1. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Systems of Equations in Three Variables Identifying Solutions Solving Systems in Three Variables Dependency, Inconsistency, and Geometric Considerations 3.4

2. Slide 3- 2 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Identifying Solutions A solution of a system of three equations in three variables is an ordered triple (x, y, z) that makes all three equations true. A linear equation in three variables is an equation equivalent to one in the form Ax + By + Cz = D, where A, B, C, and D are real numbers. We refer to the form Ax + By + Cz = D as standard form for a linear equation in three variables.

3. Slide 3- 3 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Determine whether (2, –1, 3) is a solution of the system

4. Slide 3- 4 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution We substitute (2, –1, 3) into the three equations, using alphabetical order: x + y + z = 4 2 + (–1) + 3 4 4 = 43 = 3 TRUE The triple makes all three equations true, so it is a solution of the system. 2x – 2y – z = 3 2(2) – 2(–1) – 3 3 – 4x + y + 2z = –3 – 4(2) + (–1) + 2(3) –3 –3 = –3 TRUE

5. Slide 3- 5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems in Three Variables The elimination method allows us to manipulate a system of three equations in three variables so that a simpler system of two equations in two variables is formed. Once that simpler system has been solved, we can substitute into one of the three original equations and solve for the third variable.

6. Slide 3- 6 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solution Solve the following system of equations: We select any two of the three equations and work to get one equation in two variables. Let’s add equations (1) and (2): (1) (2) (3) (1) (2) (4) 2x + 3y = 8 Adding to eliminate z

7. Slide 3- 7 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Next, we select a different pair of equations and eliminate the same variable. Let’s use (2) and (3) to again eliminate z. (5) x – y + 3z = 8 Multiplying equation (2) by 3 4x + 5y = 14. 3x + 6y – 3z = 6 Now we solve the resulting system of equations (4) and (5). That will give us two of the numbers in the solution of the original system, 4x + 5y = 14 2x + 3y = 8 (5) (4)

8. Slide 3- 8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley We multiply both sides of equation (4) by –2 and then add to equation (5): Substituting into either equation (4) or (5) we find that x = 1. 4x + 5y = 14 –4x – 6y = –16, –y = –2 y = 2 Now we have x = 1 and y = 2. To find the value for z, we use any of the three original equations and substitute to find the third number z.

9. Slide 3- 9 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Let’s use equation (1) and substitute our two numbers in it: We have obtained the ordered triple (1, 2, 3). It should check in all three equations. x + y + z = 6 1 + 2 + z = 6 z = 3. The solution is (1, 2, 3).

10. Slide 3- 10 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Three Linear Equations To use the elimination method to solve systems of three linear equations: 1. Write all the equations in standard form Ax + By+ Cz = D. 2. Clear any decimals or fractions. 3.Choose a variable to eliminate. Then select two of the three equations and work to get one equation in which the selected variable is eliminated.

11. Slide 3- 11 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Three Linear Equations (continued) 4. Next, use a different pair of equations and eliminate the same variable that you did in step (3). 5. Solve the system of equations that resulted from steps (3) and (4). 6.Substitute the solution from step (5) into one of the original three equations and solve for the third variable. Then check.

12. Slide 3- 12 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Dependency, Inconsistency, and Geometric Considerations The graph of a linear equation in three variables is a plane. Solutions are points common to the planes of each system. Since three planes can have an infinite number of points in common or no points at all in common, we need to generalize the concept of consistency in three dimensions.

13. Slide 3- 13 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

14. Slide 3- 14 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Consistency A system of equations that has at least one solution is said to be consistent. A system of equations that has no solution is said to be inconsistent.

15. Slide 3- 15 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solution Solve the following system of equations: The variable x is missing in equation (1). Let’s add equations (2) and (3) to get another equation with x missing: (1) (2) (3) (2) (3) (4) y + 2z = 6 Adding

16. Slide 3- 16 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equations (1) and (4) form a system in y and z. Solve as before: Multiplying equation (1) by –1 0 = 4. Since we ended up with a false equation, or contradiction, we know that the system has no solution. It is inconsistent. y + 2z = 6 y + 2z = 2–y – 2z = –2 y + 2z = 6

17. Slide 3- 17 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Recall that when dependent equations appeared in Section 3.1, the solution sets were always infinite in size and were written in set-builder notation. There, all systems of dependent equations were consistent. This is not always the case for systems of three or more equations. The following figures illustrate some possibilities geometrically.

18. Slide 3- 18 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley