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Objectives: The Learner Will…,

Use the elimination method to solve a system of equations, which uses ‘opposites’ to eliminate one of the variables. 7.3 The Elimination Method. Objectives: The Learner Will…,. NCSCOS. 1.01, 4.03. 3 x – 5y = − 16. x + 2y = 18. 2 x + 5 y = 31. - x + 4 y = 12.

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Objectives: The Learner Will…,

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  1. Use the elimination method to solve a system of equations, which uses ‘opposites’ to eliminate one of the variables. 7.3 The Elimination Method Objectives: The Learner Will…, NCSCOS • 1.01, 4.03

  2. 3x – 5y = −16 x + 2y = 18 2x + 5y = 31 -x + 4y = 12 7.3 The Elimination Method Solve by elimination: 5x = 15 x = 3 6y = 30 y = 5 3(3) – 5y = −16 x + 2(5) = 18 9 – 5y = −16 x + 10 = 18 – 5y = −25 x = 8 y = 5 (8, 5) (3, 5)

  3. 3x + 4y = 7 2x – 4y = 13 7.3 The Elimination Method Solve by eliminating: + 5x = 20 x = 4 3(4)+ 4y = 7 2(4) – 4y = 13 8 – 4y = 13 12+ 4y = 7 – 4y = 5 4y = -5 y = -1.25 y = -1.25 (4, -1.25)

  4. 4x + 5y = 6 3x – 5y = 8 7.3 The Elimination Method Solve by eliminating: 7x = 14 x = 2 3(2) – 5y = 8 4(2) + 5y = 6 6 – 5y = 8 8 + 5y = 6 −5y = 2 5y = −2 2 5 (2, – ) 2 5 2 5 y = – y = –

  5. -3x + 4y = 12 3x – 6y = 18 7.3 The Elimination Method Solve by eliminating: -2y = 30 y = -15 3x – 6(-15) = 18 −3x + 4(-15) = 12 −3x – 60 = 12 3x + 90 = 18 −3x = 72 3x = -72 x = −24 (-24, -15) x = −24

  6. 6x + 2y = 5 3x + 2y = 11 7.3 The Elimination Method eliminate ‘x’  6x eliminate ‘y’  2y Solve by eliminating: 6x + 2y = 5 -3x – 2y = -11 (-1)( ) ( )(-1) 3x = -6 x = -2 6(-2) + 2y = 5 3(-2) + 2y = 11 -12 + 2y = 5 -6 + 2y = 11 2y = 17 2y = 17 y = 8.5 y = 8.5 (-2, 8.5)

  7. 2x + 3y = 1 5x + 7y = 3 7.3 The Elimination Method Solve by eliminating: eliminate ‘x’  10x eliminate ‘y’  21y (-5)( ) ( )(-5) -10x – 15y = -5 (2)( ) ( )(2) 10x + 14y = 6 -y = 1 y = -1 2x + 3(-1) = 1 5x + 7(-1) = 3 2x – 3 = 1 5x – 7 = 3 5x = 10 2x = 4 (2, -1) x = 2 x = 2

  8. 2x + 3y = 1 5x + 7y = 3 7.3 The Elimination Method eliminate ‘x’  10x eliminate ‘y’  21y Solve by eliminating: (-7)( ) ( )(-7) −14x – 21y = −7 (3)( ) ( )(3) 15x + 21y = 9 x = 2 5(2) + 7y = 3 2(2) + 3y = 1 10 + 7y = 3 4 + 3y = 1 7y = −7 3y = −3 y = −1 y = −1 (2, −1)

  9. 3x – 2y = 6 5x + 7y = 41 7.3 The Elimination Method eliminate ‘x’  15x eliminate ‘y’  14y Solve by eliminating: (7)( ) ( )(7) 21x – 14y = 42 (2)( ) ( )(2) 10x + 14y = 82 31x = 124 x = 4 3(4) – 2y = 6 5(4) + 7y = 41 12 – 2y = 6 20 + 7y = 41 −2y = −6 7y = 21 y = 3 y = 3 (4, 3)

  10. 3x – 2y = 6 5x + 7y = 41 7.3 The Elimination Method eliminate ‘x’  15x eliminate ‘y’  14y Solve by eliminating: (5)( ) ( )(5) 15x – 10y = 30 (−3)( ) ( )(−3) −15x – 21y = −123 −31y = −93 y = 3 3x – 2(3) = 6 5x + 7(3) = 41 3x – 6 = 6 5x + 21 = 41 3x = 12 x = 4 5x = 20 x = 4 (4, 3)

  11. x + 3y = 2 3x – 4y = –16 7.3 The Elimination Method eliminate ‘x’  3x eliminate ‘y’  12y Solve by eliminating: 3 2 (4)( ) ( )(4) 6x + 12y = 8 (3) ( ) ( )(3) 9x – 12y = −48 15x = −40 8 3 x = − 8 3 3(− ) – 4y = −16 –4 + 3y = 2 −8 – 4y = −16 −4y = −8 3y = 6 8 3 (- , 2) y = 2 y = 2

  12. x + 3y = 2 3x – 4y = –16 7.3 The Elimination Method eliminate ‘x’  3x eliminate ‘y’  12y Solve by eliminating: 3 2 (-2)( ) ( )(-2) -3x – 6y = –4 3x – 4y = −16 –10y = −20 3 2 x + 3(2) = 2 y = 2 3 2 x + 6 = 2 3x – 4(2) = −16 3 2 3x – 8 = −16 x = –4 3x = −8 8 3 3x = −8 (- , 2) 8 3 x = – 8 3 x = –

  13. 3x + y = 4 5x – 7y = 11 7.3 The Elimination Method eliminate ‘x’  15x eliminate ‘y’  7y Solve by eliminating: (7)( ) ( )(7) 21x + 7y = 28 5x – 7y = 11 26x = 39 x = 1.5 3(1.5) + y = 4 5(1.5) – 7 = 11 4.5 + y = 4 7.5 – 7y = 11 y = −0.5 –7y = 3.5 (1.5, -0.5) y = −0.5

  14. 2x – y = 7 5x + 4y = 11 7.3 The Elimination Method Solve by eliminating: eliminate ‘x’  10x eliminate ‘y’  4y (4)( ) ( )(4) 8x – 4y = 28 5x + 4y = 11 13x = 39 x = 3 2(3) – y = 7 5(3) + 4y = 11 6 – y = 7 15 + 4y = 11 –y = 1 4y = –4 y = –1 y = –1 (3, -1)

  15. 4x + 2y = 8 5x + 6y = 3 7.3 The Elimination Method eliminate ‘x’  20x eliminate ‘y’  6y Solve by eliminating: (-3)( ) ( )(-3) -12x – 6y = -24 5x + 6y = 3 -7x = -21 x = 3 4(3) + 2y = 8 5(3) + 6y = 3 12 + 2y = 8 15 + 6y = 3 2y = −4 y = −2 6y = −12 (3, −2) y = −2

  16. 3x – 4y = 10 5x + 7y = 3 7.3 The Elimination Method eliminate ‘x’  15x eliminate ‘y’  28y Solve by eliminating: (7)( ) ( )(7) 21x – 28y = 70 (4)( ) ( )(4) 20x + 28y = 12 41x = 82 x = 2 3(2) – 4y = 10 5(2) + 7y = 3 6 – 4y = 10 10 + 7y = 3 −4y = 4 7y = –7 y = –1 (2, –1) y = –1

  17. 3x – 2y = 2 4x – 7y = 33 7.3 The Elimination Method eliminate ‘x’  12x eliminate ‘y’  14y Solve by eliminating: (-7)( ) ( )(-7) -21x + 14y = -14 (2)( ) ( )(2) 8x – 14y = 66 −13x = 52 x = −4 3(-4) – 2y = 2 4(-4) – 7y = 33 -12 – 2y = 2 −16 – 7y = 33 -2x = 14 −7x = 49 x = −7 (−4, −7) x = −7

  18. 2d + 125m = 95.75 4d + 350m= 226.50 7.3 The Elimination Method Jason went on two trips ‘Out-West’, using the same rental car company called, Airport Rent-A-Car. The first time, Jason drove 125 miles on a 2-day trip which cost him a total of $95.75. And the second time, he drove 350 miles on a 4-day trip, which cost a total of $226.50. Find the daily fee and per-mile cost for both trips.

  19. 2d + 125m = 95.75 4d + 350m= 226.50 7.3 The Elimination Method daily rate ‘d’ mileage ‘m’ Solve by elimination: (-2)( ) ( )(-2) -4d – 250m = -191.50 4d + 350m = 226.50 100m = 35 m = 0.35 4d + 350(.35) = 226.50 4d + 122.50 = 226.50 Cost per-mile: $0.35 4d = 104 Daily rental fee $26.00 d = 26

  20. 6b + 11f = 956 9b + 5f= 698 7.3 The Elimination Method A company ordered bookcases and file cabinets, which arrived in two shipments. One shipment contained 6 bookcases and 11 file cabinets and cost $956. A second shipment contained 9 bookcases and 5 file cabinets and cost $698. Find the cost of the bookcases and file cabinets.

  21. 6b + 11f = 956 9b + 5f= 698 7.3 The Elimination Method book cases ‘b’ file cabinets ‘f’ Solve by elimination: (3)( ) ( )(3) 18b + 33f = 2868 (-2)( ) ( )(-2) -18b – 10f = -1396 23f = 1472 f = 64 6b + 11(64) = 956 6b + 704 = 956 6b = 252 file cabinets $64 b = 42 bookcases $42

  22. ax + by = e cx + dy = f 7.3 The Elimination Method Solve for x and y: (d)( ) ( )(d) ax + by = e (c)( ) ( )(c) ax + by = e (-b)( ) ( )(-b) cx + dy = f (-a)( ) ( )(-a) cx + dy = f adx + bdy = de acx + bcy = ce -bcx - bdy = -bf -acx - ady = -af adx – bcx = de – bf bcy – ady = ce – af x(ad – bc) = de – bf y(bc – ad) = ce – af (ad – bc) (ad – bc) (bc – ad) (bc – ad) x = de – bf y = ce – af ad – bc bc – ad

  23. Summary of Methods for Solving Systems Example 6x + y = 10 Suggested Method y = 5 Why 7.3 The Elimination Method Rules and Properties Substitution The value of one variable is known and can easily be substituted into the other equation.

  24. Summary of Methods for Solving Systems Example 2x – 5y = –20 Suggested Method 4x + 5y = 14 Why 7.3 The Elimination Method Rules and Properties Elimination eliminate ‘x’  8 eliminate ‘y’  5

  25. Summary of Methods for Solving Systems Example 9a – 2b = –11 Suggested Method 8a + 4b = 25 Why 7.3 The Elimination Method Rules and Properties Elimination eliminate ‘a’  72 eliminate ‘b’  4

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