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2. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Linear Equations in Three Variables 4 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.

3. Slide 9- 3 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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.

4. Slide 9- 4 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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

5. Slide 9- 5 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)

6. Slide 9- 6 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.

7. Slide 9- 7 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).

8. Slide 9- 8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Three 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.

9. Slide 9- 9 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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).

10. Slide 9- 10 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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

11. Slide 9- 11 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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)

12. Slide 9- 12 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example At a movie theatre, Kara buys one popcorn, two drinks and 2 candy bars, all for $12. Rebecca buys two popcorns, three drinks, and one candy bar for $17. Leah buys one popcorn, one drink and three candy bars for $11. Find the individual cost of one popcorn, one drink and one candy bar. Understand We have three unknowns and three relationships, and we are to find the cost of each. Plan Select a variable for each unknown, translate the relationship to a system of three equations, and then solve the system.

13. Slide 9- 13 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Execute: p = popcorn, d = drink, and c = candy Relationship 1: one popcorn, two drinks and two candy bars, cost $12 Translation: p + 2d + 2c = 12 Relationship 2: two popcorns, three drinks, and one candy bar cost $17 Translation: 2p + 3d + c = 17 Relationship 3: one popcorn, one drink and three candy bars cost $11 Translation: p + d + 3c = 11

14. Slide 9- 14 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Our system: Choose to eliminate p: Start with equations 1 and 3. Choose to eliminate p: Start with equations 1 and 2.

15. Slide 9- 15 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Use equations 4 and 5 to eliminate d. Substitute for c in d – c = 1 Substitute for c and d in p + d + 3c = 11 p + 2.5 + 3(1.5) = 11 p + 7 = 11 p = 4 The cost of one candy bar is $1.50. The cost of one drink is $2.50. The cost of one popcorn is $4.00.

16. Slide 9- 16 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determine if (2, –5, 3) is a solution to the given system. a) Yes b) No

17. Slide 9- 17 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determine if (2, –5, 3) is a solution to the given system. a) Yes b) No

18. Slide 9- 18 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (–2, 2, –5) b) (–5, 2, –2) c) (–2,  5, 2) d) no solution

19. Slide 9- 19 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (–2, 2, –5) b) (–5, 2, –2) c) (–2,  5, 2) d) no solution

20. Slide 9- 20 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A company makes 3 types of cable. Cable A requires 3 black, 3 white, and 2 red wires. B requires 1 black, 2 white, and 1 red. C requires 2 black, 1 white, and 2 red. They used 100 black, 110 white and 90 red wires. How many of each cable were made? a) 10 cable A, 30 cable B, 20 cable C b) 20 cable A, 30 cable B, 10 cable C c) 10 cable A, 103 cable B, 20 cable C d) 10 cable A, 30 cable B, 93 cable C

21. Slide 9- 21 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A company makes 3 types of cable. Cable A requires 3 black, 3 white, and 2 red wires. B requires 1 black, 2 white, and 1 red. C requires 2 black, 1 white, and 2 red. They used 100 black, 110 white and 90 red wires. How many of each cable were made? a) 10 cable A, 30 cable B, 20 cable C b) 20 cable A, 30 cable B, 10 cable C c) 10 cable A, 103 cable B, 20 cable C d) 10 cable A, 30 cable B, 93 cable C

22. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Linear Equations Using Matrices 9.5 1.Write a system of equations as an augmented matrix. 2.Solve a system of linear equations by transforming its augmented matrix to echelon form.

23. Slide 9- 23 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Matrix: A rectangular arrow of numbers. The following are examples of matrices: The individual numbers are called elements.

24. Slide 9- 24 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The rows of a matrix are horizontal, and the columns are vertical. column 1column 2column 3 row 1 row 2 row 3

25. Slide 9- 25 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Augmented Matrix: A matrix made up of the coefficients and the constant terms of a system. The constant terms are separated from the coefficients by a dashed vertical line. Let’s write this system as an augmented matrix:

26. Slide 9- 26 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Row Operations The solution of a system is not affected by the following row operations in its augmented matrix: 1. Any two rows may be interchanged. 2. The elements of any row may be multiplied (or divided) by any nonzero real number. 3. Any row may be replaced by a row resulting from adding the elements of that row (or multiples of that row) to a multiple of the elements of any other row.

27. Slide 9- 27 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Echelon form: An augmented matrix whose coefficient portion has 1s on the diagonal from upper left to lower right and 0s below the 1s.

28. Slide 9- 28 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Example Solve the following linear system by transforming its augmented matrix into echelon form. We write the augmented matrix. We perform row operations to transform the matrix into echelon form. 3R 2 + R 1

29. Slide 9- 29 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The resulting matrix represents the system: R 2  10 continued Since y = 1, we can solve for x using substitution. The solution is (2, 1). 3x + 1 = 7 3x + y = 7 3x = 6 x = 2

30. Slide 9- 30 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solution Write the augmented matrix. Use the echelon method to solve

31. Slide 9- 31 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Our goal is to transform into the form Interchange Row 1 and Row 2 –2R 1 + R 2 continued

32. Slide 9- 32 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley –R 2 + R 3 (1/3)R 2 R 1 + R 3 (1/3)R 3 continued

33. Slide 9- 33 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley z = –1. Substitute z into y – z = 2 y –(  1) = 2 y + 1 = 2 y = 1 continued Substitute y and z into x – y + z = 1 x – 1 – 1 = 1 x – 2 = 1 x = 3 The solution is (3, 1, –1).

34. Slide 9- 34 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Replace R 2 in with R 1 + R 2. a)b) c)d)

35. Slide 9- 35 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Replace R 2 in with R 1 + R 2. a)b) c)d)

36. Slide 9- 36 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve by transforming the augmented matrix into echelon form. a) (2, 1,  3) b) (  3, 1, 2) c) (  3, 2, 1) d) No solution

37. Slide 9- 37 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve by transforming the augmented matrix into echelon form. a) (2, 1,  3) b) (  3, 1, 2) c) (  3, 2, 1) d) No solution