Systemsof3 equations
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This document discusses solving systems of three linear equations in three variables using the elimination method. It provides an example of using elimination to solve the system of equations x - 3y + 6z = 21, 3x + 2y - 5z = -30, and 2x - 5y + 2z = -6. The steps are: 1) rewrite the system as two smaller systems with two equations each, 2) eliminate the same variable from each smaller system, 3) solve the resulting system of two equations for the two remaining variables, 4) substitute back into one of the original equations to find the third variable, and 5) check that the solution satisfies all three original equations. The solution to the example system is (-
Systemsof3 equations
- 1. Solving Systems of Three Linear Equations in Three Variables The Elimination Method SPI 3103.3.8 Solve systems of three linear equations in three variables.
- 2. Solutions of a system with 3 equations The solution to a system of three linear equations in three variables is an ordered triple. (x, y, z) The solution must be a solution of all 3 equations.
- 3. Is (–3, 2, 4) a solution of this system? 3x + 2y + 4z = 11 2x – y + 3z = 4 5x – 3y + 5z = –1 3(–3) + 2(2) + 4(4) = 11 2(–3) – 2 + 3(4) = 4 5(–3) – 3(2) + 5(4) = –1 Yes, it is a solution to the system because it is a solution to all 3 equations.
- 4. Methods Used to Solve Systems in 3 Variables 1. Substitution 2. Elimination 3. Cramer’s Rule 4. Gauss-Jordan Method ….. And others
- 5. Why not graphing? While graphing may technically be used as a means to solve a system of three linear equations in three variables, it is very tedious and very difficult to find an accurate solution. The graph of a linear equation in three variables is a plane.
- 6. This lesson will focus on the Elimination Method.
- 7. Use elimination to solve the following system of equations. x – 3y + 6z = 21 3x + 2y – 5z = –30 2x – 5y + 2z = –6
- 8. Step 1 Rewrite the system as two smaller systems, each containing two of the three equations.
- 9. x – 3y + 6z = 21 3x + 2y – 5z = –30 2x – 5y + 2z = –6 x – 3y + 6z = 21 x – 3y + 6z = 21 3x + 2y – 5z = –30 2x – 5y + 2z = –6
- 10. Step 2 Eliminate THE SAME variable in each of the two smaller systems. Any variable will work, but sometimes one may be a bit easier to eliminate. I choose x for this system.
- 11. (x – 3y + 6z = 21) 3x + 2y – 5z = –30 –3x + 9y – 18z = –63 3x + 2y – 5z = –30 11y – 23z = –93 (x – 3y + 6z = 21) 2x – 5y + 2z = –6 –2x + 6y – 12z = –42 2x – 5y + 2z = –6 y – 10z = –48 (–3) (–2)
- 12. Step 3 Write the resulting equations in two variables together as a system of equations. Solve the system for the two remaining variables.
- 13. 11y – 23z = –93 y – 10z = –48 11y – 23z = –93 –11y + 110z = 528 87z = 435 z = 5 y – 10(5) = –48 y – 50 = –48 y = 2 (–11)
- 14. Step 4 Substitute the value of the variables from the system of two equations in one of the ORIGINAL equations with three variables.
- 15. x – 3y + 6z = 21 3x + 2y – 5z = –30 2x – 5y + 2z = –6 I choose the first equation. x – 3(2) + 6(5) = 21 x – 6 + 30 = 21 x + 24 = 21 x = –3
- 16. Step 5 CHECK the solution in ALL 3 of the original equations. Write the solution as an ordered triple.
- 17. x – 3y + 6z = 21 3x + 2y – 5z = –30 2x – 5y + 2z = –6 –3 – 3(2) + 6(5) = 21 3(–3) + 2(2) – 5(5) = –30 2(–3) – 5(2) + 2(5) = –6 The solution is (–3, 2, 5).
- 18. is very helpful to neatly organize you k on your paper in the following mann (x, y, z)
- 19. Try this one. x – 6y – 2z = –8 –x + 5y + 3z = 2 3x – 2y – 4z = 18 (4, 3, –3)
- 20. Here’s another one to try. –5x + 3y + z = –15 10x + 2y + 8z = 18 15x + 5y + 7z = 9 (1, –4, 2)