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System of Equations and Inequalities Solving Practice

Explore graphing systems of inequalities, finding vertices, and maximizing/minimizing functions within feasible regions. Practice solving real-world problems using systems of equations.

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System of Equations and Inequalities Solving Practice

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  1. Splash Screen

  2. 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 Lesson Menu

  3. Graph the system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum an minimum values of the given function for this region. 1 ≤ x ≤ 4, y≥x, y≥ 2x + 3; f(x, y) = 3x – 2y • maximum: f(4, 4) = 4minimum: f(4, 11) = –10 • B. maximum: f(4, 11) = 10minimum: f(1, 1) = 1 • C. maximum: f(4, 4) = 4minimum: f(1, 5) = –7 • D. maximum: f(4, 11) = 10minimum: f(4, 4) = 4 5-Minute Check 1

  4. Graph the system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum an minimum values of the given function for this region. 1 ≤ x ≤ 4, y≥x, y≥ 2x + 3; f(x, y) = 3x – 2y • maximum: f(4, 4) = 4minimum: f(4, 11) = –10 • B. maximum: f(4, 11) = 10minimum: f(1, 1) = 1 • C. maximum: f(4, 4) = 4minimum: f(1, 5) = –7 • D. maximum: f(4, 11) = 10minimum: f(4, 4) = 4 5-Minute Check 1

  5. A company profits $3 for every widget it manufactures and $2 for every plinket it manufactures. It must make at least one widget and one plinket each hour, but cannot make more than 7 total widgets and plinkets in any hour. What is the most profit the company can make in any hour? A. $21 B. $20 C. $18 D. $5 5-Minute Check 2

  6. A company profits $3 for every widget it manufactures and $2 for every plinket it manufactures. It must make at least one widget and one plinket each hour, but cannot make more than 7 total widgets and plinkets in any hour. What is the most profit the company can make in any hour? A. $21 B. $20 C. $18 D. $5 5-Minute Check 2

  7. 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. CCSS

  8. 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. Then/Now

  9. ordered triple Vocabulary

  10. Solve the system of equations. 5x + 3y + 2z = 2 2x + y – z = 5 x + 4y + 2z = 16 2x + y – z = 5 (+)4x + 2y – 2z = 10 Second equation Multiply by 2. A System with One Solution Step 1 Use elimination to make a system of two equations in two variables. 5x + 3y + 2z = 2 5x + 3y + 2z = 2 First equation 9x + 5y= 12 Add to eliminate z. Example 1

  11. A System with One Solution 5x + 3y + 2z = 2 First equation (–) x + 4y + 2z = 16 Third equation 4x – y= –14 Subtract to eliminate z. Notice that the z terms in each equation have been eliminated. The result istwo equations with the same two variables, x and y. Example 1

  12. 4x – y = –14 (+) 20x – 5y = –70 Multiply by 5. A System with One Solution Step 2 Solve the system of two equations. 9x + 5y = 12 9x + 5y = 12 29x = –58 Add to eliminate y. x = –2 Divide by 29. Example 1

  13. A System with One Solution Substitute –2 for x in one of the two equations with two variables and solve for y. 4x– y = –14 Equation with two variables 4(–2) – y = –14 Replace x with –2. –8 – y = –14 Multiply. y = 6 Simplify. The result is x = –2 and y = 6. Example 1

  14. A System with One Solution Step 3 Solve for z using one of the original equations with three variables. 2x+ y – z = 5 Original equation with three variables 2(–2) + 6 – z = 5 Replace x with –2 and y with 6. –4 + 6 – z = 5 Multiply. z = –3 Simplify. Answer: Example 1

  15. A System with One Solution Step 3 Solve for z using one of the original equations with three variables. 2x+ y – z = 5 Original equation with three variables 2(–2) + 6 – z = 5 Replace x with –2 and y with 6. –4 + 6 – z = 5 Multiply. z = –3 Simplify. Answer: The solution is (–2, 6, –3). You can check this solution in the other two original equations. Example 1

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

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

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

  19. Multiply by 6. x + 2y – 4z = 7 6x + 12y – 24z = 42 No Solution and Infinite Solutions 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. 6x + 3y – 9z = 15 (–)6x + 3y – 9z = 15 9y – 15z = 27 Divide by the GCF, 3. 3y – 5z = 9 Answer: Example 2

  20. Multiply by 6. x + 2y – 4z = 7 6x + 12y – 24z = 42 No Solution and Infinite Solutions 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. 6x + 3y – 9z = 15 (–)6x + 3y – 9z = 15 9y – 15z = 27 Divide by the GCF, 3. 3y – 5z = 9 Answer: The planes intersect in a line. So, there is an infinite number of solutions. Example 2

  21. B. Solve the system of equations.3x – y – 2z = 46x – 2y – 4z = 119x – 3y – 6z = 12 Multiply by 2. 3x – y – 2z = 4 6x – 2y + 4z = 8 No Solution and Infinite Solutions Eliminate x in the first two equations. 6x – 2y – 4z = 11 (–) 6x – 2y – 4z = 11 0 = –3 Answer: Example 2

  22. B. Solve the system of equations.3x – y – 2z = 46x – 2y – 4z = 119x – 3y – 6z = 12 Multiply by 2. 3x – y – 2z = 4 6x – 2y + 4z = 8 No Solution and Infinite Solutions Eliminate x in the first two equations. 6x – 2y – 4z = 11 (–) 6x – 2y – 4z = 11 0 = –3 Answer: The equation 0 = –3 is never true.So, there is no solution of this system. Example 2

  23. A. What is the solution to the system of equations shown below? x + y – 2z = 3–3x – 3y + 6z = –92x + y – z = 6 A. (1, 2, 0) B. (2, 2, 0) C. infinite number of solutions D. no solution Example 2

  24. A. What is the solution to the system of equations shown below? x + y – 2z = 3–3x – 3y + 6z = –92x + y – z = 6 A. (1, 2, 0) B. (2, 2, 0) C. infinite number of solutions D. no solution Example 2

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

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

  27. SPORTSThere 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? Write and Solve a System of Equations Explore Read 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 Example 3

  28. Write and Solve a System of Equations Plan There 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 Example 3

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

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

  31. 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 = 24,500 Equation with two variables m + 14,400 = 24,500 b = 14,400 m = 10,100 Subtract 14,400 from each side. Example 3

  32. 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 = u Equation with three variables 10,100 + 14,400 = um = 10,100, b = 14,400 24,500 = u Add. Answer: Example 3

  33. 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 = u Equation with three variables 10,100 + 14,400 = um = 10,100, b = 14,400 24,500 = u Add. Answer: There are 24,500 upper level, 10,100 middle level, and 14,400 bottom level seats. Example 3

  34. Write and Solve a System of Equations Check Check 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 Example 3

  35. 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 Example 2

  36. 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 Example 2

  37. End of the Lesson

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