Warm-Up
Solve using the linear combination method.
7x + y =10
3x -2y = -3
(1,3)
3.6 Solving Systems of Linear
Equations in 3 Variables
10/6/15
A system of linear equations
in 3 variables
Looks something like this:
A solution is an ordered triple (x,y,z) that
makes all 3 equations true.
Here is a system of three linear
equations in three variables:
 x  2 y  3z  3

2 x  5 y  4 z  13
5 x  4 y  z  5

Is the ordered triple (2, -1, 1) a
solution?
2  2(1)  3(1)  2  2  3  3

2(2)  5(1)  4(1)  4  5  4  13
5(2)  4(1)  1  10  4  1  5

Steps for solving in 3 variables
1. Use the linear combination method to
rewrite the linear system in 3 variables as a
linear system in 2 variables.
2. Solve the new linear system for both of its
variables.
3. Substitute the values found in Step 2 into
one of the original equations and solve for
the remaining variable.
Solve the
system
x + 3y − z = −11
2x + y + z = 1
z’s are easy to cancel!
3x +4y = −10
2. 2x + y + z =1
5x −2y +3z = 21
Must cancel z’s again!
−6x −3y −3z = −3
5x −2y +3z = 21
−x −5y
= 18
1.
x + 3y − z = −11
2x + y + z = 1
5x − 2y + 3z = 21
3. 3x +4y = −10
−x −5y = 18
Solve for x & y
3x + 4y = −10
−3x −15y = 54
3x +4(−4)= −10
x=2
2(2) +(−4) +z =1
4 −4+x =1
z=1
−11y = 44
y = −4
(2, −4, 1)
Solve the
system
−x +2y +z = 3
2x + 2y +z = 5
4x +4y +2z = 6
1. −x + 2y +z = 3
2x + 2y + z = 5
z’s are easy to cancel
− x + 2y +z = 3
−2x −2y −z = −5
−3x
= −2
x = 2/3
2. 2x +2y + z = 5
4x +4y +2z= 6
Cancel the z’s again
−4x −4y −2z = −10
4x +4y +2z = 6
0 = −4
Doesn’t make sense
No solution