1. Solving Systems of Linear Equations in Three Variables
§ 4.4
Solving Systems of Linear Equations in Three Variables

2. Systems in Three Variables
An equation with 2 variables raised to the power of 1 represents a line.
An equation with 3 variables raised to the power of 1 represents a plane.
A system of 3 equations in 3 variables would be represented as 3 planes.

3. Systems in Three Variables
There are 4 ways in which 3 planes could intersect.
No common intersection
Intersection in a single point
Intersection in a line
Intersection in a plane (all three are identical planes)

4. No Common Intersection
Each of these systems are inconsistent.

5. Intersection in a Point

6. Intersection in a Line

7. Intersection in a Plane
This system is both consistent AND dependent.

8. Methods for Solving a System
There are four methods that you have previously used to solve systems of equations.
Trial and error (Requires even greater luck with 3 variables)
Graphing the equations
Substitution method
Elimination method
The first two methods can be used, but are extremely time-consuming or difficult with three variables.

9. Solving a System Using Substitution
Example:
Solve the following system by the substitution method.
4x + y – z = 8
x – y + 2z = 3
3x – y + z = 6
Solve the second equation for x.
x = y – 2z + 3
Substitute this value into both of the other equations.
4(y – 2z + 3) + y – z = 8
3(y – 2z + 3) – y + z = 6
Continued.

10. Solving a System Using Substitution
Example continued:
By simplifying the equations, we get
4(y – 2z + 3) + y – z = 8
4y – 8z y – z = 8
5y – 9z = – 4
3(y – 2z + 3) – y + z = 6
3y – 6z + 9 – y + z = 6
2y – 5z = – 3
Continued.

11. Solving a System Using Substitution
Example continued:
5y – 9z = – 4
2y – 5z = – 3
There is now a system of 2 equations in 2 variables.
Since using the substitution method to solve this new system would produce fractional coefficients, we’ll now use elimination here.
Continued.

12. Solving a System Using Substitution
Example continued:
Multiply the first equation by 2 and the second equation by – 5.
2(5y – 9z = – 4) or 10y – 18z = – 8
– 5(2y – 5z = – 3) or –10y + 25z = 15
Now combine the two equations to eliminate the y variable.
7z = 7
z = 1
Continued.

13. Solving a System Using Substitution
Example continued:
Now substitute this into one of the two equations with only 2 variables.
10y – 18(1) = – 8
10y = – = 10
y = 1
Now substitute both z and y into one of the original equations.
4x + 1 – 1 = 8
4x = 8
x = 2
Continued.

14. Solving a System Using Substitution
Example continued:
So our tentative solution is (2, 1, 1). Substitute these values into each of the original equations to check them.
4(2) + 1 – 1 = 8 true
2 – 1 + 2(1) = 3 true
3(2) – = 6 true
So the solution is (2, 1, 1).

15. Solving a System Using Addition
To use the addition method, you pick two of the equations and multiply them by numbers so that one of the variables will be eliminated.
Then you pick 2 other equations and eliminate the same variable.
Now you have 2 equations with the same 2 variables. You can use the techniques from the previous section.

16. Solving a System Using Addition
Example:
Solve the following system by the addition method.
2x + 2y + z = 1
– x + y + 2z = 3
x + 2y + 4z = 0
Leave the first equation alone, and multiply the second equation by 2, since combining these two equations will eliminate the variable x.
– 2x + 2y + 4z = 6
Continued.

17. Solving a System Using Addition
Example continued:
Combine the two equations together.
4y + 5z = 7
Now combine the 2nd and 3rd equations together to eliminate x (no need to multiply by any number).
3y + 6z = 3
We’ll now work toward combining these last two equations in a fashion that will eliminate one of the two variables.
Continued.

18. Solving a System Using Addition
Example continued:
Multiply the first equation by 3 and the second equation by – 4 to combine them.
12y + 15z = 21
– 12y – 24z = – 12
– 9z = 9
z = – 1
Continued.

19. Solving a System Using Addition
Example continued:
Substitute this variable into one of the two equations with only two variables.
4y + 5(– 1) = 7
4y – 5 = 7
4y = 12
y = 3
Continued.

20. Solving a System Using Addition
Example continued:
Now substitute both of the values into one of the original equations.
2x + 2(3) + (– 1) = 1
2x + 6 – 1 = 1
2x = – 4
x = – 2
Continued.

21. Solving a System Using Addition
Example continued:
So our tentative solution is (– 2, 3, – 1).
Substitute the values into the original equations to check them.
2(– 2) + 2(3) + (– 1) = 1 true
– (– 2) (– 1) = 3 true
– 2 + 2(3) + 4(– 1) = 0 true
So the solution is (– 2, 3, – 1).

22. Solving a System Using Addition
Example:
Solve the following system of equations using the addition method.
– 6x + 12y z = – 6
2x – 4y – z = 2
– x + 2y + 0.5z = – 1
Multiply the second equation by 3 and combine it with the first equation.
– 6x + 12y + 3z = – 6
6x – 12y – 3z = 6
0 = 0
Continued.

23. Solving a System Using Addition
Example continued:
Since this last equation will always be true, for any values substituted for x, y, and z, the first two planes are actually the same plane.
We still don’t know how they are related to the third plane.
Multiply the third equation by 2 and combine it with the second equation.
2x – 4y – z = 2
– 2x + 4y + z = – 2
0 = 0
Continued.

24. Solving a System Using Addition
Example continued:
So the third plane is coincident also.
Therefore, our solution is a plane. We can use any of the equations to describe how the set of points would look.
For example, our solution could be described as the set of all points such that – 6x + 12y + 3z = – 6
Continued.