The techniques for solving a
system of linear equations in three variables are similar to those used on systems of
linear equations in two variables. We eliminate variables by either substitution or
addition.
Example 1
A linear system with a single solution
Solve the system:
(1) x + y  z = 1
(2) 2x  2y + 3z = 8
(3) 2x  y + 2z = 9
Solution
We can eliminate z from Eqs. (1) and (2) by multiplying Eq. (1) by 3 and adding it
to Eq. (2):

3x 
+ 3y 
 3z 
= 3 
Eq. (1) multiplied by 1 

2x 
 2y 
+ 3z 
= 8 
Eq.(2) 
(4) 
5x 
+ y 

= 5 
Now we must eliminate the same variable, z, from another pair of equations.
Eliminate z from (1) and (3):

2x 
+ 2y 
 2z 
= 2 
Eq. (1) multiplied by 2 

2x 
 y 
+ 2z 
= 9 

Eq. (3) 
(5) 
4x 
+ y 

= 7 

Equations (4) and (5) give us a system with two variables. We now solve this
system. Eliminate y by multiplying Eq. (5) by 1 and adding the equations:
5x + y 
= 5 
Eq. (4) 
4x  y 
= 7 
Eq. (5) multiplied by 1 
x 
= 2 

Now that we have x, we can replace x by 2 in Eq. (5) to find y:
4x + y
4(2) + y
8 + y
y 
= 7 = 7
= 7
= 15 
Now replace x by 2 and y by 15 in Eq. (1) to find z:
x + y  z 2 + 15  z
13  z
z
z 
= 1 = 1
= 1
= 14
= 14 
Check that (2, 15, 14) satisfies all three of the original equations. The solution set
is {(2, 15, 14)}.
Helpful Hint
Note that we could have chosen
to eliminate x, y, or z first
in Example 1.You should solve
this same system by eliminating
x first and then by eliminating
y first. To eliminate x
first, multiply the first equation
by 2 and add it with the second
and third equations.
In the next example we use a combination of addition and substitution.
