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2. 4.2 Solving Systems of Linear Equations in Three Variables

3. 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 of 3 variables would be represented as 3 planes. Systems in Three Variables

4. 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) Systems in Three Variables

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

6. Intersection in a Point

7. Intersection in a Line

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

9. 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. Methods for Solving a System

10. 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 Example Continued

11. By simplifying the equations, we get 4y – 8z + 12 + y – z = 5y – 9z + 12 = 8 or 5y – 9z = – 4 3y – 6z + 9 – y + z = 2y – 5z + 9 = 6 or 2y – 5z = – 3 There is now a system of 2 equations of 2 variables. Since using substitution method to solve this new system would produce fraction coefficients, we’ll now use elimination here. Example (cont) Continued

12. 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 Example (cont) Continued

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

14. 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) – 1 + 1 = 6 true So the solution is (2, 1, 1). Example (cont)

15. To use the elimination 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. Solving a System Using Elimination

16. Solve the following system by the elimination 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 + z = 1 – 2x + 2y + 4z = 6 Example Continued

17. Combine the two equations together. 4y + 5z = 7 Now combine the 2 nd and 3 rd equations together (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. Example (cont) Continued

18. 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 Example (cont) Continued

19. Substitute this variable into one of the two equations with only two variables. 4y + 5(– 1) = 7 4y – 5 = 7 4y = 12 y = 3 Example (cont) Continued

20. 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 Example (cont) Continued

21. 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) + 3 + 2(–1) = 3 true –2 + 2(3) + 4(–1) = 0 true So the solution is (–2, 3, –1). Example (cont)

22. Solve the following system of equations using the elimination method. – 6x + 12y + 3z = – 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 Example (cont) Continued

23. 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 Example (cont) Continued

24. 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 Example (cont) Continued