Solving Systems of Three Linear
Equations in Three Variables
The Elimination Method
Solutions of a system with 3 equations
The solution to a system of three linear
equations in three variables is an ordered
triple.
(x, y, z)
The solution must be a solution of all 3
equations.
Is (–3, 2, 4) a solution of this system?
3x + 2y + 4z = 11 3(–3) + 2(2) + 4(4) = 11 P
2x – y + 3z = 4
2(–3) – 2 + 3(4) = 4 P
5x – 3y + 5z = –1 5(–3) – 3(2) + 5(4) = –1P
Yes, it is a solution to the system because it
is a solution to all 3 equations.
This lesson will focus on the
Elimination Method.
Use elimination to solve the following
system of equations.
x – 3y + 6z = 21
3x + 2y – 5z = –30
2x – 5y + 2z = –6
Step 1
Rewrite the system as two smaller
systems, each containing two of the
three equations.
x – 3y + 6z = 21
3x + 2y – 5z = –30
2x – 5y + 2z = –6
x – 3y + 6z = 21
3x + 2y – 5z = –30
x – 3y + 6z = 21
2x – 5y + 2z = –6
Step 2
Eliminate THE SAME variable in each
of the two smaller systems.
Any variable will work, but sometimes
one may be a bit easier to eliminate.
I choose x for this system.
(x – 3y + 6z = 21) (–3)
3x + 2y – 5z = –30
(x – 3y + 6z = 21)(–2)
2x – 5y + 2z = –6
–3x + 9y – 18z = –63
3x + 2y – 5z = –30
–2x + 6y – 12z = –42
2x – 5y + 2z = –6
11y – 23z = –93
y – 10z = –48
Step 3
Write the resulting equations in two
variables together as a system of
equations.
Solve the system for the two
remaining variables.
11y – 23z = –93 (–11)
y – 10z = –48
11y – 23z = –93
–11y + 110z = 528
87z = 435
z=5
y – 10(5) = –48
y – 50 = –48
y=2
Step 4
Substitute the value of the variables
from the system of two equations in
one of the ORIGINAL equations with
three variables.
x – 3y + 6z = 21
3x + 2y – 5z = –30
2x – 5y + 2z = –6
I choose the first equation.
x – 3(2) + 6(5) = 21
x – 6 + 30 = 21
x + 24 = 21
x = –3
Step 5
CHECK the solution in ALL 3 of the
original equations.
Write the solution as an ordered
triple.
x – 3y + 6z = 21
3x + 2y – 5z = –30
2x – 5y + 2z = –6
P
–3 – 3(2) + 6(5) = 21
3(–3) + 2(2) – 5(5) = –30
2(–3) – 5(2) + 2(5) = –6
The solution is (–3, 2, 5).
P
P
It is very helpful to neatly organize your
work on your paper in the following manner.
(x, y, z)
Solve the system.
1.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
2(2)+(-4)+z=1
4-4+z=1
z=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
-11y=44
y=- 4
3x+4(-4)=-10
x=2
(2, - 4, 1)
Solve the system.
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
-x+2y+z=3
2x+2y+z=5
4x+4y+2z=6
2.2x+2y+z=5
4x+4y+2z=6
Cancel z’s again.
-4x-4y-2z=-10
4x+4y+2z=6
0=- 4
Doesn’t make sense!
No solution
Solve the system.
1.-2x+4y+z=1
3x-3y-z=2
z’s are easy to cancel!
x+y=3
3x-3y-z=2
5x-y-z=8
Cancel z’s again.
-3x+3y+z=-2
5x-y-z=8
2x+2y=6
2.
-2x+4y+z=1
3x-3y-z=2
5x-y-z=8
3. x+y=3
2x+2y=6
Cancel the x’s.
-2x-2y=-6
2x+2y=6
0=0
This is true.
¸ many solutions
Try this one.
x – 6y – 2z = –8
–x + 5y + 3z = 2
3x – 2y – 4z = 18
(4, 3, –3)
Here’s another one to try.
–5x + 3y + z = –15
10x + 2y + 8z = 18
15x + 5y + 7z = 9
(1, –4, 2)
Application
Courtney has a total of 256 points on three Algebra tests. His
score on the first test exceeds his score on the second by 6
points. His total score before taking the third test was 164
points. What were Courtney’s test scores on the three tests?
Explore
Problems like this one can be solved using a system of equations
in three variables. Solving these systems is very similar to
solving systems of equations in two variables. Try solving the
problem
Let f = Courtney’s score on the first test
Let s = Courtney’s score on the second test
Let t = Courtney’s score on the third test.
Plan
Write the system of equations from the information given.
f + s + t = 256
f–s=6
f + s = 164
The total of the scores is 256.
The difference between the 1st and 2nd is 6
points.
The total before taking the third test is the sum
of the first and second tests..
Solve
Now solve. First use elimination on the last two equations to
solve for f.
f–s=6
f + s = 164
2f = 170
f = 85
The first test score is 85.
Solve
Then substitute 85 for f in one of the original equations to solve
for s.
f + s = 164
85 + s = 164
s = 79
The second test score is 79.
Solve
Next substitute 85 for f and 79 for s in f + s + t = 256.
f + s + t = 256
85 + 79 + t = 256
164 + t = 256
t = 92
The third test score is 92.
Courtney’s test scores were 85, 79, and 92.
Examine
Now check your results against the original problem.
Is the total number of points on the three tests 256
points?
85 + 79 + 92 = 256 ✔
Is one test score 6 more than another test score?
79 + 6 = 85 ✔
Do two of the tests total 164 points?
85 + 79 =164 ✔
Our answers are correct.
Solutions?
You know that a system of two linear equations doesn’t
necessarily have a solution that is a unique ordered pair.
Similarly, a system of three linear equations in three variables
doesn’t always have a solution that is a unique ordered triple.
Graphs
The graph of each equation in a system of three linear equations
in three variables is a plane. Depending on the constraints
involved, one of the following possibilities occurs.
Graphs
1. The three planes intersect
at one point. So the
system has a unique
solution.
2. The three planes intersect in
a line. There are an infinite
number of solutions to the
system.
Graphs
3. Each of the diagrams below shows three planes that
have no points in common. These systems of
equations have no solutions.
Ex. 1: Solve this system of equations
x  2y  z  9
3 y  z  1
3z  12
Solve the third equation, 3z = 12
3z = 12
z=4
Substitute 4 for z in the second
equation 3y – z = -1 to find y.
3y – (4) = -1
3y = 3
y=1
Substitute 4 for z and 1 for y in the
first equation, x + 2y + z = 9 to find x.
x + 2y + z = 9
x + 2(1) + 4 = 9
x+6=9
x = 3 Solution is (3, 1, 4)
Check:
1st 3 + 2(1) +4 = 9 ✔
2nd 3(1) -4 = 1 ✔
3rd 3(4) = 12 ✔
Ex. 2: Solve this system of equations
2x  y  z  3
x  3 y  2 z  11
3x  2 y  4 z  1
Set the first two equations together
and multiply the first times 2.
2(2x – y + z = 3)
4x – 2y +2z = 6
x + 3y -2z = 11
5x + y = 17
Set the next two equations together
and multiply the first times 2.
2(x + 3y – 2z = 11)
2x + 6y – 4z = 22
3x - 2y + 4z = 1
5x + 4y = 23
Next take the two equations that only
have x and y in them and put them
together. Multiply the first times -1 to
change the signs.
Ex. 2: Solve this system of equations
2x  y  z  3
x  3 y  2 z  11
3x  2 y  4 z  1
Next take the two equations that only
have x and y in them and put them
together. Multiply the first times -1 to
change the signs.
-1(5x + y = 17)
-5x - y = -17
5x + 4y = 23
3y = 6
y=2
Now you have y = 2. Substitute y into one
of the equations that only has an x and y
in it.
5x + y = 17
5x + 2 = 17
5x = 15
x=3
Now you have x and y. Substitute values
back into one of the equations that you
started with.
2x – y + z = 3
2(3) - 2 + z = 3
6–2+z=3
4+z=3
z = -1
Ex. 2: Check your work!!!
2x  y  z  3
x  3 y  2 z  11
3x  2 y  4 z  1
Solution is (3, 2, -1)
Check:
1st 2x – y + z =
2(3) – 2 – 1 = 3 ✔
2nd x + 3y – 2z = 11
3 + 3(2) -2(-1) = 11 ✔
3rd 3x – 2y + 4z
3(3) – 2(2) + 4(-1) = 1 ✔
Ex. 2: Solve this system of equations
2x  y  z  3
x  3 y  2 z  11
3x  2 y  4 z  1
Next take the two equations that only
have x and y in them and put them
together. Multiply the first times -1 to
change the signs.
-1(5x + y = 17)
-5x - y = -17
5x + 4y = 23
3y = 6
y=2
Now you have y = 2. Substitute y into one
of the equations that only has an x and y
in it.
5x + y = 17
5x + 2 = 17
5x = 15
x=3
Now you have x and y. Substitute values
back into one of the equations that you
started with.
2x – y + z = 3
2(3) - 2 + z = 3
6–2+z=3
4+z=3
z = -1