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Objectives

Use the elimination method to solve a system of equations. Choose an appropriate method to solve a system of equations. 7.3 The Elimination Method. Objectives. NCSCOS. 1.01, 4.03. 3 x – 5y = − 16. x + 2y = 18. 2 x + 5 y = 31. - x + 4 y = 12. 7.3 The Elimination Method.

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Objectives

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  1. Use the elimination method to solve a system of equations. Choose an appropriate method to solve a system of equations. 7.3 The Elimination Method Objectives 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 3x – 5y = −16 y = 5 3(3) – 5y = −16 x + 2y = 18 9 – 5y = −16 x + 2(5) = 18 – 5y = −25 x + 10 = 18 y = 5 x = 8 (8, 5) (3, 5)

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

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

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

  6. 2x + 3y = 1 5x + 7y = 3 7.3 The Elimination Method Solve by eliminating: (y) (-7)( ) ( )(-7) 2x+ 3y = 1 −14x – 21y = −7 (3)( ) ( )(3) 5x + 7y = 3 15x + 21y = 9 x = 2 2x + 3y = 1 5x + 7y = 3 2(2) + 3y = 1 5(2) + 7y = 3 4 + 3y = 1 10 + 7y = 3 3y = −3 7y = −7 y = −1 y = −1 (2, −1)

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

  8. 6x + 2y = 5 3x + 2y = 11 7.3 The Elimination Method Solve by eliminating: (y) 6x + 2y = 5 6x + 2y = 5 -3x – 2y = -11 (-1)( ) ( )(-1) 3x + 2y = 11 3x = -6 x = -2 3x + 2y = 11 6x + 2y = 5 3(-2) + 2y = 11 6(-2) + 2y = 5 -12 + 2y = 5 -6 + 2y = 11 2y = 17 2y = 17 (-2, 8.5) y = 8.5 y = 8.5

  9. 3x + y = 4 5x – 7y = 11 7.3 The Elimination Method Solve by eliminating: (y) (7)( ) ( )(7) 3x+ y = 4 21x + 7y = 28 5x – 7y = 11 5x – 7y = 11 26x = 39 x = 1.5 5x – 7y = 11 5(1.5) – 7 = 11 3x + y = 4 3(1.5) + y = 4 7.5 – 7y = 11 –7y = 3.5 4.5 + y = 4 y = −0.5 y = −0.5 (1.5, -0.5)

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

  11. 4x + 2y = 8 5x + 6y = 3 7.3 The Elimination Method Solve by eliminating: (y) (-3)( ) ( )(-3) 4x+ 2y = 8 -12x – 6y = -24 5x + 6y = 3 5x + 6y = 3 -7x = -21 x = 3 5x + 6y = 3 4x + 2y = 8 5(3) + 6y = 3 4(3) + 2y = 8 15 + 6y = 3 12 + 2y = 8 2y = −4 6y = −12 (3, −2) y = −2 y = −2

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

  13. 3x – 2y = 6 5x + 7y = 41 7.3 The Elimination Method Solve by eliminating: (x) (5)( ) ( )(5) 3x – 2y = 6 15x – 10y = 30 −15x – 21y = −123 (−3)( ) ( )(−3) 5x + 7y = 41 −31y = −93 y = 3 5x + 7y = 41 3x – 2y = 6 5x + 7(3) = 41 3x – 2(3) = 6 3x – 6 = 6 5x + 21 = 41 3x = 12 5x = 20 (4, 3) x = 4 x = 4

  14. 3x – 4y = 10 5x + 7y = 3 7.3 The Elimination Method Solve by eliminating: (y) (7)( ) ( )(7) 3x – 4y = 10 21x – 28y = 70 (4)( ) ( )(4) 20x + 28y = 12 5x + 7y = 3 41x = 82 x = 2 5x + 7y = 3 3x – 4y = 10 5(2) + 7y = 3 3(2) – 4y = 10 10 + 7y = 3 6 – 4y = 10 7y = –7 −4y = 4 y = –1 (2, –1) y = –1

  15. 3x – 2y = 2 4x – 7y = 33 7.3 The Elimination Method Solve by eliminating: (y) (-7)( ) ( )(-7) -21x + 14y = -14 3x – 2y = 2 (2)( ) ( )(2) 8x – 14y = 66 4x – 7y = 33 −13x = 52 x = −4 3x – 2y = 2 4x – 7y = 33 3(-4) – 2y = 2 4(-4) – 7y = 33 -12 – 2y = 2 −16 – 7y = 33 -2x = 14 −7x = 49 x = −7 (−4, −7) x = −7

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

  17. 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.

  18. 2d + 125m = 95.75 4d + 350m= 226.50 7.3 The Elimination Method Solve by elimination: (-2)( ) ( )(-2) 2d + 125m = 95.75 4d + 350m = 226.50 4d + 350m = 226.50 4d + 350(.35) = 226.50 -4d – 250m = -191.50 4d + 122.50 = 226.50 4d + 350m = 226.50 4d = 104 100m = 35 d = 26 100 100 Cost per-mile: $0.35 m = 0.35 Daily rental fee $26.00

  19. 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.

  20. 6b + 11f = 956 9b + 5f= 698 7.3 The Elimination Method Solve by elimination: (3)( ) ( )(3) 6b + 11f = 956 (-2)( ) ( )(-2) 9b + 5f = 698 6b + 11f = 956 6b + 11(64) = 956 18b + 33f = 2868 6b + 704 = 956 -18b – 10f = -1396 6b = 252 23f = 1472 b = 42 23 23 f = 64 file cabinets $64 bookcases $42

  21. ax + by = e cx + dy = f 7.3 The Elimination Method Solve for x and y using elimination: (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

  22. 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.

  23. 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 5y and–5y are opposites and are easily eliminated.

  24. 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 b can easily be eliminated by multiplying the first equation by –2.

  25. Summary of Methods for Solving Systems Example 324p + 456t = 225 Suggested Method 178p – 245t = 150 Why 7.3 The Elimination Method Rules and Properties Graphics Calculator The coefficients are large numbers, so other methods may be cumbersome.

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