as a linear system in
two variables.
x+y+z=2
+x+y–z=2
2x + 2y = 4
New Equation 1
x+y–z=2
Equation 2
+ 2x + 2y + z = 4
Equation 3
3x + 3y = 6
New Equation 2
Step 2: Solve the new linear system for
both of its variables.
3(2x + 2y) = 4(3)
3 (New equation 1)
-2(3x + 3y) = 6(-2)
-2 (New equation 2)
6x + 6y = 12
-6x – 6y = -12
0=0
0 = 0 is an identity therefore there are
infinitely many solutions to this system
of equations.
Home work page 181 12 –30 every third one,
35 – 39 odd,
45 – 71 every other odd.