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2. Lesson Menu Five-Minute Check (over Lesson 3–3) CCSS Then/Now New Vocabulary Example 1: A System with One Solution Example 2:No Solution and Infinite Solutions Example 3:Real-World Example: Write and Solve a System of Equations

3. CCSS Content Standards A.CED.3 Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context. Mathematical Practices 3 Construct viable arguments and critique the reasoning of others.

4. Then/Now You solved linear equations with two variables. Solve systems of linear equations in three variables. Solve real-world problems using systems of linear equations in three variables.

5. Vocabulary ordered triple

6. Step 1 Use elimination to make a system of two equations in two variables. Step 2 Solve the system of two equations. Step 3Solve for the third variable using one of the original equations with three variables. Steps for solving systems of equations in three variables:

7. Example 1 A System with One Solution 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

8. Example 1 A System with One Solution 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.

9. Example 1 A System with One Solution 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.

10. Example 1 A System with One Solution 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.

11. Example 1 A System with One Solution 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.

12. Example 1 What is the solution to the system of equations shown below? 2x + 3y – 3z = 16 x + y + z = –3 x – 2y – z = –1 A. B.(–3, –2, 2) C.(1, 2, –6) D.(–1, 2, –4)

13. Example 2 No Solution and Infinite Solutions 6x + 3y – 9z=15(–)6x + 3y – 9z=15 0= 0 Eliminate y in the first and third equations. A. Solve the system of equations. 2x + y – 3z = 5 x + 2y – 4z = 7 6x + 3y – 9z = 15 2x + y – 3z=56x + 3y – 9z=15 Multiply by 3.

14. Example 2 No Solution and Infinite Solutions 6x + 3y – 9z=15(–)6x + 3y – 9z=15 9y – 15z=27 The equation 0 = 0 is always true. This indicates that the first and third equations represent the same plane. Check to see if this plane intersects the second plane. x + 2y – 4z=76x + 12y – 24z=42 Multiply by 6. Divide by the GCF, 3.3y – 5z=9 Answer: The planes intersect in a line. So, there is an infinite number of solutions.

15. Example 2 0= –3 Eliminate x in the first two equations. B. Solve the system of equations. 3x – y – 2z = 4 6x – 2y – 4z = 11 9x – 3y – 6z = 12 3x – y – 2z=46x – 2y + 4z= 8 Multiply by 2. 6x – 2y – 4z=11(–) 6x – 2y – 4z= 11 Answer: The equation 0 = –3is never true. So, there is no solution of this system. No Solution and Infinite Solutions

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

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

18. Example 3 Write and Solve a System of Equations SPORTS There are 49,000 seats in a sports stadium. Tickets for the seats in the upper level sell for $25, the ones in the middle level cost $30, and the ones in the bottom level are $35 each. The number of seats in the middle and bottom levels together equals the number of seats in the upper level. When all of the seats are sold for an event, the total revenue is $1,419,500. How many seats are there in each level? ExploreRead the problem and define the variables. u = number of seats in the upper level m = number of seats in the middle level b = number of seats in the bottom level

19. Example 3 Write and Solve a System of Equations PlanThere are 49,000 seats. u + m + b = 49,000 When all the seats are sold, the revenue is 1,419,500. Seats cost $25, $30, and $35. 25u + 30m + 35b = 1,419,500 The number of seats in the middle and bottom levels together equal the number of seats in the upper level. m + b = u

20. Example 3 Write and Solve a System of Equations SolveSubstitute u = m + b in each of the first two equations. (m + b) + m + b = 49,000Replace u with m + b. 2m + 2b=49,000Simplify. m + b=24,500Divide by 2. 25(m + b) + 30m + 35b = 1,419,500Replace u with m + b. 25m + 25b + 30m + 35b=1,419,500Distributive Property 55m + 60b=1,419,500Simplify.

21. Example 3 Write and Solve a System of Equations Now, solve the system of two equations in two variables. –5b= –72,000 m + b =24,50055m + 55b =1,347,500 Multiply by 55. 55m + 60b =1,419,500(–) 55m + 60b =1,419,500 b=14,400

22. Example 3 Write and Solve a System of Equations Substitute 14,400 for b in one of the equations with two variables and solve for m. m + b = 24,500Equation with two variables m + 14,400=24,500b = 14,400 m =10,100Subtract 14,400 from each side.

23. Example 3 Write and Solve a System of Equations Substitute 14,400 for b and 10,100 for m in one of the original equations with three variables. m + b = uEquation with three variables 10,100 + 14,400=um = 10,100, b = 14,400 24,500 =uAdd. Answer:There are 24,500 upper level, 10,100 middle level, and 14,400 bottom level seats.

24. Example 3 Write and Solve a System of Equations CheckCheck to see if all the criteria are met. 24,500 + 10,100 + 14,400 = 49,000 The number of seats in the middle and bottom levels equals the number of seats in the upper level. 10,100 + 14,400 = 24,500 When all of the seats are sold, the revenue is $1,419,500. 24,500($25) + 10,100($30) + 14,400($35) = $1,419,500

25. Example 2 BUSINESS The school store sells pens, pencils, and paper. The pens are $1.25 each, the pencils are $0.50 each, and the paper is $2 per pack. Yesterday the store sold 25 items and earned $32. The number of pens sold equaled the number of pencils sold plus the number of packs of paper sold minus 5. How many of each item did the store sell? A.pens: 5; pencils: 10; paper: 10 B.pens: 8; pencils: 7; paper: 10 C.pens: 10; pencils: 7; paper: 8 D.pens: 11; pencils: 2; paper: 12

26. End of the Lesson