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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 1 1 Chapter 4 Systems of Linear Equations and Inequalities

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 2 Systems of Equations and Inequalities 4.1Solving Systems of Linear Equations in Two Variables 4.2Solving Systems of Linear Equations in Three Variables 4.3Solving Applications Using Systems of Equations CHAPTER 4

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 3 Solving Systems of Linear Equations in Three Variables 1.Determine if an ordered triple is a solution for a system of equations. 2.Understand the types of solution sets for systems of three equations. 3.Solve a system of three linear equations using the elimination method. 4.2

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 4 Example Determine whether (2, –1, 3) is a solution of the system Solution In all three equations, replace x with 2, y with –1, and z with 3. x + y + z = 4 2 + (–1) + 3 = 4 4 = 4 TRUE 3 = 3 TRUE 2x – 2y – z = 3 2(2) – 2(–1) – 3 = 3 – 4x + y + 2z = –3 – 4(2) + (–1) + 2(3) = –3 –3 = –3 TRUE Because (2, 1, 3) satisfies all three equations in the system, it is a solution for the system.

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 5 Types of Solution Sets A Single Solution: If the planes intersect at a single point, that ordered triple is the solution to the system.

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 6 Types of Solution Sets Infinite Number of Solutions: If the three planes intersect along a line, the system has an infinite number of solutions, which are the coordinates of any point along that line. Infinite Number of Solutions: If all three graphs are the same plane, the system has an infinite number of solutions, which are the coordinates of any point in the plane.

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 7 Types of Solution Sets No Solution: If all of the planes are parallel, the system has no solution. No Solution: Pairs of planes also can intersect, as shown. However, because all three planes do not have a common intersection, the system has no solution.

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 8 Example Solution Solve the system using elimination. 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

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 9 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)

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 10 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.

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 11 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).

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 12 Solving Systems of Three Linear Equations Using Elimination 1. Write each equation in the form Ax + By+ Cz = D. 2. Eliminate one variable from one pair of equations using the elimination method. 3. If necessary, eliminate the same variable from another pair of equations. 4. Steps 2 and 3 result in two equations with the same two variables. Solve these equations using the elimination method. 5. To find the third variable, substitute the values of the variables found in step 4 into any of the three original equations that contain the third variable. 6. Check the ordered triple in all three of the original equations.

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 13 Example Solution Solve the system using elimination. The equations are in standard form. (1) (2) (3) (2) (3) (4) 3x + 2y = 4 Adding Eliminate z from equations (2) and (3).

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 14 Eliminate z from equations (1) and (2). Multiplying equation (2) by 6 (1) (2) (3) Adding 15x + 15y = 15 Eliminate x from equations (4) and (5). 3x + 2y = 4 15x + 15y = 15 Multiplying top by 5 15x – 10y = 20 15x + 15y = 15 Adding 5y = 5 y = 1 Using y = 1, find x from equation 4 by substituting. 3x + 2y = 4 3x + 2( 1) = 4 x = 2 continued

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 15 continued Substitute x = 2 and y = 1 to find z. x + y + z = 2 2 – 1 + z = 2 1 + z = 2 z = 1 The solution is the ordered triple (2, 1, 1). (1) (2) (3)

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 16 Example Solution Solve the system using elimination. The equations are in standard form. (1) (2) (3) (1) (2) (4) 5y 4z = 0 Adding Eliminate x from equations (1) and (2).

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 17 Eliminate x from equations (1) and (3). (1) (2) (3) 5y + 4z = 2 (5) Eliminate y from equations (4) and (5). 5y 4z = 0 5y + 4z = 2 0 = 2 All variables are eliminated and the resulting equation is false, which means that this system has no solution; it is inconsistent. continued

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