1. Solving Systems of Three Linear Equations in Three Variables
The Elimination Method

2. 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.

3. Is (–3, 2, 4) a solution of this system?
3x + 2y + 4z = 11
2x – y + 3z = 4
5x – 3y + 5z = –1
3(–3) + 2(2) + 4(4) = 11
2(–3) – 2 + 3(4) = 4
5(–3) – 3(2) + 5(4) = –1 P P P
Yes, it is a solution to the system because it is a solution to all 3 equations.

4. This lesson will focus on the
Elimination Method.

5. Use elimination to solve the following system of equations.
x – 3y + 6z = 21
3x + 2y – 5z = –30
2x – 5y + 2z = –6

6. Step 1
Rewrite the system as two smaller systems, each containing two of the three equations.

7. x – 3y + 6z = 21
3x + 2y – 5z = –30
2x – 5y + 2z = –6
x – 3y + 6z = 21 x – 3y + 6z = 21
3x + 2y – 5z = –30 2x – 5y + 2z = –6

8. 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.

9. (x – 3y + 6z = 21)
3x + 2y – 5z = –30
–3x + 9y – 18z = –63
11y – 23z = –93
(x – 3y + 6z = 21)
2x – 5y + 2z = –6
–2x + 6y – 12z = –42
y – 10z = –48
(–3)
(–2)

10. Step 3
Write the resulting equations in two variables together as a system of equations.
Solve the system for the two remaining variables.

11. 11y – 23z = –93 (–11) y – 10z = –48 –11y + 110z = 528 87z = 435 z = 5

12. Step 4
Substitute the value of the variables from the system of two equations in one of the ORIGINAL equations with three variables.

13. x – 3y + 6z = 21
3x + 2y – 5z = –30
2x – 5y + 2z = –6
I choose the first equation.
x – 3(2) + 6(5) = 21
x – = 21
x + 24 = 21
x = –3

14. Step 5
CHECK the solution in ALL 3 of the original equations.
Write the solution as an ordered triple.

15. P P P The solution is (–3, 2, 5). x – 3y + 6z = 21
–3 – 3(2) + 6(5) = 21
3(–3) + 2(2) – 5(5) = –30
2(–3) – 5(2) + 2(5) = –6 P P
The solution is (–3, 2, 5).

16. It is very helpful to neatly organize your
work on your paper in the following manner.
(x, y, z)

17. Solve the system. x+3y-z=-11 2x+y+z=1 5x-2y+3z=21 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
-x-5y=18
2(2)+(-4)+z=1
4-4+z=1
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)
z=1

18. Solve the system. -x+2y+z=3 2x+2y+z=5 4x+4y+2z=6 2x+2y+z=5 -x+2y+z=3
Cancel z’s again.
-4x-4y-2z=-10
0=- 4
Doesn’t make sense!
No solution
-x+2y+z=3
2x+2y+z=5
z’s are easy to cancel!
-2x-2y-z=-5
-3x=-2
x=2/3

19. Solve the system. -2x+4y+z=1 3x-3y-z=2 5x-y-z=8 -2x+4y+z=1 3x-3y-z=2
z’s are easy to cancel!
x+y=3
5x-y-z=8
Cancel z’s again.
-3x+3y+z=-2
2x+2y=6
3. x+y=3
2x+2y=6
Cancel the x’s.
-2x-2y=-6
0=0
This is true.
¸ many solutions

20. (4, 3, –3) Try this one. x – 6y – 2z = –8 –x + 5y + 3z = 2

21. (1, –4, 2) Here’s another one to try. –5x + 3y + z = –15

22. 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?

23. 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.

24. 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..

25. 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.

26. 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.

27. Solve Next substitute 85 for f and 79 for s in f + s + t = 256.
The third test score is 92.
Courtney’s test scores were 85, 79, and 92.

28. Examine Now check your results against the original problem.
Is the total number of points on the three tests 256 points?
= 256 ✔
Is one test score 6 more than another test score?
= 85 ✔
Do two of the tests total 164 points?
=164 ✔
Our answers are correct.

29. 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.

30. 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.

31. Graphs
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.

32. Graphs
3. Each of the diagrams below shows three planes that have no points in common. These systems of equations have no solutions.

33. Ex. 1: Solve this system of equations
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 ✔
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

34. Ex. 2: Solve this system of equations
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.
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

35. Ex. 2: Solve this system of equations
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) z = 3
6 – 2 + z = 3
4 + z = 3
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

36. Ex. 2: Check your work!!! 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 ✔

37. Ex. 2: Solve this system of equations
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) z = 3
6 – 2 + z = 3
4 + z = 3
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