1. Algebra 2 Systems With Three variables Lesson 3-5

2. Goals Goal To solve systems in three variables using elimination. Rubric Level 1 – Know the goals. Level 2 – Fully understand the goals. Level 3 – Use the goals to solve simple problems. Level 4 – Use the goals to solve more advanced problems. Level 5 – Adapts and applies the goals to different and more complex problems.

3. Vocabulary None

4. Essential Question Big Idea: Solving Equations and Inequalities How is solving a system of three equations in three variables similar to solving a system of two equations in two variables?

5. A Global Positioning System (GPS) gives locations using the three coordinates of latitude, longitude, and elevation. You can represent any location in three-dimensional space using a three-dimensional coordinate system, sometimes called coordinate space. Three-Dimensional Space

6. Coordinate Space Each point in coordinate space can be represented by an ordered triple of the form (x, y, z). The system is similar to the coordinate plane but has an additional coordinate based on the z-axis. Notice that the axes form three planes that intersect at the origin.

7. Graph the point in three-dimensional space. A(3, –2, 1) From the origin, move 3 units forward along the x-axis, 2 units left, and 1 unit up. y x z  A(3, –2, 1) Graphing Points in Three Dimensions

8. Graph the point in three-dimensional space. C(–1, 0, 2) From the origin, move 1 unit back along the x-axis, 2 units up. Notice that this point lies in the xz-plane because the y-coordinate is 0. y x z C(–1,0, 2)  Graphing Points in Three Dimensions

9. Linear Equation in Three Dimensions Recall that the graph of a linear equation in two dimensions is a straight line. In three- dimensional space, the graph of a linear equation is a plane. Because a plane is defined by three points, you can graph linear equations in three dimensions by graphing the three intercepts.

10. Graph the linear equation 2x – 3y + z = –6 in three-dimensional space. Step 1 Find the intercepts: x-intercept: 2x – 3(0) + (0) = –6 x = –3 y-intercept: 2(0) – 3y + (0) = –6 z-intercept: 2(0) – 3(0) + z = –6 y = 2 z = –6 Linear Equation in Three Dimensions

11. Step 2 Plot the points (–3, 0, 0), (0, 2, 0), and (0, 0, –6). Sketch a plane through the three points. y x z  (–3, 0, 0) (0, 2, 0) (0, 0, –6)   Linear Equation in Three Dimensions

12. System of Three Equations Systems of three equations with three variables are often called 3-by-3 systems. In general, to find a single solution to any system of equations, you need as many equations as you have variables.

13. The graph of a linear equation in three variables is a plane. When you graph a system of three linear equations in three dimensions, the result is three planes that may or may not intersect. The solution to the system is the set of points where all three planes intersect. These systems may have one, infinitely many, or no solution. System of Three Equations

14. There are 4 ways in which 3 planes could intersect. 1.No common intersection (no solution) 2.Intersection in a single point (one solution) 3.Intersection in a line (infinite solutions) 4.Intersection in a plane (all three are identical planes – infinite solutions) System of Three Equations

15. Each of these systems are inconsistent. No solution. 1. No Common Intersection

16. 2. Intersection in a Point Consistent, Independent system. One solution.

17. 3. Intersection in a Line Consistent, Dependent system. Infinite solutions.

18. All three equations are the same plane. 4. Intersection in a Plane Consistent, Dependent system. Infinite solutions.

19. Solving a System of Three Equations Identifying the exact solution from a graph of a 3-by-3 system can be very difficult. However, you can use the method of elimination to reduce a 3-by-3 system to a 2-by-2 system and then use the methods that you learned earlier to complete the solution.

20. To solve a linear system with three unknowns, first eliminate a variable from any two of the equations. Then eliminate the same variable from a different pair of equations. Eliminate a second variable using the resulting two equations in two variables to get an equation with just one variable whose value you can now determine. Find the values of the remaining variables by substitution. The solution of the system is written as an ordered triple. Solving a System of Three Equations

21. Use elimination to solve the system of equations. Step 1 Eliminate one variable. 5x – 2y – 3z = –7 2x – 3y + z = –16 3x + 4y – 2z = 7 In this system, z is a reasonable choice to eliminate first because the coefficient of z in the second equation is 1 and z is easy to eliminate from the other equations. 1 2 3 Example:

22. 5x – 2y – 3z = –7 11x – 11y = –55 3(2x –3y + z = –16) 5x – 2y – 3z = –7 6x – 9y + 3z = –48 1 2 1 4 3x + 4y – 2z = 7 7x – 2y = –25 2(2x –3y + z = –16) 3x + 4y – 2z = 7 4x – 6y + 2z = –32 3 2 Multiply equation - by 3, and add to equation. 1 2 Multiply equation - by 2, and add to equation. 3 2 5 Use equations and to create a second equation in x and y. 3 2 Example: continued

23. 11x – 11y = –55 7x – 2y = –25 You now have a 2-by-2 system. 4 5 Example: continued

24. –2(11x – 11y = –55) 55x = –165 11(7x – 2y = –25) –22x + 22y = 110 77x – 22y = –275 4 5 1 1 Multiply equation - by –2, and equation - by 11 and add. 4 5 Step 2 Eliminate another variable. Then solve for the remaining variable. You can eliminate y by using methods from Lesson 3-2. x = –3 Solve for x. Example: continued

25. 11x – 11y = –55 11(–3) – 11y = –55 4 1 1 Step 3 Use one of the equations in your 2-by-2 system to solve for y. y = 2 Substitute –3 for x. Solve for y. Example: continued

26. 2x – 3y + z = –16 2(–3) – 3(2) + z = –16 2 1 1 Step 4 Substitute for x and y in one of the original equations to solve for z. z = –4 Substitute –3 for x and 2 for y. Solve for y. The solution is (–3, 2, –4). Example: continued

27. Use elimination to solve the system of equations. Step 1 Eliminate one variable. –x + y + 2z = 7 2x + 3y + z = 1 –3x – 4y + z = 4 1 2 3 In this system, z is a reasonable choice to eliminate first because the coefficient of z in the second equation is 1. Example:

28. –x + y + 2z = 7 –5x – 5y = 5 –2(2x + 3y + z = 1) –4x – 6y – 2z = –2 1 2 1 4 5x + 9y = –1 –2(–3x – 4y + z = 4) –x + y + 2z = 7 6x + 8y – 2z = –8 1 3 Multiply equation - by –2, and add to equation. 1 2 1 3 5 –x + y + 2z = 7 Use equations and to create a second equation in x and y. 1 3 Example: continued

29. You now have a 2-by-2 system. 4 5 –5x – 5y = 5 5x + 9y = –1 Example: continued

30. 4y = 4 4 5 1 Add equation to equation. 4 5 Step 2 Eliminate another variable. Then solve for the remaining variable. You can eliminate x by using methods from Lesson 3-2. Solve for y. –5x – 5y = 5 5x + 9y = –1 y = 1 Example: continued

31. –5x – 5(1) = 5 4 1 1 Step 3 Use one of the equations in your 2-by-2 system to solve for x. x = –2 Substitute 1 for y. Solve for x. –5x – 5y = 5 –5x – 5 = 5 –5x = 10 Example: continued

32. 2(–2) +3(1) + z = 1 2x +3y + z = 1 2 1 1 Step 4 Substitute for x and y in one of the original equations to solve for z. z = 2 Substitute –2 for x and 1 for y. Solve for z. The solution is (–2, 1, 2). –4 + 3 + z = 1 Example: continued

33. 5.1 - 33 Your Turn: Solve the system. (1) (2) (3) Solution Eliminate z by adding equations (2) and (3) to get (4)

34. 5.1 - 34 Your Turn: To eliminate z from another pair of equations, multiply both sides of equation (2) by 6 and add the result to equation (1). Multiply (2) by 6. (1) Make sure equation (5) has the same two variables as equation (4). (5)

35. 5.1 - 35 Your Turn: Multiply (4) by – 5. (5) To eliminate x from equations (4) and (5), multiply both sides of equation (4) by 5 and add the result to equation (5). Solve the resulting equation for y. Add. Divide by 5.

36. 5.1 - 36 Your Turn: (4) with y = – 1. Using y = – 1, find x from equation (4) by substitution.

37. 5.1 - 37 Your Turn: (3) with x = 2, y = – 1. Substitute 2 for x and – 1 for y in equation (3) to find z. The solution set is {(2, –1,1)}.

38. Solve the following system by the elimination method. 2x + 2y + z = 1 – x + y + 2z = 3 x + 2y + 4z = 0 Solution: Leave the first equation alone, and multiply the second equation by 2, since combining these two equations will eliminate the variable x. 2x + 2y + z = 1 – 2x + 2y + 4z = 6 Continued. Your Turn:

39. Combine the two equations together. 4y + 5z = 7 Now combine the 2 nd and 3 rd equations together (no need to multiply by any number). 3y +6z = 3 Continued. Your Turn: continued

40. Multiply the first equation by 3 and the second equation by – 4 to combine them. 12y + 15z = 21 – 12y – 24z = – 12 – 9z = 9 z = – 1 Continued. Your Turn: continued

41. Substitute this variable into one of the two equations with only two variables. 4y + 5(– 1) = 7 4y – 5 = 7 4y = 12 y = 3 Continued. Your Turn: continued

42. Now substitute both of the values into one of the original equations. 2x + 2(3) + (– 1) = 1 2x + 6 – 1 = 1 2x = – 4 x = – 2 Your Turn: continued So the solution is (– 2, 3, – 1).

43. Step 1 Use elimination to make a system of two equations in two variables. Solve the system of equations. 5x + 3y + 2z = 2 2x + y – z = 5 x + 4y + 2z = 16 5x + 3y + 2z=25x + 3y + 2z= 2First equation 9x + 5y = 12Add to eliminate z. Multiply by 2. 2x + y – z=5(+)4x + 2y – 2z=10Second equation Your Turn:

44. 5x + 3y + 2z= 2First equation 4x – y =–14Subtract to eliminate z. (–) x + 4y + 2z= 16Third equation Notice that the z terms in each equation have been eliminated. The result is two equations with the same two variables x and y. Your Turn: continued

45. Step 2 Solve the system of two equations. 9x + 5y =129x + 5y = 12 29x = –58Add to eliminate y. Multiply by 5 4x – y =–14(+) 20x – 5y =–70 x=–2Divide by 29. Your Turn: continued

46. Substitute –2 for x in one of the two equations with two variables and solve for y. 4x – y=–14Equation with two variables 4(–2) – y=–14Replace x with –2. –8 – y=–14Multiply. y=6Simplify. The result is x = –2 and y = 6. Your Turn: continued

47. Step 3Solve for z using one of the original equations with three variables. 2x + y – z=5Original equation with three variables 2(–2) + 6 – z=5Replace x with –2 and y with 6. –4 + 6 – z=5Multiply. z=–3Simplify. Answer: The solution is (–2, 6, –3). You can check this solution in the other two original equations. Your Turn: continued

48. The table shows the number of each type of ticket sold and the total sales amount for each night of the school play. Find the price of each type of ticket. OrchestraMezzanineBalconyTotal Sales Fri2003040$1470 Sat2506050$1950 Sun150300$1050 Example: Application

49. Step 1 Let x represent the price of an orchestra seat, y represent the price of a mezzanine seat, and z represent the present of a balcony seat. Write a system of equations to represent the data in the table. 200x + 30y + 40z = 1470 250x + 60y + 50z = 1950 150x + 30y = 1050 1 2 3 Friday’s sales. Saturday’s sales. Sunday’s sales. A variable is “missing” in the last equation; however, the same solution methods apply. Elimination is a good choice because eliminating z is straightforward. Example: continued

50. 5(200x + 30y + 40z = 1470) –4(250x + 60y + 50z = 1950) 1 Step 2 Eliminate z. Multiply equation by 5 and equation by –4 and add. 12 2 1000x + 150y + 200z = 7350 –1000x – 240y – 200z = –7800 y = 5 By eliminating z, due to the coefficients of x, you also eliminated x providing a solution for y. Example: continued

51. 150x + 30y = 1050 150x + 30(5) = 1050 3 Substitute 5 for y. x = 6 Solve for x. Step 3 Use equation to solve for x. 3 Example: continued

52. 200x + 30y + 40z = 1470 1 Substitute 6 for x and 5 for y. 1 z = 3 Solve for x. Step 4 Use equations or to solve for z. 21 200(6) + 30(5) + 40z = 1470 The solution to the system is (6, 5, 3). So, the cost of an orchestra seat is $6, the cost of a mezzanine seat is $5, and the cost of a balcony seat is $3. Example: continued

53. Jada’s chili won first place at the winter fair. The table shows the results of the voting. How many points are first-, second-, and third-place votes worth? Name 1st Place 2nd Place 3rd Place Total Points Jada 31415 Maria 24014 Al 22313 Winter Fair Chili Cook-off Your Turn:

54. Step 1 Let x represent first-place points, y represent second-place points, and z represent third- place points. Write a system of equations to represent the data in the table. 3x + y + 4z = 15 2x + 4y = 14 2x + 2y + 3z = 13 1 2 3 Jada’s points. Maria’s points. Al’s points. A variable is “missing” in one equation; however, the same solution methods apply. Elimination is a good choice because eliminating z is straightforward. Solution:

55. 3(3x + y + 4z = 15) –4(2x + 2y + 3z = 13) 1 Step 2 Eliminate z. Multiply equation by 3 and equation by –4 and add. 31 3 9x + 3y + 12z = 45 –8x – 8y – 12z = –52 x – 5y = –7 4 2 –2(x – 5y = –7) 4 2x + 4y = 14 –2x + 10y = 14 2x + 4y = 14 y = 2 Multiply equation by –2 and add to equation. 2 4 Solve for y. Solution:

56. 2x + 4y = 14 Step 3 Use equation to solve for x. 2 2 2x + 4(2) = 14 x = 3 Solve for x. Substitute 2 for y. Solution:

57. Step 4 Substitute for x and y in one of the original equations to solve for z. z = 1 Solve for z. 2x + 2y + 3z = 13 3 2(3) + 2(2) + 3z = 13 6 + 4 + 3z = 13 The solution to the system is (3, 2, 1). The points for first-place is 3, the points for second-place is 2, and 1 point for third-place. Solution:

58. Other Solutions The systems so far have had unique solutions (one solution). However, 3-by-3 systems may have no solution or an infinite number of solutions.

59. Classify the system and determine the number of solutions. 2x – 6y + 4z = 2 –3x + 9y – 6z = –3 5x – 15y + 10z = 5 1 2 3 Example:

60. 3(2x – 6y + 4z = 2) 2(–3x + 9y – 6z = –3) First, eliminate x. 1 2 6x – 18y + 12z = 6 –6x + 18y – 12z = –6 0 = 0 Multiply equation by 3 and equation by 2 and add. 21 Example: continued

61. 5(2x – 6y + 4z = 2) –2(5x – 15y + 10z = 5) 1 3 10x – 30y + 20z = 10 –10x + 30y – 20z = –10 0 = 0 Multiply equation by 5 and equation by –2 and add. 3 1 Because 0 is always equal to 0, the equation is an identity. Therefore, the system is consistent, dependent and has an infinite number of solutions. Example: continued

62. 1. 2. 2x – y + 2z = 5 –3x +y – z = –1 x – y + 3z = 2 9x – 3y + 6z = 3 12x – 4y + 8z = 4 –6x + 2y – 4z = 5 inconsistent; none consistent; dependent; infinite Classify each system and determine the number of solutions. Your Turn:

63. EXTRA PRACTICE

64. A.(0, 6, 1) B.(1, 0, –2) C.infinite number of solutions D.no solution 1. What is the solution to the system of equations shown below? 3x + y – z = 5 –15x – 5y + 5z = 11 x + y + z = 2

65. 2. What is the solution to the system of equations shown below? 2x + 3y – 3z = 16 x + y + z = –3 x – 2y – z = –1 A. (–1, 2, –4) B.(–3, –2, 2) C. infinite number of solutions D. no solution

66. A.(1, 2, 0) B.(2, 2, 0) C.infinite number of solutions D.no solution 3. What is the solution to the system of equations shown below? x + y – 2z = 3 –3x – 3y + 6z = –9 2x + y – z = 6

67. Essential Question Big Idea: Solving Equations and Inequalities How is solving a system of three equations in three variables similar to solving a system of two equations in two variables? Use elimination, the same method you used for a system of two equations in two variables.

68. Assignment Section 3-5, Pg. 181 – 183; #1 – 3 all, 6 - 12 even, 16 – 20 even (use elimination), 24 – 38 even.