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# 4.4 system of linear equations 2

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### 4.4 system of linear equations 2

1. 1. Systems of Linear Equations With Three Variables
2. 2. Systems of Linear Equations With Three VariablesTo solve for three unknowns, we need three pieces ofnumerical information about the unknowns,
3. 3. Systems of Linear Equations With Three VariablesTo solve for three unknowns, we need three pieces ofnumerical information about the unknowns, i.e. threesequations.
4. 4. Systems of Linear Equations With Three VariablesTo solve for three unknowns, we need three pieces ofnumerical information about the unknowns, i.e. threesequations. The standard method for solving systems oflinear equations is the elimination method.
5. 5. Systems of Linear Equations With Three VariablesTo solve for three unknowns, we need three pieces ofnumerical information about the unknowns, i.e. threesequations. The standard method for solving systems oflinear equations is the elimination method.We use elimination method to extract a system of twoequations with two unknowns from the system ofthree equations.
6. 6. Systems of Linear Equations With Three VariablesTo solve for three unknowns, we need three pieces ofnumerical information about the unknowns, i.e. threesequations. The standard method for solving systems oflinear equations is the elimination method.We use elimination method to extract a system of twoequations with two unknowns from the system ofthree equations. Solve the system of 2 equations and plugthe answers back to get the third answer.
7. 7. Systems of Linear Equations With Three VariablesTo solve for three unknowns, we need three pieces ofnumerical information about the unknowns, i.e. threesequations. The standard method for solving systems oflinear equations is the elimination method.We use elimination method to extract a system of twoequations with two unknowns from the system ofthree equations. Solve the system of 2 equations and plugthe answers back to get the third answer.This is also the general method for solving a system of Nequations with N unknowns.
8. 8. Systems of Linear Equations With Three VariablesTo solve for three unknowns, we need three pieces ofnumerical information about the unknowns, i.e. threesequations. The standard method for solving systems oflinear equations is the elimination method.We use elimination method to extract a system of twoequations with two unknowns from the system ofthree equations. Solve the system of 2 equations and plugthe answers back to get the third answer.This is also the general method for solving a system of Nequations with N unknowns. We use elimination method toextract a system of (N – 1) equations with (N – 1) unknowns.from the system of N equations.
9. 9. Systems of Linear Equations With Three VariablesTo solve for three unknowns, we need three pieces ofnumerical information about the unknowns, i.e. threesequations. The standard method for solving systems oflinear equations is the elimination method.We use elimination method to extract a system of twoequations with two unknowns from the system ofthree equations. Solve the system of 2 equations and plugthe answers back to get the third answer.This is also the general method for solving a system of Nequations with N unknowns. We use elimination method toextract a system of (N – 1) equations with (N – 1) unknowns.from the system of N equations. Then we extract a system of(N – 2) equations with (N – 2) unknowns from the (N – 1)equations.
10. 10. Systems of Linear Equations With Three VariablesTo solve for three unknowns, we need three pieces ofnumerical information about the unknowns, i.e. threesequations. The standard method for solving systems oflinear equations is the elimination method.We use elimination method to extract a system of twoequations with two unknowns from the system ofthree equations. Solve the system of 2 equations and plugthe answers back to get the third answer.This is also the general method for solving a system of Nequations with N unknowns. We use elimination method toextract a system of (N – 1) equations with (N – 1) unknowns.from the system of N equations. Then we extract a system of(N – 2) equations with (N – 2) unknowns from the (N – 1)equations. Continue this process until we get to and solve asystem of 2 equations.
11. 11. Systems of Linear Equations With Three VariablesTo solve for three unknowns, we need three pieces ofnumerical information about the unknowns, i.e. threesequations. The standard method for solving systems oflinear equations is the elimination method.We use elimination method to extract a system of twoequations with two unknowns from the system ofthree equations. Solve the system of 2 equations and plugthe answers back to get the third answer.This is also the general method for solving a system of Nequations with N unknowns. We use elimination method toextract a system of (N – 1) equations with (N – 1) unknowns.from the system of N equations. Then we extract a system of(N – 2) equations with (N – 2) unknowns from the (N – 1)equations. Continue this process until we get to and solve asystem of 2 equations. Then plug the answers back to getthe other answers.
12. 12. Systems of Linear Equations With Three VariablesExample A. We bought the following items:2 hamburgers, 3 orders of fries and 3 sodas cost \$13.1 hamburger, 2 orders of fries and 2 sodas cost \$8.3 hamburgers, 2 fries, 3 sodas cost \$13.Find the price of each item.
13. 13. Systems of Linear Equations With Three VariablesExample A. We bought the following items:2 hamburgers, 3 orders of fries and 3 sodas cost \$13.1 hamburger, 2 orders of fries and 2 sodas cost \$8.3 hamburgers, 2 fries, 3 sodas cost \$13.Find the price of each item.Let x = cost of a hamburger, y = cost of an order of fries,z = cost of a soda.
14. 14. Systems of Linear Equations With Three VariablesExample A. We bought the following items:2 hamburgers, 3 orders of fries and 3 sodas cost \$13.1 hamburger, 2 orders of fries and 2 sodas cost \$8.3 hamburgers, 2 fries, 3 sodas cost \$13.Find the price of each item.Let x = cost of a hamburger, y = cost of an order of fries,z = cost of a soda. 2x + 3y + 3z = 13 E1
15. 15. Systems of Linear Equations With Three VariablesExample A. We bought the following items:2 hamburgers, 3 orders of fries and 3 sodas cost \$13.1 hamburger, 2 orders of fries and 2 sodas cost \$8.3 hamburgers, 2 fries, 3 sodas cost \$13.Find the price of each item.Let x = cost of a hamburger, y = cost of an order of fries,z = cost of a soda. 2x + 3y + 3z = 13 E1 x + 2y + 2z = 8 E2
16. 16. Systems of Linear Equations With Three VariablesExample A. We bought the following items:2 hamburgers, 3 orders of fries and 3 sodas cost \$13.1 hamburger, 2 orders of fries and 2 sodas cost \$8.3 hamburgers, 2 fries, 3 sodas cost \$13.Find the price of each item.Let x = cost of a hamburger, y = cost of an order of fries,z = cost of a soda.{ 2x + 3y + 3z = 13 E1 x + 2y + 2z = 8 E2 3x + 2y + 3z = 13 E3
17. 17. Systems of Linear Equations With Three VariablesExample A. We bought the following items:2 hamburgers, 3 orders of fries and 3 sodas cost \$13.1 hamburger, 2 orders of fries and 2 sodas cost \$8.3 hamburgers, 2 fries, 3 sodas cost \$13.Find the price of each item.Let x = cost of a hamburger, y = cost of an order of fries,z = cost of a soda.{ 2x + 3y + 3z = 13 E1 x + 2y + 2z = 8 E2 3x + 2y + 3z = 13 E3Select x to eliminate since there is 1x in E2.
18. 18. Systems of Linear Equations With Three VariablesExample A. We bought the following items:2 hamburgers, 3 orders of fries and 3 sodas cost \$13.1 hamburger, 2 orders of fries and 2 sodas cost \$8.3 hamburgers, 2 fries, 3 sodas cost \$13.Find the price of each item.Let x = cost of a hamburger, y = cost of an order of fries,z = cost of a soda.{ 2x + 3y + 3z = 13 E1 x + 2y + 2z = 8 E2 3x + 2y + 3z = 13 E3Select x to eliminate since there is 1x in E2. –2*E 2 + E1:
19. 19. Systems of Linear Equations With Three VariablesExample A. We bought the following items:2 hamburgers, 3 orders of fries and 3 sodas cost \$13.1 hamburger, 2 orders of fries and 2 sodas cost \$8.3 hamburgers, 2 fries, 3 sodas cost \$13.Find the price of each item.Let x = cost of a hamburger, y = cost of an order of fries,z = cost of a soda.{ 2x + 3y + 3z = 13 E1 x + 2y + 2z = 8 E2 3x + 2y + 3z = 13 E3Select x to eliminate since there is 1x in E2. –2*E 2 + E1: –2x – 4y – 4z = -16
20. 20. Systems of Linear Equations With Three VariablesExample A. We bought the following items:2 hamburgers, 3 orders of fries and 3 sodas cost \$13.1 hamburger, 2 orders of fries and 2 sodas cost \$8.3 hamburgers, 2 fries, 3 sodas cost \$13.Find the price of each item.Let x = cost of a hamburger, y = cost of an order of fries,z = cost of a soda.{ 2x + 3y + 3z = 13 E1 x + 2y + 2z = 8 E2 3x + 2y + 3z = 13 E3 Select x to eliminate since there is 1x in E2. –2*E 2 + E1: –2x – 4y – 4z = -16+) 2x + 3y + 3z = 13
21. 21. Systems of Linear Equations With Three VariablesExample A. We bought the following items:2 hamburgers, 3 orders of fries and 3 sodas cost \$13.1 hamburger, 2 orders of fries and 2 sodas cost \$8.3 hamburgers, 2 fries, 3 sodas cost \$13.Find the price of each item.Let x = cost of a hamburger, y = cost of an order of fries,z = cost of a soda.{ 2x + 3y + 3z = 13 E1 x + 2y + 2z = 8 E2 3x + 2y + 3z = 13 E3 Select x to eliminate since there is 1x in E2. –2*E 2 + E1: –2x – 4y – 4z = -16+) 2x + 3y + 3z = 13 0– y – z =–3
22. 22. Systems of Linear Equations With Three VariablesExample A. We bought the following items:2 hamburgers, 3 orders of fries and 3 sodas cost \$13.1 hamburger, 2 orders of fries and 2 sodas cost \$8.3 hamburgers, 2 fries, 3 sodas cost \$13.Find the price of each item.Let x = cost of a hamburger, y = cost of an order of fries,z = cost of a soda.{ 2x + 3y + 3z = 13 E1 x + 2y + 2z = 8 E2 3x + 2y + 3z = 13 E3 Select x to eliminate since there is 1x in E2. –2*E 2 + E1: –3*E 2 + E3: –2x – 4y – 4z = -16+) 2x + 3y + 3z = 13 0– y – z =–3
23. 23. Systems of Linear Equations With Three VariablesExample A. We bought the following items:2 hamburgers, 3 orders of fries and 3 sodas cost \$13.1 hamburger, 2 orders of fries and 2 sodas cost \$8.3 hamburgers, 2 fries, 3 sodas cost \$13.Find the price of each item.Let x = cost of a hamburger, y = cost of an order of fries,z = cost of a soda.{ 2x + 3y + 3z = 13 E1 x + 2y + 2z = 8 E2 3x + 2y + 3z = 13 E3 Select x to eliminate since there is 1x in E2. –2*E 2 + E1: –3*E 2 + E3: –2x – 4y – 4z = -16 –3x – 6y – 6z = –24+) 2x + 3y + 3z = 13 0– y – z =–3
24. 24. Systems of Linear Equations With Three VariablesExample A. We bought the following items:2 hamburgers, 3 orders of fries and 3 sodas cost \$13.1 hamburger, 2 orders of fries and 2 sodas cost \$8.3 hamburgers, 2 fries, 3 sodas cost \$13.Find the price of each item.Let x = cost of a hamburger, y = cost of an order of fries,z = cost of a soda.{ 2x + 3y + 3z = 13 E1 x + 2y + 2z = 8 E2 3x + 2y + 3z = 13 E3 Select x to eliminate since there is 1x in E2. –2*E 2 + E1: –3*E 2 + E3: –2x – 4y – 4z = -16 –3x – 6y – 6z = –24+) 2x + 3y + 3z = 13 +) 3x + 2y + 3z = 13 0– y – z =–3
25. 25. Systems of Linear Equations With Three VariablesExample A. We bought the following items:2 hamburgers, 3 orders of fries and 3 sodas cost \$13.1 hamburger, 2 orders of fries and 2 sodas cost \$8.3 hamburgers, 2 fries, 3 sodas cost \$13.Find the price of each item.Let x = cost of a hamburger, y = cost of an order of fries,z = cost of a soda.{ 2x + 3y + 3z = 13 E1 x + 2y + 2z = 8 E2 3x + 2y + 3z = 13 E3 Select x to eliminate since there is 1x in E2. –2*E 2 + E1: –3*E 2 + E3: –2x – 4y – 4z = -16 –3x – 6y – 6z = –24+) 2x + 3y + 3z = 13 +) 3x + 2y + 3z = 13 0– y – z =–3 0 – 4y – 3z = –11
26. 26. Systems of Linear Equations With Three VariablesExample A. We bought the following items:2 hamburgers, 3 orders of fries and 3 sodas cost \$13.1 hamburger, 2 orders of fries and 2 sodas cost \$8.3 hamburgers, 2 fries, 3 sodas cost \$13.Find the price of each item.Let x = cost of a hamburger, y = cost of an order of fries,z = cost of a soda.{ 2x + 3y + 3z = 13 E1 x + 2y + 2z = 8 E2 3x + 2y + 3z = 13 E3 Select x to eliminate since there is 1x in E2. –2*E 2 + E1: –3*E 2 + E3: –2x – 4y – 4z = -16 –3x – 6y – 6z = –24+) 2x + 3y + 3z = 13 +) 3x + 2y + 3z = 13 0– y – z =–3 0 – 4y – 3z = –11 Group these two equations into a system.
27. 27. Systems of Linear Equations With Three VariablesHence, weve reduced the original system to two equationswith two unknowns: –y–z =–3{–4y – 3z = –11
28. 28. Systems of Linear Equations With Three VariablesHence, weve reduced the original system to two equationswith two unknowns: –y–z =–3 (-1){–4y – 3z = –11
29. 29. Systems of Linear Equations With Three VariablesHence, weve reduced the original system to two equationswith two unknowns: E4{ –y–z =–3 –4y – 3z = –11 (-1) { y +z = 3 4y + 3z = 11 E5
30. 30. Systems of Linear Equations With Three VariablesHence, weve reduced the original system to two equationswith two unknowns: E4{ –y–z =–3 –4y – 3z = –11 (-1) { y +z = 3 4y + 3z = 11 E5To eliminate z we –3*E 4 + E5:
31. 31. Systems of Linear Equations With Three VariablesHence, weve reduced the original system to two equationswith two unknowns: E4{ –y–z =–3 –4y – 3z = –11 (-1) { y +z = 3 4y + 3z = 11 E5To eliminate z we –3*E 4 + E5: –3y – 3z = –9
32. 32. Systems of Linear Equations With Three VariablesHence, weve reduced the original system to two equationswith two unknowns: E4{ –y–z =–3 –4y – 3z = –11 (-1) { y +z = 3 4y + 3z = 11 E5 To eliminate z we –3*E 4 + E5: –3y – 3z = –9+) 4y + 3z = 11
33. 33. Systems of Linear Equations With Three VariablesHence, weve reduced the original system to two equationswith two unknowns: E4{ –y–z =–3 –4y – 3z = –11 (-1) { y +z = 3 4y + 3z = 11 E5 To eliminate z we –3*E 4 + E5: –3y – 3z = –9+) 4y + 3z = 11 y+0 = 2 y= 2
34. 34. Systems of Linear Equations With Three VariablesHence, weve reduced the original system to two equationswith two unknowns: E4{ –y–z =–3 –4y – 3z = –11 (-1) { y +z = 3 4y + 3z = 11 E5 To eliminate z we –3*E 4 + E5: –3y – 3z = –9+) 4y + 3z = 11 y+0 = 2 y= 2 To get z, set 2 for y in E4:
35. 35. Systems of Linear Equations With Three VariablesHence, weve reduced the original system to two equationswith two unknowns: E4{ –y–z =–3 –4y – 3z = –11 (-1) { y +z = 3 4y + 3z = 11 E5 To eliminate z we –3*E 4 + E5: –3y – 3z = –9+) 4y + 3z = 11 y+0 = 2 y= 2 To get z, set 2 for y in E4: 2+z=3
36. 36. Systems of Linear Equations With Three VariablesHence, weve reduced the original system to two equationswith two unknowns: E4{ –y–z =–3 –4y – 3z = –11 (-1) { y +z = 3 4y + 3z = 11 E5 To eliminate z we –3*E 4 + E5: –3y – 3z = –9+) 4y + 3z = 11 y+0 = 2 y= 2 To get z, set 2 for y in E4: 2+z=3 z=1
37. 37. Systems of Linear Equations With Three VariablesHence, weve reduced the original system to two equationswith two unknowns: E4{ –y–z =–3 –4y – 3z = –11 (-1) { y +z = 3 4y + 3z = 11 E5 To eliminate z we –3*E 4 + E5: –3y – 3z = –9+) 4y + 3z = 11 y+0 = 2 y= 2 To get z, set 2 for y in E4: 2+z=3 z=1 For x, set 2 for y , set 1 for z in E2: x + 2y + 2z = 8
38. 38. Systems of Linear Equations With Three VariablesHence, weve reduced the original system to two equationswith two unknowns: E4{ –y–z =–3 –4y – 3z = –11 (-1) { y +z = 3 4y + 3z = 11 E5 To eliminate z we –3*E 4 + E5: –3y – 3z = –9+) 4y + 3z = 11 y+0 = 2 y= 2 To get z, set 2 for y in E4: 2+z=3 z=1 For x, set 2 for y , set 1 for z in E2: x + 2y + 2z = 8 x + 2(2) + 2(1) = 8
39. 39. Systems of Linear Equations With Three VariablesHence, weve reduced the original system to two equationswith two unknowns: E4{ –y–z =–3 –4y – 3z = –11 (-1) { y +z = 3 4y + 3z = 11 E5 To eliminate z we –3*E 4 + E5: –3y – 3z = –9+) 4y + 3z = 11 y+0 = 2 y= 2 To get z, set 2 for y in E4: 2+z=3 z=1 For x, set 2 for y , set 1 for z in E2: x + 2y + 2z = 8 x + 2(2) + 2(1) = 8 x+6 =8
40. 40. Systems of Linear Equations With Three VariablesHence, weve reduced the original system to two equationswith two unknowns: E4{ –y–z =–3 –4y – 3z = –11 (-1) { y +z = 3 4y + 3z = 11 E5 To eliminate z we –3*E 4 + E5: –3y – 3z = –9+) 4y + 3z = 11 y+0 = 2 y= 2 To get z, set 2 for y in E4: 2+z=3 z=1 For x, set 2 for y , set 1 for z in E2: x + 2y + 2z = 8 x + 2(2) + 2(1) = 8 x+6 =8 x=2
41. 41. Systems of Linear Equations With Three VariablesHence, weve reduced the original system to two equationswith two unknowns: E4{ –y–z =–3 –4y – 3z = –11 (-1) { y +z = 3 4y + 3z = 11 E5 To eliminate z we –3*E 4 + E5: –3y – 3z = –9+) 4y + 3z = 11 y+0 = 2 y= 2 To get z, set 2 for y in E4: 2+z=3 z=1 For x, set 2 for y , set 1 for z in E2: x + 2y + 2z = 8 x + 2(2) + 2(1) = 8 x+6 =8 x=2 Hence the solution is (2, 2, 1).