1. College Algebra Sixth Edition James Stewart Lothar Redlin Saleem Watson

2. Systems of Equations and Inequalities 5

3. Systems of Linear Equations in Several Variables 5.2

4. Linear Equation A linear equation in n variables is an equation that can be put in the form a 1 x 1 + a 2 x 2 + · · · + a n x n = c where: a 1, a 2, · · ·, a n and c are real numbers. x 1, x 2, · · ·, x n are the variables.

5. Linear Equations If we have only three or four variables, we generally use x, y, z, and w instead of x 1, x 2, x 3, and x 4

6. Linear Equations Such equations are called linear because: If we have just two variables, the equation is a 1 x + a 2 y = c which is the equation of a line.

7. Linear Equations Here are some examples of equations in three variables that illustrate the difference between linear and nonlinear equations.

8. Linear Equations vs. Nonlinear Equations LinearNonlinear Reason for Nonlinearity Contains the square and the square root of a variable Contains a product of variables

9. Solving a Linear System

10. Systems of Linear Equations The following are two examples of systems of linear equations in three variables. System of Linear Equations System in Triangular Form

11. Systems of Linear Equations in Triangular Form The second system is in triangular form. That is, the variable x doesn’t appear in the second equation and the variables x and y do not appear in the third equation. System of Linear Equations System in Triangular Form

12. Solving a Linear System It’s easy to solve a system that is in triangular form using back-substitution. So, our goal in this section is: To start with a system of linear equations and change it to a system in triangular form that has the same solutions as the original system.

13. Solving a Linear System We begin by showing how to use back-substitution to solve a system that is already in triangular form.

14. E.g. 1—Solving a Triangular System Using Back-Substitution Solve the system using back- substitution:

15. From the last equation, we know that z = 3. We back-substitute this into the second equation and solve for y. y + 2(3) = 5 y = –1 E.g. 1—Solving a Triangular System Using Back-Substitution

16. Then, we back-substitute y = –1 and z = 3 into the first equation and solve for x. x – 2(–1) – (3) = 1 x = 2 The solution is: x = 2, y = –1, z = 3 We can also write the solution as the ordered triple (2, –1, 3). E.g. 1—Solving a Triangular System Using Back-Substitution

17. Changing to an Equivalent System To change a system of linear equations to an equivalent system—a system with the same solutions as the original system— we use the elimination method. This means we can use the following operations.

18. Operations that lead to an Equivalent System 1. Add a nonzero multiple of one equation to another. 2. Multiply an equation by a nonzero constant. 3. Interchange the positions of two equations.

19. Gaussian Elimination To solve a linear system, we use these operations to change the system to an equivalent triangular system. Then, we use back-substitution as in Example 1. This process is called Gaussian elimination.

20. E.g. 2—Solving a System of Three Equations in Three Variables Solve the system using Gaussian elimination. We need to change this to a triangular system.

21. E.g. 2—Solving a System of Three Equations in Three Variables So, we begin by eliminating the x-term from the second equation. Equation 2 + (–1)  Equation 1 = New Equation 2

22. This gives us a new, equivalent system that is one step closer to triangular form: E.g. 2—Solving a System of Three Equations in Three Variables

23. Now, we eliminate the x-term from the third equation. Equation 3 + (–3)  Equation 1 = New Equation 3 E.g. 2—Solving a System of Three Equations in Three Variables

24. Then, we eliminate the y-term from the third equation. Equation 3 + (–2)  Equation 2 = New Equation 3 E.g. 2—Solving a System of Three Equations in Three Variables

25. The system is now in triangular form. However, it will be easier to work with if we divide the second and third equations by the common factors of each term. E.g. 2—Solving a System of Three Equations in Three Variables

26. Now, we use back-substitution to solve the system. From the third equation, we get z = 4. We back-substitute this into the second equation and solve for y. y – (4) = 3 y = 7 E.g. 2—Solving a System of Three Equations in Three Variables

27. Then, we back-substitute y = 7 and z = 4 into the first equation and solve for x. x – 2(7) + 3(4) = 1 x = 3 The solution of the system is x = 3, y = 7, z = 4. We can write as the ordered triple (3, 7, 4). E.g. 2—Solving a System of Three Equations in Three Variables

28. Check Your Answer We must check that the answer satisfies all three equations, x = 3, y = 7, z = 4: (3) – 2(7) + 3(4) = 1 (3) + 2(7) – (4) = 13 3(3) + 2(7) – 5(4) = 3

29. The Number of Solutions of a Linear System

30. Number of Solutions of a Linear System For a system of linear equations, exactly one of the following is true. The system has exactly one solution. The system has no solution. The system has infinitely many solutions.

31. The Number of Solutions of a Linear System The graph of a linear equation in three variables is a plane in three-dimensional space. A system of three equations in three variables represents three planes in space.

32. Intersection of Three Planes The solutions of the system are the points where all three planes intersect. Three planes may intersect in a point, a line, not at all, or all three planes may coincide. For a system of three equations in three variables, the following situations arise.

33. Situation 1 The three planes intersect in a single point. The system has a one solution.

34. Situation 2 The three planes intersect at more than one point. The system has infinitely many solutions.

35. Situation 3 The three planes have no point in common. The system has no solution.

36. Inconsistent and Dependent Systems A system with no solutions is said to be inconsistent. A system with infinitely many solutions is said to be dependent.

37. Number of Solutions of a Linear System As we see in the next example, a linear system has no solution if we end up with a false equation after applying Gaussian elimination to the system.

38. E.g. 3—A System with No Solution Solve the following system. To put this in triangular form, we begin by eliminating the x-terms from the second equation and the third equation.

39. E.g. 3—A System with No Solution Equation 2 + (–2)  Equation 1 = New Equation 2 Equation 3 + (–3)  Equation 1 = New Equation 3

40. E.g. 3—A System with No Solution Now, we eliminate the y-term from the third equation. Equation 3 + (–1)  Equation 2 = New Equation 3 The system is now in triangular form. However, the third equation says 0 = 2, which is false.

41. E.g. 3—A System with No Solution No matter what values we assign x, y, and z, the third equation will never be true. This means the system has no solution.

42. E.g. 4—A System with Infinitely Many Solutions Solve the following system. To put this in triangular form, we begin by eliminating the x-terms from the second and third equations.

43. E.g. 4—A System with Infinitely Many Solutions Equation 2 + (–2)  Equation 1 = New Equation 2 Equation 3 + (–2)  Equation 1 = New Equation 3

44. E.g. 4—A System with Infinitely Many Solutions Now, we eliminate the y-term from the third equation. Equation 3 + (–2)  Equation 2 = New Equation 3

45. E.g. 4—A System with Infinitely Many Solutions The new third equation is true. However, it gives us no new information. So, we can drop it from the system.

46. E.g. 4—A System with Infinitely Many Solutions Only two equations are left. We can use them to solve for x and y in terms of z. z can take on any value, though. So, there are infinitely many solutions.

47. E.g. 4—A System with Infinitely Many Solutions To find the complete solution of the system, we begin by solving for y in terms of z, using the new second equation. 3y – 6z = 6 y – 2z = 2 y = 2z + 2

48. E.g. 4—A System with Infinitely Many Solutions Then, we solve for x in terms of z, using the first equation. x – (2z + 2) + 5z = –2 x + 3z – 2 = –2 x = –3z

49. E.g. 4—A System with Infinitely Many Solutions To describe the complete solution, we let t represent any real number. The solution is: x = –3t y = 2t + 2 z = t We can also write this as the ordered triple (–3t, 2t + 2, t).

50. Parameter In the solution of Example 4, the variable t is called a parameter. To get a specific solution, we give a specific value to the parameter t.

51. Parameters For instance, if we set t = 2, we get: x = –3(2) = –6 y = 2(2) + 2 = 6 z = 2 Thus, (–6, 6, 2) is a solution of the system.

52. Parameters Here are some other solutions of the system obtained by substituting other values for the parameter t.

53. Parameters You should check that these points satisfy the original equations. There are infinitely many choices for the parameter t. So, the system has infinitely many solutions.

54. Modeling Using Linear Systems

55. Linear systems are used to model situations that involve several varying quantities. In the next example, we consider an application to finance.

56. E.g. 5—Modeling a Financial Problem Using a Linear System Jason receives an inheritance of $50,000. His financial advisor suggests that he invest this in three mutual funds: Money-market fund Blue-chip stock fund High-tech stock fund

57. The advisor estimates that, over the next year: The money-market fund will return 5%. The blue-chip fund will return 9%. The high-tech fund will return 16%. E.g. 5—Modeling a Financial Problem Using a Linear System

58. Jason wants a total first-year return of $4000. To avoid excessive risk, he decides to invest three times as much in the money-market fund as in the high-tech stock fund. How much should he invest in each fund? E.g. 5—Modeling a Financial Problem Using a Linear System

59. Let: x = amount invested in the money-market fund y = amount invested in the blue-chip stock fund z = amount invested in the high-tech stock fund E.g. 5—Modeling a Financial Problem Using a Linear System

60. We convert each fact given in the problem into an equation. x + y + z = 50,000 Total amount invested is: $50,000 0.05x + 0.09y + 0.16z = 4000 Total investment return is: $4000 x = 3zMoney-market amount is: 3  high-tech amount E.g. 5—Modeling a Financial Problem Using a Linear System

61. Multiplying the second equation by 100 and rewriting the third gives the following system. We solve this using Gaussian elimination. E.g. 5—Modeling a Financial Problem Using a Linear System

62. Equation 2 + (–5)  Equation 1 = New Equation 2 Equation 3 + (–1)  Equation 1 = New Equation 3 E.g. 5—Modeling a Financial Problem Using a Linear System

63. Equation 2 + 4  Equation 3 = New Equation 3 E.g. 5—Modeling a Financial Problem Using a Linear System

65. Now that the system is in triangular form, we use back-substitution to find that: x = 30,000 y = 10,000 z = 10,000 E.g. 5—Modeling a Financial Problem Using a Linear System

66. This means that Jason should invest: $30,000 in the money market fund $10,000 in the blue-chip stock fund $10,000 in the high-tech stock fund E.g. 5—Modeling a Financial Problem Using a Linear System