Section 2Chapter 4. 1 Copyright © 2012, 2008, 2004 Pearson Education, Inc. Objectives 2 3 4 Systems of Linear Equations in Three Variables Understand.

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Presentation transcript:

Section 2Chapter 4

1 Copyright © 2012, 2008, 2004 Pearson Education, Inc. Objectives Systems of Linear Equations in Three Variables Understand the geometry of systems of three equations in three variables. Solve linear systems (with three equations and three variables) by elimination. Solve linear systems (with three equations and three variables) in which some of the equations have missing terms. Solve special systems. 4.2

Copyright © 2012, 2008, 2004 Pearson Education, Inc. A solution of an equation in three variables, such as 2x + 3y – z = 4 is called an ordered triple and is written (x, y, z). Slide Systems of Linear Equations in Three Variables

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Understand the geometry of systems of three equations in three variables. Objective 1 Slide

Copyright © 2012, 2008, 2004 Pearson Education, Inc. The graph of a linear equation with three variables is a plane, not a line. A number of possible solutions are shown below. Slide Understand the geometry of system of three equations in three variables.

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Graphs of Linear Systems in Three Variables 1. The three planes may meet at a single, common point that is the solution of the system. (See Figure 7a). 2. The three planes may have the points of a line in common, so that the infinite set of points that satisfy the equation of the line is the solution of the system. (See Figure 7b). 3. The three planes may coincide, so that the solution of the system is the set of all points on a plane. (See Figure 7c). 4. The planes may have no points common to all three, so that there is no solution of the system. (See Figures 7d-g). Slide Understand the geometry of system of three equations in three variables.

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve linear systems (with three equations and three variables) by elimination. Objective 2 Slide

Copyright © 2012, 2008, 2004 Pearson Education, Inc. In the steps that follow, we use the term focus variable to identify the first variable to be eliminated in the process. The focus variable will always be present in the working equation, which will be used twice to eliminate this variable. Slide Sole linear systems (with three equations and three variables) by elimination.

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solving a Linear System in Three Variables Step 1 Select a variable and an equation. A good choice for the variable, which we call the focus variable, is one that has coefficient 1 or −1. Then select an equation, one that contains the focus variable, as the working equation. Step 2 Eliminate the focus variable. Use the working equation and one of the other two equations of the original system. The result is an equation in two variables. Step 3 Eliminate the focus variable again. Use the working equation and the remaining equation of the original system. The result is another equation in two variables. Slide Solve linear systems (with three equations and three variables) by elimination.

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solving a Linear System in Three Variables (cont’d) Step 4 Write the equations in two variables that result from Steps 2 and 3 as a system, and solve it. Doing this gives the values of two of the variables. Step 5 Find the value of the remaining variable. Substitute the values of the two variables found in Step 4 into the working equation to obtain the value of the focus variable. Step 6 Check the ordered-pair solution in each of the original equations of the system. Then write the solution set. Slide Solve linear systems (with three equations and three variables) by elimination.

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve the system. Step 1 Select the variable y as the focus variable and equation (1) as the working equation. Step 2 Multiplying can be skipped as the focus variable can be eliminated when adding equations (1) and (2) Slide CLASSROOM EXAMPLE 1 Solving a System in Three Variables Solution:

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Step 3 Use the working equation (1), multiply both sides by −2 and add to equation (3) to again eliminate focus variable y. Make sure equation (5) has the same variables as equation 4. Slide CLASSROOM EXAMPLE 1 Solving a System in Three Variables (cont’d)

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Step 4 Write the equations in two variables that result in Steps 2 and 3 as a system, then solve to eliminate z. Substitute the value of x into equation (4) to sole for z. Step 5 Substitute  1 for x and 2 for z in equation (1) to find y. CLASSROOM EXAMPLE 1 Solving a System in Three Variables (cont’d) Slide

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Step 6 Check the ordered triple (  1, 1, 2) to be sure the solution satisfies all three equations in the original system. The solution set is {(  1, 1, 2)}. Write the values of x, y, and z in the correct order. Slide CLASSROOM EXAMPLE 1 Solving a System in Three Variables (cont’d)

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve linear systems (with three equations and three variables) in which some of the equations have missing terms. Objective 3 Slide

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve the system. Since equation (3) is missing y, eliminate y again from equations (1) and (2). Multiply equation (1) by 2 and add the result to equation (2). Slide CLASSROOM EXAMPLE 2 Solving a System of Equations with Missing Terms Solution:

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Use equation (4) together with equation (3) to eliminate x. Multiply equation (4) by 3 and equation (3) by  2. Then add the results. Slide CLASSROOM EXAMPLE 2 Solving a System of Equations with Missing Terms (cont’d)

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Substitute 1 for z in equation (2) to find y. Substitute  2 for y in (1) to find x. Check (4,  2, 1) in each of the original equations to verify that it is the solution set. Slide CLASSROOM EXAMPLE 2 Solving a System of Equations with Missing Terms (cont’d)

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve special systems. Objective 4 Slide

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve the system. Multiply equation (1) by 2 and add the result to equation (3). Since a false statement results, the system is inconsistent. The solution set is . Slide CLASSROOM EXAMPLE 3 Solving an Inconsistent System with Three Variables Solution:

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve the system. Since equation (2) is  3 times equation (1) and equation (3) is 2 times equation (1), the three equations are dependent. All three have the same graph. The solution set is {(x, y, z) | x – y + z = 4}. Slide CLASSROOM EXAMPLE 4 Solving a System of Dependent Equations with Three Variables Solution:

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve the system. Eliminate the fractions in equations (2) and (3). Multiply equation (2) by 4. Multiply equation (3) by 2. Slide CLASSROOM EXAMPLE 5 Solving Another Special System Solution:

Copyright © 2012, 2008, 2004 Pearson Education, Inc. Equations (1) and (4) are dependent (they have the same graph). Equations (1) and (5) are not equivalent. Since they have the same coefficients but different constant terms, their graphs have no points in common (the planes are parallel). Thus the system is inconsistent and the solution set is . Slide CLASSROOM EXAMPLE 5 Solving Another Special System (cont’d)