College Algebra Sixth Edition James Stewart  Lothar Redlin  Saleem Watson

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 a1x1 + a2x2 + · · · + anxn =c where: • a1, a2, · · · , anand c are real numbers. • x1, x2, · · · , xnare the variables.

5. Linear Equations • If we have only three or four variables, we generally use x, y, z, and winstead of x1, x2, x3, and x4

6. Linear Equations • Such equations are called linear because: • If we have just two variables, the equation is a1x +a2y =cwhich 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

9. Solving a Linear System

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

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.

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. E.g. 1—Solving a Triangular System Using Back-Substitution • 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

16. E.g. 1—Solving a Triangular System Using Back-Substitution • 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).

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

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

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

25. E.g. 2—Solving a System of Three Equations in Three Variables • 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.

26. E.g. 2—Solving a System of Three Equations in Three Variables • 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

27. E.g. 2—Solving a System of Three Equations in Three Variables • 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).

28. Check Your Answer • We must check that the answer satisfies allthreeequations, 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.