3-6 Solving Systems with 3 Variables
Use substitution, elimination, or both to solve for all three variables.
If one variable is easy to solve for, find its value and substitute it into the other
equations. Repeat substitution until all variables are solved.
2 x  3 y  z  10

1.  4 x  2 z  10

4 z  12

z is easy to solve for in the last equation. Thus, Z = 3.
Substitute z = 3 into the other two equations.
4x + 2(3) = 10
4x + 6 = 10
4x = 4
X=1
2x + 3y – 3 = -10
2x + 3y
= -7
2(1) + 3y = -7
2 + 3y = -7
3y = -9
Y = -3
Substitute x = 1 into the last equation.
Write the answer as an ordered triple: (1, -3, 3).
If no variable is easy to solve for, rewrite the system as two sets of equations with the
middle equation used twice and use double elimination.
2 x  3 y  z  3

2. 2 x  y  3z  9
 2x  4 y  z  5

Rewrite the system as two new sets of equations using the middle equation twice.
2 x  3 y  z  3
2 x  y  3 z  9
Eliminate the same variable in both.


2 x  y  3 z  9
 2x  4 y  z  5
2y +2z = 6
3y + 4z = 14
Use the new equations to create a
system and solve with elimination.
 2 y  2 z  6 multiply by  2 4 y  4 z  12



3 y  4 z  14
3 y  4 z  14
-y
=2
Y = -2
Substitute into 2 variable equation to solve for z.
2(-2) + 2z = 6
-4 + 2z = 6
2z = 10
Z=5
Substitute z = 5 and y = -2 into a 3 variable equation to solve for x.
2x + 3(-2) – 5 = -3
2x – 6 – 5 = -3
2x = 8
x=4
Write the answer as an ordered triple: (4, -2, 5)
If a variable is missing, put in the variable using 0 as the coefficient and use double
elimination.
 x y 8

3. 4 y  2 z  8
3 x  2 z  4

Rewrite the system as two new sets of equations using the middle equation twice.
 For any missing variables, use 0 as the missing variable’s coefficient to fill its place.
x  y  0z  8
x  y  0z  8
0 x  4 y  2 z  8




multiply by 0
0 x  4 y  2 z  8
0 x  0 y  0 z  0
3x  0 y  2 z  4
x+y
=8
3x + 4y
= 12
Use the new equations to create a system and solve with elimination.
x  y  8
3x  3 y  24
multiply by -3



3x  4 y  12
3x  4 y  12
y = -12
Substitute y = -14 back into
one of the original equations:
x + (-12) = 8
x = 20
Substitute x = 20 back into another
equation and solve for z:
3(20) + 2z = 4
60 + 2z = 4
2z = -58
z = -29
Write the answer as an ordered triple: (20, -12, -29)
Try on your own:
5r  2s  0

1. 3t  12
6s  5t  10

6a  b  17

3. a  3c  6
2b  9c  8

b  c  4

2. 2a  4b  c  3
3b  3


2 x  3 y  4 z  3

4. 5 x  9 y  6 z  1
1
1
1
1
 x y z 
2
3
12
3