4.4 system of linear equations 2
- 1. Systems of Linear Equations With Three Variables
- 2. Systems of Linear Equations With Three Variables To solve for three unknowns, we need three pieces of numerical information about the unknowns,
- 3. Systems of Linear Equations With Three Variables To solve for three unknowns, we need three pieces of numerical information about the unknowns, i.e. three sequations.
- 4. Systems of Linear Equations With Three Variables To solve for three unknowns, we need three pieces of numerical information about the unknowns, i.e. three sequations. The standard method for solving systems of linear equations is the elimination method.
- 5. Systems of Linear Equations With Three Variables To solve for three unknowns, we need three pieces of numerical information about the unknowns, i.e. three sequations. The standard method for solving systems of linear equations is the elimination method. We use elimination method to extract a system of two equations with two unknowns from the system of three equations.
- 6. Systems of Linear Equations With Three Variables To solve for three unknowns, we need three pieces of numerical information about the unknowns, i.e. three sequations. The standard method for solving systems of linear equations is the elimination method. We use elimination method to extract a system of two equations with two unknowns from the system of three equations. Solve the system of 2 equations and plug the answers back to get the third answer.
- 7. Systems of Linear Equations With Three Variables To solve for three unknowns, we need three pieces of numerical information about the unknowns, i.e. three sequations. The standard method for solving systems of linear equations is the elimination method. We use elimination method to extract a system of two equations with two unknowns from the system of three equations. Solve the system of 2 equations and plug the answers back to get the third answer. This is also the general method for solving a system of N equations with N unknowns.
- 8. Systems of Linear Equations With Three Variables To solve for three unknowns, we need three pieces of numerical information about the unknowns, i.e. three sequations. The standard method for solving systems of linear equations is the elimination method. We use elimination method to extract a system of two equations with two unknowns from the system of three equations. Solve the system of 2 equations and plug the answers back to get the third answer. This is also the general method for solving a system of N equations with N unknowns. We use elimination method to extract a system of (N – 1) equations with (N – 1) unknowns. from the system of N equations.
- 9. Systems of Linear Equations With Three Variables To solve for three unknowns, we need three pieces of numerical information about the unknowns, i.e. three sequations. The standard method for solving systems of linear equations is the elimination method. We use elimination method to extract a system of two equations with two unknowns from the system of three equations. Solve the system of 2 equations and plug the answers back to get the third answer. This is also the general method for solving a system of N equations with N unknowns. We use elimination method to extract 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. Systems of Linear Equations With Three Variables To solve for three unknowns, we need three pieces of numerical information about the unknowns, i.e. three sequations. The standard method for solving systems of linear equations is the elimination method. We use elimination method to extract a system of two equations with two unknowns from the system of three equations. Solve the system of 2 equations and plug the answers back to get the third answer. This is also the general method for solving a system of N equations with N unknowns. We use elimination method to extract 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 a system of 2 equations.
- 11. Systems of Linear Equations With Three Variables To solve for three unknowns, we need three pieces of numerical information about the unknowns, i.e. three sequations. The standard method for solving systems of linear equations is the elimination method. We use elimination method to extract a system of two equations with two unknowns from the system of three equations. Solve the system of 2 equations and plug the answers back to get the third answer. This is also the general method for solving a system of N equations with N unknowns. We use elimination method to extract 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 a system of 2 equations. Then plug the answers back to get the other answers.
- 12. Systems of Linear Equations With Three Variables Example 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. Systems of Linear Equations With Three Variables Example 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. Systems of Linear Equations With Three Variables Example 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. Systems of Linear Equations With Three Variables Example 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. Systems of Linear Equations With Three Variables Example 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. Systems of Linear Equations With Three Variables Example 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.
- 18. Systems of Linear Equations With Three Variables Example 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:
- 19. Systems of Linear Equations With Three Variables Example 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
- 20. Systems of Linear Equations With Three Variables Example 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. Systems of Linear Equations With Three Variables Example 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. Systems of Linear Equations With Three Variables Example 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. Systems of Linear Equations With Three Variables Example 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. Systems of Linear Equations With Three Variables Example 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. Systems of Linear Equations With Three Variables Example 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. Systems of Linear Equations With Three Variables Example 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. Systems of Linear Equations With Three Variables Hence, we've reduced the original system to two equations with two unknowns: –y–z =–3 {–4y – 3z = –11
- 28. Systems of Linear Equations With Three Variables Hence, we've reduced the original system to two equations with two unknowns: –y–z =–3 (-1) {–4y – 3z = –11
- 29. Systems of Linear Equations With Three Variables Hence, we've reduced the original system to two equations with two unknowns: E4 { –y–z =–3 –4y – 3z = –11 (-1) { y +z = 3 4y + 3z = 11 E5
- 30. Systems of Linear Equations With Three Variables Hence, we've reduced the original system to two equations with 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:
- 31. Systems of Linear Equations With Three Variables Hence, we've reduced the original system to two equations with 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
- 32. Systems of Linear Equations With Three Variables Hence, we've reduced the original system to two equations with 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. Systems of Linear Equations With Three Variables Hence, we've reduced the original system to two equations with 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. Systems of Linear Equations With Three Variables Hence, we've reduced the original system to two equations with 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. Systems of Linear Equations With Three Variables Hence, we've reduced the original system to two equations with 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. Systems of Linear Equations With Three Variables Hence, we've reduced the original system to two equations with 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. Systems of Linear Equations With Three Variables Hence, we've reduced the original system to two equations with 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. Systems of Linear Equations With Three Variables Hence, we've reduced the original system to two equations with 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. Systems of Linear Equations With Three Variables Hence, we've reduced the original system to two equations with 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. Systems of Linear Equations With Three Variables Hence, we've reduced the original system to two equations with 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. Systems of Linear Equations With Three Variables Hence, we've reduced the original system to two equations with 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).