3 3 systems of linear equations 1

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3 3 systems of linear equations 1

  1. 1. Systems of Linear Equations Back to 123a-Home
  2. 2. If 2 hamburgers cost 4$, then one hamburger is 2$. Systems of Linear Equations
  3. 3. If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. Systems of Linear Equations
  4. 4. If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. Systems of Linear Equations
  5. 5. If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… Systems of Linear Equations
  6. 6. If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… A system of linear equations is a collection two or more linear equations in two or more variables. Systems of Linear Equations
  7. 7. If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… A system of linear equations is a collection two or more linear equations in two or more variables. A solution for the system is a collection of numbers, one for each variable, that works for all equations in the system. Systems of Linear Equations
  8. 8. If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… A system of linear equations is a collection two or more linear equations in two or more variables. A solution for the system is a collection of numbers, one for each variable, that works for all equations in the system. For example, 2x + y = 7 x + y = 5 is a system { Systems of Linear Equations
  9. 9. If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… A system of linear equations is a collection two or more linear equations in two or more variables. A solution for the system is a collection of numbers, one for each variable, that works for all equations in the system. For example, 2x + y = 7 x + y = 5 is a system and that x = 2 and y = 3 is a solution because the pair fits both equations. { Systems of Linear Equations 2(2) + (3) = 7 (2) + (3) = 5 ck:
  10. 10. If 2 hamburgers cost 4$, then one hamburger is 2$. We need one piece of information to solve for one unknown quantity. But if a hamburger and a drink cost $4 then we won’t be able to pin down the price of each item-not enough information. In general, to find two unknowns, two pieces of information are needed, for three unknowns, three pieces of information, etc… A system of linear equations is a collection two or more linear equations in two or more variables. A solution for the system is a collection of numbers, one for each variable, that works for all equations in the system. For example, 2x + y = 7 x + y = 5 is a system and that x = 2 and y = 3 is a solution because the pair fits both equations. This solution is written as (2, 3). { Systems of Linear Equations 2(2) + (3) = 7 (2) + (3) = 5 ck:
  11. 11. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Example A. Systems of Linear Equations
  12. 12. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. Example A. Systems of Linear Equations
  13. 13. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 Example A. E1 Systems of Linear Equations
  14. 14. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ Example A. E1 E2 Systems of Linear Equations
  15. 15. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ The difference between the order is one hamburger and the difference between the cost is $2, so the hamburger is $2. Example A. E1 E2 Systems of Linear Equations
  16. 16. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ The difference between the order is one hamburger and the difference between the cost is $2, so the hamburger is $2. In symbols, subtract the equations, i.e. E1 – E2: Example A. 2x + y = 7 ) x + y = 5 E1 E2 Systems of Linear Equations
  17. 17. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ The difference between the order is one hamburger and the difference between the cost is $2, so the hamburger is $2. In symbols, subtract the equations, i.e. E1 – E2: Example A. 2x + y = 7 ) x + y = 5 x + 0 = 2 so x = 2 E1 E2 Systems of Linear Equations
  18. 18. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ The difference between the order is one hamburger and the difference between the cost is $2, so the hamburger is $2. In symbols, subtract the equations, i.e. E1 – E2: Example A. substituting 2 for x back into E2, we get: 2 + y = 5 or y = 3 2x + y = 7 ) x + y = 5 x + 0 = 2 so x = 2 E1 E2 Systems of Linear Equations
  19. 19. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $5. Let x = cost of a hamburger, y = cost of a salad. We can translate the information into the system 2x + y = 7 x + y = 5{ The difference between the order is one hamburger and the difference between the cost is $2, so the hamburger is $2. In symbols, subtract the equations, i.e. E1 – E2: Example A. substituting 2 for x back into E2, we get: 2 + y = 5 or y = 3 Hence the solution is (2 , 3) or, $2 for a hamburger, $3 for a salad. 2x + y = 7 ) x + y = 5 x + 0 = 2 so x = 2 E1 E2 Systems of Linear Equations
  20. 20. Example B. Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost?
  21. 21. Example B. Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad.
  22. 22. Example B. Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
  23. 23. 4x + 3y = 18 3x + 2y = 13 Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system: Example B. { E1 E2 Systems of Linear Equations
  24. 24. 4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. E1 E2 Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
  25. 25. 4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. E1 E2 Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
  26. 26. 4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. E1 E2 Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
  27. 27. 4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. Multiplying E1 by (–2), E2 by 3, E1 E2 Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
  28. 28. 4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. Multiplying E1 by (–2), E2 by 3, E1 E2 – 8x – 6y = –36E1*(–2) : Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
  29. 29. 4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. Multiplying E1 by (–2), E2 by 3, E1 E2 – 8x – 6y = –36 9x + 6y = 39 E1*(–2) : E2*(3): Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system:
  30. 30. 4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. Multiplying E1 by (–2), E2 by 3, and add the results, the y-terms are eliminated. E1 E2 – 8x – 6y = –36 9x + 6y = 39 E1*(–2) : E2*(3): + ) Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system: We adjust the y’s into opposite signs then add to eliminate them.
  31. 31. 4x + 3y = 18 3x + 2y = 13 Example B. { Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate x or y. Suppose we chose to eliminate the y-terms, we find the LCM of 3y and 2y first. It's 6y. Multiplying E1 by (–2), E2 by 3, and add the results, the y-terms are eliminated. E1 E2 – 8x – 6y = –36 9x + 6y = 39 E1*(–2) : E2*(3): + ) x = 3 Systems of Linear Equations Suppose four hamburgers and three salads is $18, three hamburger and two salads is $13. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate the information into the system: So a hamburger is 3$. We adjust the y’s into opposite signs then add to eliminate them.
  32. 32. To find y, set 3 for x in E2: 3x + 2y = 13 Systems of Linear Equations
  33. 33. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 3x + 2y = 13 Systems of Linear Equations
  34. 34. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 3x + 2y = 13 Systems of Linear Equations
  35. 35. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 3x + 2y = 13 Systems of Linear Equations
  36. 36. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). 3x + 2y = 13 Systems of Linear Equations
  37. 37. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 Systems of Linear Equations
  38. 38. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. Systems of Linear Equations
  39. 39. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. 1. Line up the x’s & y’s to the left and the numbers to the right. Systems of Linear Equations
  40. 40. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. 1. Line up the x’s & y’s to the left and the numbers to the right. 2. Select the variable x or y to eliminate. Systems of Linear Equations
  41. 41. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. 1. Line up the x’s & y’s to the left and the numbers to the right. 2. Select the variable x or y to eliminate. 3. Find the LCM of the terms with that variable, and convert these terms to the LCM (by multiplication). Systems of Linear Equations
  42. 42. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. 1. Line up the x’s & y’s to the left and the numbers to the right. 2. Select the variable x or y to eliminate. 3. Find the LCM of the terms with that variable, and convert these terms to the LCM (by multiplication). 4. Add or subtract the adjusted equations and solve the resulting equation. Systems of Linear Equations
  43. 43. To find y, set 3 for x in E2: and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2 Hence the solution is (3, 2). Therefore a hamburger cost $3 and a salad cost $2. 3x + 2y = 13 The above method is called the elimination (addition) method. We summarize the steps below. 1. Line up the x’s & y’s to the left and the numbers to the right. 2. Select the variable x or y to eliminate. 3. Find the LCM of the terms with that variable, and convert these terms to the LCM (by multiplication). 4. Add or subtract the adjusted equations and solve the resulting equation. 5. Substitute the answer back into any equation to solve for the other variable. Systems of Linear Equations
  44. 44. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 {Example C. Systems of Linear Equations
  45. 45. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. {Example C. Systems of Linear Equations
  46. 46. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. {Example C. Systems of Linear Equations
  47. 47. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 {Example C. 15x + 12y = 63*E1: Systems of Linear Equations
  48. 48. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 8x – 12y = - 52 3*E1: 4*E2: Systems of Linear Equations
  49. 49. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: Systems of Linear Equations
  50. 50. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46 Systems of Linear Equations
  51. 51. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Systems of Linear Equations
  52. 52. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 Systems of Linear Equations
  53. 53. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 Systems of Linear Equations
  54. 54. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Systems of Linear Equations
  55. 55. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Hence the solution is (-2, 3). Systems of Linear Equations
  56. 56. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Hence the solution is (-2, 3). In all the above examples, we obtain exactly one solution in each case. Systems of Linear Equations
  57. 57. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Hence the solution is (-2, 3). In all the above examples, we obtain exactly one solution in each case. However, there are two other possibilities. Systems of Linear Equations
  58. 58. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Hence the solution is (-2, 3). In all the above examples, we obtain exactly one solution in each case. However, there are two other possibilities. Systems of Linear Equations The system might not have any solution
  59. 59. 5x + 4y = 2 2x – 3y = -13 Solve E1 E2 Eliminate the y. The LCM of the y-terms is 12y. Multiply E1 by 3 and E2 by 4. {Example C. 15x + 12y = 6 ) 8x – 12y = - 52 3*E1: 4*E2: Add: 23x + 0 = - 46  x = - 2 Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y = 3 Hence the solution is (-2, 3). In all the above examples, we obtain exactly one solution in each case. However, there are two other possibilities. Systems of Linear Equations The system might not have any solution or the system might have infinite many solutions.
  60. 60. Two Special Cases: I. Contradictory (Inconsistent) Systems Systems of Linear Equations
  61. 61. Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 (E1) (E2) Systems of Linear Equations
  62. 62. Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 (E1) (E2) Remove the x-terms by subtracting the equations. Systems of Linear Equations
  63. 63. Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 E1 – E2 : x + y = 2 ) x + y = 3 (E1) (E2) Remove the x-terms by subtracting the equations. Systems of Linear Equations
  64. 64. Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 E1 – E2 : x + y = 2 ) x + y = 3 0 = -1 (E1) (E2) Remove the x-terms by subtracting the equations. Systems of Linear Equations
  65. 65. Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 E1 – E2 : x + y = 2 ) x + y = 3 0 = -1 (E1) (E2) Remove the x-terms by subtracting the equations. This is impossible! Such systems are said to be contradictory or inconsistent. Systems of Linear Equations
  66. 66. Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = 2 x + y = 3 E1 – E2 : x + y = 2 ) x + y = 3 0 = -1 (E1) (E2) Remove the x-terms by subtracting the equations. This is impossible! Such systems are said to be contradictory or inconsistent. These system don’t have solution. Systems of Linear Equations
  67. 67. II. Dependent Systems { x + y = 3 2x + 2y = 6 Example E. (E1) (E2) Systems of Linear Equations
  68. 68. II. Dependent Systems { x + y = 3 2x + 2y = 6 Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations
  69. 69. II. Dependent Systems { x + y = 3 2x + 2y = 6 2*E1 2x + 2y = 6 Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations
  70. 70. II. Dependent Systems { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations
  71. 71. II. Dependent Systems { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations
  72. 72. II. Dependent Systems { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations
  73. 73. { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Such systems are called dependent or redundant systems. Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations II. Dependent Systems
  74. 74. { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Such systems are called dependent or redundant systems. Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations II. Dependent Systems Finally, as before with linear equations, clear the denominators of fractional equations first.
  75. 75. { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Such systems are called dependent or redundant systems. Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations II. Dependent Systems Finally, as before with linear equations, clear the denominators of fractional equations first. For example, change { x – y = 3 x – y = –1 3 2 2 3 1 2 1 4
  76. 76. { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Such systems are called dependent or redundant systems. Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations II. Dependent Systems Finally, as before with linear equations, clear the denominators of fractional equations first. For example, change { x – y = 3 x – y = –1 3 2 2 3 1 2 1 4 ( )*6 ( )*4
  77. 77. { x + y = 3 2x + 2y = 6 2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0 = 0 This means the equations are actually the same and it has infinitely many solutions such as (3, 0), (2, 1), (1, 2) etc… Such systems are called dependent or redundant systems. Example E. (E1) (E2) Remove the x-terms by multiplying E1 by 2 then subtract E2. Systems of Linear Equations II. Dependent Systems Finally, as before with linear equations, clear the denominators of fractional equations first. For example, change { x – y = 3 x – y = –1 3 2 2 3 1 2 1 4 ( )*6 9x – 4y = 18 ( )*4 { 2x – y = –4 then solve.
  78. 78. Exercise. A. Select the solution of the system. {x + y = 3 2x + y = 4 Systems of Linear Equations {x + y = 3 2x + y = 5 {x + y = 3 2x + y = 6 {x + y = 3 2x + y = 31. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0) 2. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0) 3. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0) 4. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0) B. Solve the following systems. 5. {y = 3 2x + y = 3 6. {x = 1 2x + y = 4 7. {y = 1 2x + y = 5 8. {x = 2 2x + y = 6 C. Add or subtract vertically. (These are warm ups) 8x –9x+) 9. 8 –9–) 10. –8y –9y–) 11. –8 9–) 12. –4y –5y+) 13. +4x –5x–) 14.
  79. 79. Systems of Linear Equations 18.{–x + 2y = –12 2x + y = 4 D. Solve the following system by the elimination method. 15.{x + y = 3 2x + y = 4 16. 17.{x + 2y = 3 2x – y = 6 {x + y = 3 2x – y = 6 19. {3x + 4y = 3 x – 2y = 6 20. { x + 3y = 3 2x – 9y = –4 21.{–3x + 2y = –1 2x + 3y = 5 22. {2x + 3y = –1 3x + 4y = 2 23. {4x – 3y = 3 3x – 2y = –4 24.{5x + 3y = 2 2x + 4y = –2 25. {3x + 4y = –10 –5x + 3y = 7 26. {–4x + 9y = 1 5x – 2y = 8 { x – y = 3 x – y = –1 3 2 2 3 1 2 1 4 27. { x + y = 1 x – y = –1 1 2 1 5 3 4 1 6 28. 29.{ x + 3y = 4 2x + 6y = 8 E. Which system is inconsistent and which is dependent? 30.{2x – y = 2 8x – 4y = 6

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