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7 January 2011

7 January 2011. Precalculus. Solve the system of equations x + 3y = 18 and –x + 2y = 7. (5,3) (-3,5) (5,-3) (3,5) (-3,-5). 360. 0 of 30. Notes: Solving Systems of Equations 1/7. Only two options are available if you have a system of equations with three variables. Three Variables.

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7 January 2011

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  1. 7 January 2011 Precalculus

  2. Solve the system of equationsx + 3y = 18 and –x + 2y = 7 • (5,3) • (-3,5) • (5,-3) • (3,5) • (-3,-5) 360 0 of 30

  3. Notes: Solving Systems of Equations 1/7 Only two options are available if you have a system of equations with three variables. Three Variables

  4. Notes: Solving Systems of Equations 1/7 Only two options are available if you have a system of equations with three variables. - Elimination Three Variables

  5. Notes: Solving Systems of Equations 1/7 Only two options are available if you have a system of equations with three variables. - Elimination - Substitution Three Variables

  6. Notes: Solving Systems of Equations 1/7 Solve using elimination x – 2y + z = 15 2x + 3y – 3z = 1 4x + 10y – 5z = -3 Elimination

  7. Notes: Solving Systems of Equations 1/7 x – 2y + z = 15 2x + 3y – 3z = 1 4x + 10y – 5z = -3 *HINT* the first equation has x by itself use that to your advantage Use Elimination here Elimination

  8. Notes: Solving Systems of Equations 1/7 x – 2y + z = 15 2x + 3y – 3z = 1 4x + 10y – 5z = -3 *HINT* the first equation has x by itself use that to your advantage -2(x – 2y + z = 15) = Elimination

  9. Notes: Solving Systems of Equations 1/7 x – 2y + z = 15 2x + 3y – 3z = 1 4x + 10y – 5z = -3 *HINT* the first equation has x by itself use that to your advantage -2(x – 2y + z = 15) = Elimination

  10. Notes: Solving Systems of Equations 1/7 x – 2y + z = 15 2x + 3y – 3z = 1 4x + 10y – 5z = -3 *HINT* the first equation has x by itself use that to your advantage -2(x – 2y + z = 15) = -2x + 4y – 2z = -30 Elimination

  11. Notes: Solving Systems of Equations 1/7 x – 2y + z = 15 2x + 3y – 3z = 1 4x + 10y – 5z = -3 *HINT* the first equation has x by itself use that to your advantage -2(x – 2y + z = 15) = -2x + 4y – 2z = -30 2x + 3y – 3z = 1__ Elimination

  12. Notes: Solving Systems of Equations 1/7 x – 2y + z = 15 2x + 3y – 3z = 1 4x + 10y – 5z = -3 *HINT* the first equation has x by itself use that to your advantage -2(x – 2y + z = 15) = -2x + 4y – 2z = -30 2x + 3y – 3z = 1__ 7y – 5z = -29 Elimination

  13. Notes: Solving Systems of Equations 1/7 x – 2y + z = 15 2x + 3y – 3z = 1 4x + 10y – 5z = -3 *HINT* the first equation has x by itself use that to your advantage -2(x – 2y + z = 15) = -2x + 4y – 2z = -30 2x + 3y – 3z = 1__ 7y – 5z = -29 we will use this later Elimination

  14. Notes: Solving Systems of Equations 1/7 x – 2y + z = 15 2x + 3y – 3z = 1 4x + 10y – 5z = -3 *HINT* the first equation has x by itself use that to your advantage -4(x – 2y + z = 15 Elimination Use Elimination here

  15. Notes: Solving Systems of Equations 1/7 x – 2y + z = 15 2x + 3y – 3z = 1 4x + 10y – 5z = -3 *HINT* the first equation has x by itself use that to your advantage -4(x – 2y + z = 15) = Elimination

  16. Notes: Solving Systems of Equations 1/7 x – 2y + z = 15 2x + 3y – 3z = 1 4x + 10y – 5z = -3 *HINT* the first equation has x by itself use that to your advantage -4(x – 2y + z = 15) = -4x + 8y – 4z = -60 Elimination

  17. Notes: Solving Systems of Equations 1/7 x – 2y + z = 15 2x + 3y – 3z = 1 4x + 10y – 5z = -3 *HINT* the first equation has x by itself use that to your advantage -4(x – 2y + z = 15) = -4x + 8y – 4z = -60 4x + 10y – 5z = -3 Elimination

  18. Notes: Solving Systems of Equations 1/7 x – 2y + z = 15 2x + 3y – 3z = 1 4x + 10y – 5z = -3 *HINT* the first equation has x by itself use that to your advantage -4(x – 2y + z = 15) = -4x + 8y – 4z = -60 4x + 10y – 5z = -3 18y – 9z = -63 Elimination

  19. Notes: Solving Systems of Equations 1/7 x – 2y + z = 15 2x + 3y – 3z = 1 4x + 10y – 5z = -3 *HINT* the first equation has x by itself use that to your advantage -4(x – 2y + z = 15) = -4x + 8y – 4z = -60 4x + 10y – 5z = -3 18y – 9z = -63 we will use this later Elimination

  20. Notes: Solving Systems of Equations 1/7 Line up the two equations you already found 7y – 5z = -29 18y – 9z = -63 Elimination

  21. Notes: Solving Systems of Equations 1/7 Multiply the first equation by whatever number is in front of the z term of the second equation 7y – 5z = -29 18y – 9z = -63 Elimination

  22. Notes: Solving Systems of Equations 1/7 Multiply the first equation by whatever number is in front of the z term of the second equation 7y – 5z = -29 18y – 9z = -63 -9(7y – 5z = -29) = -63y + 45z = 261 Elimination

  23. Notes: Solving Systems of Equations 1/7 Multiply the second equation by the opposite of whatever number is in front of the z term of the first equation 7y – 5z = -29 18y – 9z = -63 -9(7y – 5z = -29) = -63y + 45z = 261 5(18y – 9z = -63) = 90y – 45z = -315 Elimination

  24. Notes: Solving Systems of Equations 1/7 Multiply the second equation by the opposite of whatever number is in front of the z term of the first equation 7y – 5z = -29 18y – 9z = -63 -9(7y – 5z = -29) = -63y + 45z = 261 5(18y – 9z = -63) = 90y – 45z = -315 27y = -54 y = -2 Elimination

  25. Notes: Solving Systems of Equations 1/7 Using one of the equations you already found, substitute -2 for y 7y – 5z = -29 7(-2) – 5z = -29 y = -2 Elimination

  26. Notes: Solving Systems of Equations 1/7 Using the equations you already found, substitute -2 for y 7y – 5z = -29 7(-2) – 5z = -29 -14 – 5z = -29 y = -2 Elimination

  27. Notes: Solving Systems of Equations 1/7 Using the equations you already found, substitute -2 for y 7y – 5z = -29 7(-2) – 5z = -29 -14 – 5z = -29 - 5z = -15 y = -2 Elimination

  28. Notes: Solving Systems of Equations 1/7 Using the equations you already found, substitute -2 for y 7y – 5z = -29 7(-2) – 5z = -29 -14 – 5z = -29 - 5z = -15 z = 3 y = -2, z = 3 Elimination

  29. Notes: Solving Systems of Equations 1/7 Going all the way back to one of the original three equations, substitute in the two values you know x – 2y + z = 15 y = -2, z = 3 Elimination

  30. Notes: Solving Systems of Equations 1/7 Going all the way back to one of the original three equations, substitute in the two values you know x – 2y + z = 15 x – 2(-2) + 3 = 15 y = -2, z = 3 Elimination

  31. Notes: Solving Systems of Equations 1/7 Going all the way back to one of the original three equations, substitute in the two values you know x – 2y + z = 15 x – 2(-2) + 3 = 15 x + 4 + 3 = 15 y = -2, z = 3 Elimination

  32. Notes: Solving Systems of Equations 1/7 Going all the way back to one of the original three equations, substitute in the two values you know x – 2y + z = 15 x – 2(-2) + 3 = 15 x + 4 + 3 = 15 x + 7 = 15 y = -2, z = 3 Elimination

  33. Notes: Solving Systems of Equations 1/7 Going all the way back to one of the original three equations, substitute in the two values you know x – 2y + z = 15 x – 2(-2) + 3 = 15 x + 4 + 3 = 15 x + 7 = 15 x = 8 y = -2, z = 3 Elimination

  34. Notes: Solving Systems of Equations 1/7 Going all the way back to one of the original three equations, substitute in the two values you know x – 2y + z = 15 x – 2(-2) + 3 = 15 x + 4 + 3 = 15 x + 7 = 15 x = 8 x = 8, y = -2, z = 3 Elimination

  35. Notes: Solving Systems of Equations 1/7 x = 8, y = -2, z = 3 Final Answer  (8, -2, 3) (Yay!! We’re done with 1 problem!!) Elimination

  36. Notes: Solving Systems of Equations 1/7 x = 4, y = 5, z = -2 Final Answer  (4,5,-2) (Yay!!! Two questions answered!!!) Substitution

  37. Notes: Solving Systems of Equations 1/7 Solve for an easy variable 4x + 8z = 0  this is easy 3x – 2y + z = 0 -2x + y – z = -1 ignore these two Substitution

  38. Notes: Solving Systems of Equations 1/7 Solve for an easy variable 4x + 8z = 0  this is easy Substitution

  39. Notes: Solving Systems of Equations 1/7 Solve for an easy variable 4x + 8z = 0  this is easy 4x = -8z Substitution

  40. Notes: Solving Systems of Equations 1/7 Solve for an easy variable 4x + 8z = 0  this is easy 4x = -8z x = -2z Substitution

  41. Notes: Solving Systems of Equations 1/7 Replace every x with -2z in the original equations 3x– 2y + z = 0 -2x + y – z = -1 Substitution

  42. Notes: Solving Systems of Equations 1/7 Replace every x with -2z in the original equations 3(-2z) – 2y + z = 0 -2(-2z) + y – z = -1 Substitution

  43. Notes: Solving Systems of Equations 1/7 Simplify both equations 3(-2z) – 2y + z = 0  -6z – 2y + z = 0 -2(-2z) + y – z = -1 Substitution

  44. Notes: Solving Systems of Equations 1/7 Simplify both equations 3(-2z) – 2y + z = 0  -6z – 2y + z = 0 - 2y – 5z = 0 -2(-2z) + y – z = -1 Substitution

  45. Notes: Solving Systems of Equations 1/7 Simplify both equations 3(-2z) – 2y + z = 0  -6z – 2y + z = 0 - 2y – 5z = 0 -2(-2z) + y – z = -1  4z + y – z = -1 Substitution

  46. Notes: Solving Systems of Equations 1/7 Simplify both equations 3(-2z) – 2y + z = 0  -6z – 2y + z = 0 - 2y – 5z = 0 -2(-2z) + y – z = -1  4z + y – z = -1 y + 3 = -1 Substitution

  47. Notes: Solving Systems of Equations 1/7 Solve y + 3z = -1 for y y = -3z – 1 Substitute y = -3z – 1 into -2y – 5z = 0 -2y – 5z = 0 -2(-3z – 1) – 5z = 0 6z + 2 – 5z = 0 z = -2 z = -2 Substitution

  48. Notes: Solving Systems of Equations 1/7 Substitute z = -2 into the following two equations y = -3z – 1 x = -2z y = -3(-2) – 1 z = -2 Substitution

  49. Notes: Solving Systems of Equations 1/7 Substitute z = -2 into the following two equations y = -3z – 1 x = -2z y = -3(-2) – 1 y = 6 – 1 z = -2 Substitution

  50. Notes: Solving Systems of Equations 1/7 Substitute z = -2 into the following two equations y = -3z – 1 x = -2z y = -3(-2) – 1 y = 6 – 1 y = 5 y = 5, z = -2 Substitution

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