Systems of Linear Equations Chapter 5 Systems of Linear Equations

Systems of Linear Equations in Three Variables 5.2 Systems of Linear Equations in Three Variables

5.2 Systems of Linear Equations in Three Variables Objectives 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. Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables An Equation in Three Variables A solution of an equation in three variables, such as 4x + y – 5z = 1 is called an ordered triple and is written (x, y, z). For example, the ordered triple (3, –6, 1) is a solution to the equation, because 4(3) + (–6) – 5(1) = 12 – 6 – 5 = 1. Verify that another solution of this equation is (–2, 4, –1). In the rest of this chapter, the term linear equation is extended to equa- tions of the form Ax + By + Cz + . . . + Dw = K, where not all the coefficients A, B, C, . . . D equal zero. Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables Possible Solutions 1. The three planes may meet at a single, common point that is the solution of the system. Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables Possible Solutions 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. Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables Possible Solutions 3. The three planes may coincide so that the solution of the system is the set of all points on a plane. Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables Possible Solutions 4. The planes may have no points common to all three so that there is no solution of the system. Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables 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, usually the 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. Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables Solving a Linear System in Three Variables Step 4 Write the equation 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-triple solution in each of the original equations of the system. Then write the solution set. Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables EXAMPLE 1 Solving a System in Three Variables Solve the system. 2x + 3y – z = 5 (1) –3x + 2y – 4z = 2 (2) x – 4y + 3z = –9 (3) 7x + 5y = 6 (4) Eliminate a variable from the sum of two equations. The choice of the variable to eliminate is arbitrary. We will eliminate z. 6x + 9y – 3z = 15 Multiply each side of (1) by 3. x – 4y + 3z = –9 (3) 7x + 5y = 6 Add. (4) Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables EXAMPLE 1 Solving a System in Three Variables Solve the system. 2x + 3y – z = 5 (1) –3x + 2y – 4z = 2 (2) x – 4y + 3z = –9 (3) 7x + 5y = 6 (4) –5x – 10y = –30 (5) Eliminate the same variable, z, from any other two equations. –9x + 6y – 12z = 6 Multiply each side of (2) by 3. 4x – 16y + 12z = –36 Multiply each side of (3) by 4. –5x – 10y = –30 Add. (5) Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables EXAMPLE 1 Solving a System in Three Variables Solve the system. 2x + 3y – z = 5 (1) –3x + 2y – 4z = 2 (2) x – 4y + 3z = –9 (3) 7x + 5y = 6 (4) –5x – 10y = –30 (5) Eliminate a different variable and solve. 14x + 10y = 12 Multiply each side of (4) by 2. –5x – 10y = –30 (5) 9x = –18 Add. (6) x = –2 Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables EXAMPLE 1 Solving a System in Three Variables Solve the system. 2x + 3y – z = 5 (1) –3x + 2y – 4z = 2 (2) x – 4y + 3z = –9 (3) 7x + 5y = 6 (4) –5x – 10y = –30 (5) Find a second value by substituting –2 for x in (4) or (5). 7x + 5y = 6 (4) 7(–2) + 5y = 6 Let x = –2. –14 + 5y = 6 5y = 20 y = 4 Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables EXAMPLE 1 Solving a System in Three Variables Solve the system. 2x + 3y – z = 5 (1) –3x + 2y – 4z = 2 (2) x – 4y + 3z = –9 (3) 7x + 5y = 6 (4) –5x – 10y = –30 (5) Find the value of the remaining variable by substituting –2 for x and 4 for y into any of the three original equations. 2x + 3y – z = 5 (1) 2(–2) + 3(4) – z = 5 Let x = –2 and y = 4. 8 – z = 5 z = 3 Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables EXAMPLE 1 Solving a System in Three Variables Solve the system. 2x + 3y – z = 5 (1) –3x + 2y – 4z = 2 (2) x – 4y + 3z = –9 (3) Check that the ordered triple (–2, 4, 3) is the solution of the system. The ordered triple must satisfy all three original equations. (1) (2) (3) 2x + 3y – z = 5 –3x + 2y – 4z = 2 x – 4y + 3z = –9 2(–2) + 3(4) – 3 = 5 –3(–2) + 2(4) – 4(3) = 2 (–2) – 4(4) + 3(3) = –9 –4 + 12 – 3 = 5 6 + 8 – 12 = 2 –2 – 16 + 9 = –9 5 = 5 2 = 2 –9 = –9 True True True Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables EXAMPLE 1 Solving a System in Three Variables Solve the system. 2x + 3y – z = 5 (1) –3x + 2y – 4z = 2 (2) x – 4y + 3z = –9 (3) Because (–2, 4, 3) satisfies all three equations of the system, the solution set is { (–2, 4, 3) }. Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables EXAMPLE 2 Solving a System of Equations with Missing Terms Solve the system. 5x + 4y = 23 (1) –5y – 2z = 3 (2) –8x + 9z = –2 (3) 25x – 8z = 127 (4) Since equation (3) is missing the variable y, a good way to begin the solution is to eliminate y again by using equations (1) and (2). 25x + 20y = 115 Multiply each side of (1) by 5. –20y – 8z = 12 Multiply each side of (2) by 4. 25x – 8z = 127 Add. (4) Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables EXAMPLE 2 Solving a System of Equations with Missing Terms Solve the system. 5x + 4y = 23 (1) –5y – 2z = 3 (2) –8x + 9z = –2 (3) 25x – 8z = 127 (4) Now use equations (3) and (4) to eliminate z and solve. –64x + 72z = –16 Multiply each side of (3) by 8. 225x – 72z = 1143 Multiply each side of (4) by 9. 161x = 1127 Add. (5) x = 7 Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables EXAMPLE 2 Solving a System of Equations with Missing Terms Solve the system. 5x + 4y = 23 (1) –5y – 2z = 3 (2) –8x + 9z = –2 (3) 25x – 8z = 127 (4) Substituting into equation (1) gives Substituting into equation (2) gives 5x + 4y = 23 –5y – 2z = 3 5(7) + 4y = 23 –5(–3) – 2z = 3 35 + 4y = 23 15 – 2z = 3 4y = –12 –2z = –12 y = –3. z = 6. Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables EXAMPLE 2 Solving a System of Equations with Missing Terms Solve the system. 5x + 4y = 23 (1) –5y – 2z = 3 (2) –8x + 9z = –2 (3) Thus, x = 7, y = –3, and z = 6. Check these values in each of the original equations of the system to verify that the solution set of the system is { (7, –3, 6) }. Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables EXAMPLE 3 Solving an Inconsistent System with Three Variables Solve the system. 3x – 2y + z = 8 (1) 4x – y – z = –2 (2) –6x + 4y – 2z = 3 (3) 7x – 3y = 6 (4) Eliminate z by adding equations (1) and (2). 3x – 2y + z = 8 (1) 4x – y – z = –2 (2) 7x – 3y = 6 Add. (4) Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables EXAMPLE 3 Solving an Inconsistent System with Three Variables Solve the system. 3x – 2y + z = 8 (1) 4x – y – z = –2 (2) –6x + 4y – 2z = 3 (3) 7x – 3y = 6 (4) Now, eliminate z again using equations (1) and (3). The resulting false statement indicates that equations (1) and (3) have no common solution. Thus, the system is inconsistent and the solution set is Ø. The graph of this system would show these two planes parallel to one another. 6x – 4y + 2z = 16 Multiply each side of (1) by 2. –6x + 4y – 2z = 3 (3) 0 = 19 False Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables Parallel Planes NOTE If a false statement results when adding as in Example 3, it is not necessary to go any further with the solution. Since two of the three planes are parallel, it is not possible for the three planes to have any common points. Copyright © 2010 Pearson Education, Inc. All rights reserved.

5.2 Systems of Linear Equations in Three Variables EXAMPLE 4 Solving a System of Dependent Equations with Three Variables Solve the system. 3x – 4y + z = 12 (1) x – y + z = 3 (2) 3 4 1 6x – 8y + 2z = 24 (3) Multiplying each side of equation (1) by 2 gives equation (3). Multiplying each side of equation (2) by 8 also gives equation (3). Because of this, the equations are dependent. All three equations have the same graph. The solution is written { (x, y, z) | 3x – 4y + z = 12 }. Although any one of the three equations could be used to write the solution set, we use the equation with coefficients that are integers with no common factor (except 1), as we did in Section 5.1. Copyright © 2010 Pearson Education, Inc. All rights reserved.