Solving Linear Systems in Three Variables

1. Solving Linear Systems in Three Variables 3-6 Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2

2. x = 4y + 10 6x – 5y = 9 x = 3y – 1 3x – y = 8 4x + 2y = 4 2x – y =1 6x – 12y = –4 6x – 2y = 2 Warm Up Solve each system of equations algebraically. Classify each system and determine the number of solutions. (2, –2) 1. 2. (–1,–3) 4. 3. inconsistent; none consistent, independent; one

3. Objectives Represent solutions to systems of equations in three dimensions graphically. Solve systems of equations in three dimensions algebraically.

4. 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.

5. Recall from Lesson 3-5 that 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.

8. Identifying the exact solution from a graph of a 3-by-3 system can be very difficult. However, you can use the methods of elimination and substitution to reduce a 3-by-3 system to a 2-by-2 system and then use the methods that you learned in Lesson 3-2.

9. 1 2 3 Example 1: Solving a Linear System in Three Variables Use elimination to solve the system of equations. 5x – 2y – 3z = –7 2x – 3y + z = –16 3x + 4y – 2z = 7 Step 1 Eliminate one variable. 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.

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

11. 4 5 Example 1 Continued 11x – 11y = –55 You now have a 2-by-2 system. 7x – 2y = –25

12. 4 5 4 5 Multiply equation - by –2, and equation - by 11 and add. Example 1 Continued Step 2 Eliminate another variable. Then solve for the remaining variable. You can eliminate y by using methods from Lesson 3-2. –2(11x – 11y = –55) –22x + 22y = 110 1 11(7x – 2y = –25) 77x – 22y = –275 55x = –165 1 x = –3 Solve for x.

13. 4 Example 1 Continued Step 3 Use one of the equations in your 2-by-2 system to solve for y. 11x – 11y = –55 1 11(–3) – 11y = –55 Substitute –3 for x. 1 Solve for y. y = 2

14. 2 Example 1 Continued Step 4 Substitute for x and y in one of the original equations to solve for z. 2x – 3y + z = –16 Substitute –3 for x and 2 for y. 1 2(–3) – 3(2) + z = –16 z = –4 Solve for y. 1 The solution is (–3, 2, –4).

15. 1 2 3 Check It Out! Example 1 Use elimination to solve the system of equations. –x + y + 2z = 7 2x + 3y + z = 1 –3x – 4y + z = 4 Step 1 Eliminate one variable. In this system, z is a reasonable choice to eliminate first because the coefficient of z in the second equation is 1.

16. Multiply equation - by –2, and add to equation . 3 5 1 1 4 3 2 1 3 1 1 2 Multiply equation - by –2, and add to equation . Use equations and to create a second equation in x and y. Check It Out! Example 1 Continued –x + y + 2z = 7 –x + y + 2z = 7 –2(2x + 3y + z = 1) –4x – 6y – 2z = –2 –5x – 5y = 5 1 –x + y + 2z = 7 –x + y + 2z = 7 6x + 8y – 2z = –8 –2(–3x – 4y + z = 4) 5x + 9y = –1

17. 4 5 Check It Out! Example 1 Continued –5x – 5y = 5 You now have a 2-by-2 system. 5x + 9y = –1

18. 4 5 4 5 Add equation to equation . Check It Out! Example 1 Continued Step 2 Eliminate another variable. Then solve for the remaining variable. You can eliminate x by using methods from Lesson 3-2. –5x – 5y = 5 5x + 9y = –1 Solve for y. 4y = 4 1 y = 1

19. 4 Check It Out! Example 1 Step 3 Use one of the equations in your 2-by-2 system to solve for x. –5x – 5y = 5 Substitute 1 for y. –5x – 5(1) = 5 1 Solve for x. –5x – 5 = 5 1 –5x = 10 x = –2

20. 2 2x +3y + z = 1 Check It Out! Example 1 Step 4 Substitute for x and y in one of the original equations to solve for z. Substitute –2 for x and 1 for y. 2(–2) +3(1) + z = 1 Solve for z. 1 –4 + 3 + z = 1 z = 2 1 The solution is (–2, 1, 2).

21. You can also use substitution to solve a 3-by-3 system. Again, the first step is to reduce the 3-by-3 system to a 2-by-2 system.

22. Example 2: Business Application 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.

23. 1 2 3 Example 2 Continued 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 Friday’s sales. 250x + 60y + 50z = 1950 Saturday’s sales. 150x + 30y = 1050 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.

24. 1 2 1 2 Multiply equation by 5 and equation by –4 and add. Example 2 Continued Step 2 Eliminate z. 1000x + 150y + 200z = 7350 5(200x + 30y + 40z = 1470) –1000x– 240y– 200z= –7800 –4(250x + 60y + 50z = 1950) y = 5 By eliminating z, due to the coefficients of x, you also eliminated x providing a solution for y.

25. 3 3 Step 3 Use equation to solve for x. Example 2 Continued 150x + 30y= 1050 Substitute 5 for y. 150x + 30(5)= 1050 Solve for x. x = 6

26. Step 4 Use equations or to solve for z. 1 2 1 Example 2 Continued 1 200x + 30y + 40z = 1470 Substitute 6 for x and 5 for y. Solve for x. 200(6) + 30(5)+ 40z = 1470 z = 3 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.

27. Check It Out! Example 2 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?

28. 1 2 3 Check It Out! Example 2 Continued 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 Jada’s points. 2x + 4y = 14 Maria’s points. 2x + 2y + 3z = 13 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.

29. Multiply equation by –2 and add to equation . 4 1 3 4 2 2 3 1 4 Multiply equation by 3 and equation by –4 and add. Check It Out! Example 2 Continued Step 2 Eliminate z. 9x + 3y + 12z = 45 3(3x + y + 4z = 15) –8x– 8y – 12z = –52 –4(2x + 2y + 3z = 13) x– 5y = –7 –2(x– 5y = –7) –2x+ 10y = 14 2x + 4y = 14 2x + 4y = 14 y = 2 Solve for y.

30. 2 2 Step 3 Use equation to solve for x. Check It Out! Example 2 Continued 2x + 4y = 14 Substitute 2 for y. 2x + 4(2)= 14 Solve for x. x = 3

31. 2x + 2y + 3z = 13 3 Check It Out! Example 2 Continued Step 4 Substitute for x and y in one of the original equations to solve for z. 2(3)+ 2(2) + 3z = 13 6+ 4 + 3z = 13 Solve for z. z = 1 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.

32. Remember! Consistent means that the system of equations has at least one solution. The systems in Examples 1 and 2 have unique solutions. However, 3-by-3 systems may have no solution or an infinite number of solutions.

33. 1 2 3 Example 3: Classifying Systems with Infinite Many Solutions or No Solutions Classify the system as consistent or inconsistent, and determine the number of solutions. 2x – 6y + 4z = 2 –3x + 9y – 6z = –3 5x – 15y + 10z = 5

34. 1 2 2 1 Multiply equation by 3 and equation by 2 and add. Example 3 Continued The elimination method is convenient because the numbers you need to multiply the equations are small. First, eliminate x. 6x– 18y + 12z = 6 3(2x – 6y + 4z = 2) 2(–3x+ 9y – 6z = –3) –6x+ 18y – 12z = –6 0 = 0 

35. Multiply equation by 5 and equation by –2 and add. 1 3 3 1 Example 3 Continued 10x– 30y + 20z = 10 5(2x – 6y + 4z = 2) –2(5x– 15y + 10z = 5) –10x+ 30y – 20z = –10  0 = 0 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.

36. 1 2 3 Check It Out! Example 3a Classify the system, and determine the number of solutions. 3x – y + 2z = 4 2x – y + 3z = 7 –9x + 3y – 6z = –12

37. 1 3 4 1 2 Multiply equation by –1 and add to equation . Check It Out! Example 3a Continued The elimination method is convenient because the numbers you need to multiply the equations by are small. First, eliminate y. 3x – y + 2z = 4 3x – y + 2z = 4 –1(2x– y + 3z = 7) –2x+ y – 3z = –7 x – z = –3

38. 4 2 3 5 5 3 2 Multiply equation by 3 and add to equation . Check It Out! Example 3a Continued 6x– 3y + 9z = 21 3(2x – y + 3z = 7) –9x + 3y – 6z = –12 –9x+ 3y – 6z = –12 –3x + 3z = 9 Now you have a 2-by-2 system. x –z = –3 –3x + 3z = 9

39. 5 4 Check It Out! Example 3a Continued Eliminate x. 3(x –z = –3) 3x– 3z = –9 –3x + 3z = 9 –3x+ 3z = 9  0= 0 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.

40. 1 2 3 Check It Out! Example 3b Classify the system, and determine the number of solutions. 2x – y + 3z = 6 2x – 4y + 6z = 10 y – z = –2

41. 3 4 1 4 5 Substitute equation in for y in equation . Check It Out! Example 3b Continued Use the substitution method. Solve for y in equation 3. y – z = –2 Solve for y. y = z – 2 2x – y + 3z = 6 2x – (z – 2) + 3z = 6 2x – z + 2 + 3z = 6 2x + 2z = 4

42. 4 2 6 6 5 Substitute equation in for y in equation . Check It Out! Example 3b Continued 2x – 4y + 6z = 10 2x – 4(z – 2) + 6z = 10 2x – 4z + 8 + 6z = 10 2x + 2z = 2 Now you have a 2-by-2 system. 2x +2z = 4 2x + 2z = 2

43. 6 5 Check It Out! Example 3b Continued Eliminate z. 2x + 2z = 4 –1(2x + 2z = 2) 0  2  Because 0 is never equal to 2, the equation is a contradiction. Therefore, the system is inconsistent and has no solutions.

44. Lesson Quiz: Part I 1. At the library book sale, each type of book is priced differently. The table shows the number of books Joy and her friends each bought, and the amount each person spent. Find the price of each type of book. hardcover: $3; paperback: $1; audio books: $4

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