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4.3 system of linear equations 1

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4.3 system of linear equations 1

1. 1. Systems of Linear Equations
2. 2. Systems of Linear EquationsIn general, we need one piece of information to solve for oneunknown quantity.
3. 3. Systems of Linear EquationsIn general, we need one piece of information to solve for oneunknown quantity. If 2 hamburgers cost 4 dollars and x is thecost of a hamburger,
4. 4. Systems of Linear EquationsIn general, we need one piece of information to solve for oneunknown quantity. If 2 hamburgers cost 4 dollars and x is thecost of a hamburger, then 2x = 4 or x = 2 \$.
5. 5. Systems of Linear EquationsIn general, we need one piece of information to solve for oneunknown quantity. If 2 hamburgers cost 4 dollars and x is thecost of a hamburger, then 2x = 4 or x = 2 \$.If there are two unknowns, we need two pieces of informationto solve them,
6. 6. Systems of Linear EquationsIn general, we need one piece of information to solve for oneunknown quantity. If 2 hamburgers cost 4 dollars and x is thecost of a hamburger, then 2x = 4 or x = 2 \$.If there are two unknowns, we need two pieces of informationto solve them, three unknowns needs three pieces ofinformation, etc…
7. 7. Systems of Linear EquationsIn general, we need one piece of information to solve for oneunknown quantity. If 2 hamburgers cost 4 dollars and x is thecost of a hamburger, then 2x = 4 or x = 2 \$.If there are two unknowns, we need two pieces of informationto solve them, three unknowns needs three pieces ofinformation, etc… These lead to the systems of equations.
8. 8. Systems of Linear EquationsIn general, we need one piece of information to solve for oneunknown quantity. If 2 hamburgers cost 4 dollars and x is thecost of a hamburger, then 2x = 4 or x = 2 \$.If there are two unknowns, we need two pieces of informationto solve them, three unknowns needs three pieces ofinformation, etc… These lead to the systems of equations.A system of linear equations is a collection two or more linearequations in two or more variables.
9. 9. Systems of Linear EquationsIn general, we need one piece of information to solve for oneunknown quantity. If 2 hamburgers cost 4 dollars and x is thecost of a hamburger, then 2x = 4 or x = 2 \$.If there are two unknowns, we need two pieces of informationto solve them, three unknowns needs three pieces ofinformation, etc… These lead to the systems of equations.A system of linear equations is a collection two or more linearequations in two or more variables.A solution for the system is a collection of numbers, one foreach variable, that works for all equations in the system.
10. 10. Systems of Linear EquationsIn general, we need one piece of information to solve for oneunknown quantity. If 2 hamburgers cost 4 dollars and x is thecost of a hamburger, then 2x = 4 or x = 2 \$.If there are two unknowns, we need two pieces of informationto solve them, three unknowns needs three pieces ofinformation, etc… These lead to the systems of equations.A system of linear equations is a collection two or more linearequations in two or more variables.A solution for the system is a collection of numbers, one foreach variable, that works for all equations in the system.For example,{2x + y = 7 x+y=5is a system.
11. 11. Systems of Linear EquationsIn general, we need one piece of information to solve for oneunknown quantity. If 2 hamburgers cost 4 dollars and x is thecost of a hamburger, then 2x = 4 or x = 2 \$.If there are two unknowns, we need two pieces of informationto solve them, three unknowns needs three pieces ofinformation, etc… These lead to the systems of equations.A system of linear equations is a collection two or more linearequations in two or more variables.A solution for the system is a collection of numbers, one foreach variable, that works for all equations in the system.For example,{2x + y = 7 x+y=5is a system. x = 2 and y = 3 is a solution because it fits bothequations the ordered pair.
12. 12. Systems of Linear EquationsIn general, we need one piece of information to solve for oneunknown quantity. If 2 hamburgers cost 4 dollars and x is thecost of a hamburger, then 2x = 4 or x = 2 \$.If there are two unknowns, we need two pieces of informationto solve them, three unknowns needs three pieces ofinformation, etc… These lead to the systems of equations.A system of linear equations is a collection two or more linearequations in two or more variables.A solution for the system is a collection of numbers, one foreach variable, that works for all equations in the system.For example,{2x + y = 7 x+y=5is a system. x = 2 and y = 3 is a solution because it fits bothequations the ordered pair. This solution is written as (2, 3).
13. 13. Systems of Linear EquationsExample A.Suppose two hamburgers and an order of fries cost \$7, andone hamburger and an order of fries cost \$5.
14. 14. Systems of Linear EquationsExample A.Suppose two hamburgers and an order of fries cost \$7, andone hamburger and an order of fries cost \$5.Let x = cost of a hamburger, y = cost of an order of fries.
15. 15. Systems of Linear EquationsExample A.Suppose two hamburgers and an order of fries cost \$7, andone hamburger and an order of fries cost \$5.Let x = cost of a hamburger, y = cost of an order of fries.We can translate these into the system:{2x + y = 7
16. 16. Systems of Linear EquationsExample A.Suppose two hamburgers and an order of fries cost \$7, andone hamburger and an order of fries cost \$5.Let x = cost of a hamburger, y = cost of an order of fries.We can translate these into the system:{2x + y = 7 x+y=5
17. 17. Systems of Linear EquationsExample A.Suppose two hamburgers and an order of fries cost \$7, andone hamburger and an order of fries cost \$5.Let x = cost of a hamburger, y = cost of an order of fries.We can translate these into the system:{2x + y = 7 x+y=5 E1 E2
18. 18. Systems of Linear EquationsExample A.Suppose two hamburgers and an order of fries cost \$7, andone hamburger and an order of fries cost \$5.Let x = cost of a hamburger, y = cost of an order of fries.We can translate these into the system:{ 2x + y = 7 x+y=5 E1 E2The difference between the order is one hamburger and thedifference between the cost is \$2,
19. 19. Systems of Linear EquationsExample A.Suppose two hamburgers and an order of fries cost \$7, andone hamburger and an order of fries cost \$5.Let x = cost of a hamburger, y = cost of an order of fries.We can translate these into the system:{ 2x + y = 7 x+y=5 E1 E2The difference between the order is one hamburger and thedifference between the cost is \$2, so the hamburger is \$2.
20. 20. Systems of Linear EquationsExample A.Suppose two hamburgers and an order of fries cost \$7, andone hamburger and an order of fries cost \$5.Let x = cost of a hamburger, y = cost of an order of fries.We can translate these into the system:{ 2x + y = 7 x+y=5 E1 E2The difference between the order is one hamburger and thedifference between the cost is \$2, so the hamburger is \$2.In symbols, subtract the equations, i.e. E1 – E2:
21. 21. Systems of Linear EquationsExample A.Suppose two hamburgers and an order of fries cost \$7, andone hamburger and an order of fries cost \$5.Let x = cost of a hamburger, y = cost of an order of fries.We can translate these into the system:{ 2x + y = 7 x+y=5 E1 E2The difference between the order is one hamburger and thedifference between the cost is \$2, so the hamburger is \$2.In symbols, subtract the equations, i.e. E1 – E2: 2x + y = 7 ) x+y=5
22. 22. Systems of Linear EquationsExample A.Suppose two hamburgers and an order of fries cost \$7, andone hamburger and an order of fries cost \$5.Let x = cost of hamburger, y = cost of an order of fries.We can translate these into the system:{ 2x + y = 7 x+y=5 E1 E2The difference between the order is one hamburger and thedifference between the cost is \$2, so the hamburger is \$2.In symbols, subtract the equations, i.e. E1 – E2: 2x + y = 7 ) x+y=5 x+0=2
23. 23. Systems of Linear EquationsExample A.Suppose two hamburgers and an order of fries cost \$7, andone hamburger and an order of fries cost \$5.Let x = cost of a hamburger, y = cost of an order of fries.We can translate these into the system:{ 2x + y = 7 x+y=5 E1 E2The difference between the order is one hamburger and thedifference between the cost is \$2, so the hamburger is \$2.In symbols, subtract the equations, i.e. E1 – E2: 2x + y = 7 ) x+y=5 x+0=2 so x=2
24. 24. Systems of Linear EquationsExample A.Suppose two hamburgers and an order of fries cost \$7, andone hamburger and an order of fries cost \$5.Let x = cost of a hamburger, y = cost of an order of fries.We can translate these into the system:{ 2x + y = 7 x+y=5 E1 E2The difference between the order is one hamburger and thedifference between the cost is \$2, so the hamburger is \$2.In symbols, subtract the equations, i.e. E1 – E2: 2x + y = 7 Substitute 2 for x back into E2, ) x+y=5 we get: 2 + y = 5 x+0=2 so x=2
25. 25. Systems of Linear EquationsExample A.Suppose two hamburgers and an order of fries cost \$7, andone hamburger and an order of fries cost \$5.Let x = cost of a hamburger, y = cost of an order of fries.We can translate these into the system:{ 2x + y = 7 x+y=5 E1 E2The difference between the order is one hamburger and thedifference between the cost is \$2, so the hamburger is \$2.In symbols, subtract the equations, i.e. E1 – E2: 2x + y = 7 Substitute 2 for x back into E2, ) x+y=5 we get: 2 + y = 5 or y = 3 x+0=2 so x=2
26. 26. Systems of Linear EquationsExample A.Suppose two hamburgers and an order of fries cost \$7, andone hamburger and an order of fries cost \$5.Let x = cost of a hamburger, y = cost of an order of fries.We can translate these into the system:{ 2x + y = 7 x+y=5 E1 E2The difference between the order is one hamburger and thedifference between the cost is \$2, so the hamburger is \$2.In symbols, subtract the equations, i.e. E1 – E2: 2x + y = 7 Substitute 2 for x back into E2, ) x+y=5 we get: 2 + y = 5 or y = 3 x+0=2 Hence the solution is (2 , 3) or, so x=2 \$2 for a hamburger, \$3 for an order of fries.
27. 27. Systems of Linear EquationsExample B.Suppose four hamburgers and three orders of fries cost \$18,and three hamburger and two order of fries cost \$13. Howmuch does each cost?
28. 28. Systems of Linear EquationsExample B.Suppose four hamburgers and three orders of fries cost \$18,and three hamburger and two order of fries cost \$13.How much does each cost?Let x = cost of a hamburger, y = cost of an order of fries.
29. 29. Systems of Linear EquationsExample B.Suppose four hamburgers and three orders of fries cost \$18,and three hamburger and two order of fries cost \$13.How much does each cost?Let x = cost of a hamburger, y = cost of an order of fries.We translate these into the system:
30. 30. Systems of Linear EquationsExample B.Suppose four hamburgers and three orders of fries cost \$18,and three hamburger and two order of fries cost \$13.How much does each cost?Let x = cost of a hamburger, y = cost of an order of fries.We translate these into the system:
31. 31. Systems of Linear EquationsExample B.Suppose four hamburgers and three orders of fries cost \$18,and three hamburger and two order of fries cost \$13.How much does each cost?Let x = cost of a hamburger, y = cost of an order of fries.We translate these into the system:{4x + 3y = 18 3x + 2y = 13 E1 E2
32. 32. Systems of Linear EquationsExample B.Suppose four hamburgers and three orders of fries cost \$18,and three hamburger and two order of fries cost \$13.How much does each cost?Let x = cost of a hamburger, y = cost of an order of fries.We translate these into the system:{4x + 3y = 18 3x + 2y = 13 E1 E2Subtracting the equations now will not eliminate the x nor y.
33. 33. Systems of Linear EquationsExample B.Suppose four hamburgers and three orders of fries cost \$18,and three hamburger and two order of fries cost \$13.How much does each cost?Let x = cost of a hamburger, y = cost of an order of fries.We translate these into the system:{ 4x + 3y = 18 3x + 2y = 13 E1 E2Subtracting the equations now will not eliminate the x nor y.We have to adjust the equations before we add or subtractthem to eliminate one of the variable.
34. 34. Systems of Linear EquationsExample B.Suppose four hamburgers and three orders of fries cost \$18,and three hamburger and two order of fries cost \$13.How much does each cost?Let x = cost of a hamburger, y = cost of an order of fries.We translate these into the system:{ 4x + 3y = 18 3x + 2y = 13 E1 E2Subtracting the equations now will not eliminate the x nor y.We have to adjust the equations before we add or subtractthem to eliminate one of the variable.Suppose we chose to eliminate the y-terms, we find the LCMof 3y and 2y first. Its 6y.
35. 35. Systems of Linear EquationsExample B.Suppose four hamburgers and three orders of fries cost \$18,and three hamburger and two order of fries cost \$13.How much does each cost?Let x = cost of a hamburger, y = cost of an order of fries.We translate these into the system:{ 4x + 3y = 18 3x + 2y = 13 E1 E2Subtracting the equations now will not eliminate the x nor y.We have to adjust the equations before we add or subtractthem to eliminate one of the variable.Suppose we chose to eliminate the y-terms, we find the LCMof 3y and 2y first. Its 6y. Then we multiply E1 by 2, E2 by (-3),
36. 36. Systems of Linear EquationsExample B.Suppose four hamburgers and three orders of fries cost \$18,and three hamburger and two order of fries cost \$13.How much does each cost?Let x = cost of a hamburger, y = cost of an order of fries.We translate these into the system:{ 4x + 3y = 18 3x + 2y = 13 E1 E2Subtracting the equations now will not eliminate the x nor y.We have to adjust the equations before we add or subtractthem to eliminate one of the variable.Suppose we chose to eliminate the y-terms, we find the LCMof 3y and 2y first. Its 6y. Then we multiply E1 by 2, E2 by (-3)E1*2: 8x + 6y = 36E2*(-3): -9x – 6y = -39
37. 37. Systems of Linear EquationsExample B.Suppose four hamburgers and three orders of fries cost \$18,and three hamburger and two order of fries cost \$13.How much does each cost?Let x = cost of a hamburger, y = cost of an order of fries.We translate these into the system:{ 4x + 3y = 18 3x + 2y = 13 E1 E2Subtracting the equations now will not eliminate the x nor y.We have to adjust the equations before we add or subtractthem to eliminate one of the variable.Suppose we chose to eliminate the y-terms, we find the LCMof 3y and 2y first. Its 6y. Then we multiply E1 by 2, E2 by (-3),and add the results, the y-terms are eliminated.E1*2: 8x + 6y = 36E2*(-3): + ) -9x – 6y = -39
38. 38. Systems of Linear EquationsExample B.Suppose four hamburgers and three orders of fries cost \$18,and three hamburger and two order of fries cost \$13.How much does each cost?Let x = cost of a hamburger, y = cost of an order of fries.We translate these into the system:{ 4x + 3y = 18 3x + 2y = 13 E1 E2Subtracting the equations now will not eliminate the x nor y.We have to adjust the equations before we add or subtractthem to eliminate one of the variable.Suppose we chose to eliminate the y-terms, we find the LCMof 3y and 2y first. Its 6y. Then we multiply E1 by 2, E2 by (-3),and add the results, the y-terms are eliminated.E1*2: 8x + 6y = 36E2*(-3): + ) -9x – 6y = -39 -x = -3  x = 3
39. 39. Systems of Linear EquationsTo find y, set 3 for x in E2: 3x + 2y = 13
40. 40. Systems of Linear EquationsTo find y, set 3 for x in E2: 3x + 2y = 13and get 3(3) + 2y = 13
41. 41. Systems of Linear EquationsTo find y, set 3 for x in E2: 3x + 2y = 13and get 3(3) + 2y = 13 9 + 2y = 13
42. 42. Systems of Linear EquationsTo find y, set 3 for x in E2: 3x + 2y = 13and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y=2
43. 43. Systems of Linear EquationsTo find y, set 3 for x in E2: 3x + 2y = 13and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y=2Hence the solution is (3, 2).
44. 44. Systems of Linear EquationsTo find y, set 3 for x in E2: 3x + 2y = 13and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y=2Hence the solution is (3, 2).Therefore a hamburger cost \$3 and an order of fries cost \$2.
45. 45. Systems of Linear EquationsTo find y, set 3 for x in E2: 3x + 2y = 13and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y=2Hence the solution is (3, 2).Therefore a hamburger cost \$3 and an order of fries cost \$2.The above method is called the elimination (addition) method.We summarize the steps below.
46. 46. Systems of Linear EquationsTo find y, set 3 for x in E2: 3x + 2y = 13and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y=2Hence the solution is (3, 2).Therefore a hamburger cost \$3 and an order of fries cost \$2.The above method is called the elimination (addition) method.We summarize the steps below.1. Select a variable to eliminate.
47. 47. Systems of Linear EquationsTo find y, set 3 for x in E2: 3x + 2y = 13and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y=2Hence the solution is (3, 2).Therefore a hamburger cost \$3 and an order of fries cost \$2.The above method is called the elimination (addition) method.We summarize the steps below.1. Select a variable to eliminate.2. Find the LCM of the terms with that variable, and convertthe corresponding term to the LCM by multiplying eachequation by a number.
48. 48. Systems of Linear EquationsTo find y, set 3 for x in E2: 3x + 2y = 13and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y=2Hence the solution is (3, 2).Therefore a hamburger cost \$3 and an order of fries cost \$2.The above method is called the elimination (addition) method.We summarize the steps below.1. Select a variable to eliminate.2. Find the LCM of the terms with that variable, and convertthe corresponding term to the LCM by multiplying eachequation by a number.3. Add or subtract the adjusted equations to eliminate theselected variable and solve the resulting equation.
49. 49. Systems of Linear EquationsTo find y, set 3 for x in E2: 3x + 2y = 13and get 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y=2Hence the solution is (3, 2).Therefore a hamburger cost \$3 and an order of fries cost \$2.The above method is called the elimination (addition) method.We summarize the steps below.1. Select a variable to eliminate.2. Find the LCM of the terms with that variable, and convertthe corresponding term to the LCM by multiplying eachequation by a number.3. Add or subtract the adjusted equations to eliminate theselected variable and solve the resulting equation.4. Substitute the answer back into any equation to solve for theother variable.
50. 50. Systems of Linear Equations 5x + 4y = 2 E1Example C. Solve { 2x – 3y = -13 E2
51. 51. Systems of Linear Equations 5x + 4y = 2 E1Example C. Solve { 2x – 3y = -13 E2Eliminate the y.
52. 52. Systems of Linear Equations 5x + 4y = 2 E1Example C. Solve { 2x – 3y = -13 E2Eliminate the y. The LCM of the y-terms is 12y.
53. 53. Systems of Linear Equations 5x + 4y = 2 E1Example C. Solve { 2x – 3y = -13 E2Eliminate the y. The LCM of the y-terms is 12y.Multiply E1 by 33*E1: 15x + 12y = 6
54. 54. Systems of Linear Equations 5x + 4y = 2 E1Example C. Solve { 2x – 3y = -13 E2Eliminate the y. The LCM of the y-terms is 12y.Multiply E1 by 3 and E2 by 4.3*E1: 15x + 12y = 64*E2: 8x – 12y = - 52
55. 55. Systems of Linear Equations 5x + 4y = 2 E1Example C. Solve { 2x – 3y = -13 E2Eliminate the y. The LCM of the y-terms is 12y.Multiply E1 by 3 and E2 by 4.3*E1: 15x + 12y = 64*E2: ) 8x – 12y = - 52Add:
56. 56. Systems of Linear Equations 5x + 4y = 2 E1Example C. Solve { 2x – 3y = -13 E2Eliminate the y. The LCM of the y-terms is 12y.Multiply E1 by 3 and E2 by 4.3*E1: 15x + 12y = 64*E2: ) 8x – 12y = - 52Add: 23x + 0 = - 46
57. 57. Systems of Linear Equations 5x + 4y = 2 E1Example C. Solve { 2x – 3y = -13 E2Eliminate the y. The LCM of the y-terms is 12y.Multiply E1 by 3 and E2 by 4.3*E1: 15x + 12y = 64*E2: ) 8x – 12y = - 52Add: 23x + 0 = - 46  x = - 2
58. 58. Systems of Linear Equations 5x + 4y = 2 E1Example C. Solve { 2x – 3y = -13 E2Eliminate the y. The LCM of the y-terms is 12y.Multiply E1 by 3 and E2 by 4.3*E1: 15x + 12y = 64*E2: ) 8x – 12y = - 52Add: 23x + 0 = - 46  x = - 2Set -2 for x in E1 and get 5(-2) + 4y = 2
59. 59. Systems of Linear Equations 5x + 4y = 2 E1Example C. Solve { 2x – 3y = -13 E2Eliminate the y. The LCM of the y-terms is 12y.Multiply E1 by 3 and E2 by 4.3*E1: 15x + 12y = 64*E2: ) 8x – 12y = - 52Add: 23x + 0 = - 46  x = - 2Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2
60. 60. Systems of Linear Equations 5x + 4y = 2 E1Example C. Solve { 2x – 3y = -13 E2Eliminate the y. The LCM of the y-terms is 12y.Multiply E1 by 3 and E2 by 4.3*E1: 15x + 12y = 64*E2: ) 8x – 12y = - 52Add: 23x + 0 = - 46  x = - 2Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y=3
61. 61. Systems of Linear Equations 5x + 4y = 2 E1Example C. Solve { 2x – 3y = -13 E2Eliminate the y. The LCM of the y-terms is 12y.Multiply E1 by 3 and E2 by 4.3*E1: 15x + 12y = 64*E2: ) 8x – 12y = - 52Add: 23x + 0 = - 46  x = - 2Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y=3Hence the solution is (-2, 3).
62. 62. Systems of Linear Equations 5x + 4y = 2 E1Example C. Solve { 2x – 3y = -13 E2Eliminate the y. The LCM of the y-terms is 12y.Multiply E1 by 3 and E2 by 4.3*E1: 15x + 12y = 64*E2: ) 8x – 12y = - 52Add: 23x + 0 = - 46  x = - 2Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y=3Hence the solution is (-2, 3).In all the above examples, we obtain exactly one solution ineach case.
63. 63. Systems of Linear Equations 5x + 4y = 2 E1Example C. Solve { 2x – 3y = -13 E2Eliminate the y. The LCM of the y-terms is 12y.Multiply E1 by 3 and E2 by 4.3*E1: 15x + 12y = 64*E2: ) 8x – 12y = - 52Add: 23x + 0 = - 46  x = - 2Set -2 for x in E1 and get 5(-2) + 4y = 2 -10 + 4y = 2 4y = 12 y=3Hence the solution is (-2, 3).In all the above examples, we obtain exactly one solution ineach case. However, there are two other possibilities; there isno solution or there are infinitely many solutions.
64. 64. Systems of Linear EquationsTwo Special Cases:II. Contradictory (Inconsistent) Systems
65. 65. Systems of Linear EquationsTwo Special Cases:II. Contradictory (Inconsistent) SystemsExample D.{ x+y= 2 x+y=3 (E1) (E2)
66. 66. Systems of Linear EquationsTwo Special Cases:II. Contradictory (Inconsistent) SystemsExample D.{ x+y= 2 x+y=3 (E1) (E2)Remove the x-terms by subtracting the equations.
67. 67. Systems of Linear EquationsTwo Special Cases:II. Contradictory (Inconsistent) SystemsExample D.{ x+y= 2 x+y=3 (E1) (E2)Remove the x-terms by subtracting the equations.E1 – E2 : x + y = 2 ) x+y=3
68. 68. Systems of Linear EquationsTwo Special Cases:II. Contradictory (Inconsistent) SystemsExample D.{ x+y= 2 x+y=3 (E1) (E2)Remove the x-terms by subtracting the equations.E1 – E2 : x + y = 2 ) x+y=3 0 = -1
69. 69. Systems of Linear EquationsTwo Special Cases:II. Contradictory (Inconsistent) SystemsExample D.{ x+y= 2 x+y=3 (E1) (E2)Remove the x-terms by subtracting the equations.E1 – E2 : x + y = 2 ) x+y=3 0 = -1This is impossible! Such systems are said to be contradictoryor inconsistent.
70. 70. Systems of Linear EquationsTwo Special Cases:II. Contradictory (Inconsistent) SystemsExample D.{ x+y= 2 x+y=3 (E1) (E2)Remove the x-terms by subtracting the equations.E1 – E2 : x + y = 2 ) x+y=3 0 = -1This is impossible! Such systems are said to be contradictoryor inconsistent.These system don’t have solution.
71. 71. Systems of Linear EquationsII. Dependent Systems {Example E. x + y = 3 2x + 2y = 6 (E1) (E2)
72. 72. Systems of Linear EquationsII. Dependent Systems {Example E. x + y = 3 2x + 2y = 6 (E1) (E2)Remove the x-terms by multiplying E1 by 2 then subtract E2.
73. 73. Systems of Linear EquationsII. Dependent Systems {Example E. x + y = 3 2x + 2y = 6 (E1) (E2)Remove the x-terms by multiplying E1 by 2 then subtract E2.2*E1: 2x + 2y = 6
74. 74. Systems of Linear EquationsII. Dependent Systems {Example E. x + y = 3 2x + 2y = 6 (E1) (E2)Remove the x-terms by multiplying E1 by 2 then subtract E2.2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6
75. 75. Systems of Linear EquationsII. Dependent Systems {Example E. x + y = 3 2x + 2y = 6 (E1) (E2)Remove the x-terms by multiplying E1 by 2 then subtract E2.2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0= 0
76. 76. Systems of Linear EquationsII. Dependent Systems {Example E. x + y = 3 2x + 2y = 6 (E1) (E2)Remove the x-terms by multiplying E1 by 2 then subtract E2.2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0= 0This means the equations are actually the same and it hasinfinitely many solutions such as (3, 0), (2, 1), (1, 2) etc…
77. 77. Systems of Linear EquationsII. Dependent Systems {Example E. x + y = 3 2x + 2y = 6 (E1) (E2)Remove the x-terms by multiplying E1 by 2 then subtract E2.2*E1 – E2 : 2x + 2y = 6 ) 2x + 2y = 6 0= 0This means the equations are actually the same and it hasinfinitely many solutions such as (3, 0), (2, 1), (1, 2) etc…Such systems are called dependent or redundant systems.
78. 78. Systems of Linear EquationsExercise. A. Select the solution of the system. { x+y=31. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0) 2x + y = 3 { x+y=32. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0) 2x + y = 43. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0) { 2x + y = 5 x+y=34. a. (0, 3) b. (1, 2) c. (2, 1) d.(3, 0) { 2x + y = 6 x+y=3B. 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 = 6C. Add or subtract vertically. (These are warm ups)9. 8x 10. 8 11. –8y 12. –8 13. –4y 14. +4x+) –9x –) –9 –) –9y –) 9 +) –5y –) –5x
79. 79. Systems of Linear EquationsD. Solve the following system by the elimination method.15.{x+y=3 2x + y = 4 16. { x+y=3 2x – y = 6 { 17. x + 2y = 3 2x – y = 618. { –x + 2y = –12 19. { 3x + 4y = 3 20. { x + 3y = 3 2x + y = 4 x – 2y = 6 2x – 9y = –421. { –3x + 2y = –1 22. { 2x + 3y = –1 23. { 4x – 3y = 3 2x + 3y = 5 3x + 4y = 2 3x – 2y = –424. { 5x + 3y = 2 25. { 3x + 4y = –10 26. { –4x + 9y = 1 2x + 4y = –2 –5x + 3y = 7 5x – 2y = 8 3 x– 2 y=3 1 x+1 y=127. 2 { 3 1 x – 1 y = –1 28. 2 { 5 3 x – 1 y = –1 2 4 4 6E. Which system is inconsistent and which is dependent?29. { x + 3y = 4 2x + 6y = 8 { 30. 2x – y = 2 8x – 4y = 6