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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. 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. Is (–3, 2, 4) a solution of this system? 3x + 2y + 4z = 11 3(–3) + 2(2) + 4(4) = 11 P 2x – y + 3z = 4 2(–3) – 2 + 3(4) = 4 P 5x – 3y + 5z = –1 5(–3) – 3(2) + 5(4) = –1P Yes, it is a solution to the system because it is a solution to all 3 equations. Methods Used to Solve Systems in 3 Variables 1. Substitution 2. Elimination 3. Cramer’s Rule 4. Gauss-Jordan Method ….. And others 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. This lesson will focus on the Elimination Method. Use elimination to solve the following system of equations. x – 3y + 6z = 21 3x + 2y – 5z = –30 2x – 5y + 2z = –6 Step 1 Rewrite the system as two smaller systems, each containing two of the three equations. 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 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. (x – 3y + 6z = 21) (–3) (x – 3y + 6z = 21) (–2) 3x + 2y – 5z = –30 2x – 5y + 2z = –6 –3x + 9y – 18z = –63 –2x + 6y – 12z = –42 3x + 2y – 5z = –30 2x – 5y + 2z = –6 11y – 23z = –93 y – 10z = –48 Step 3 Write the resulting equations in two variables together as a system of equations. Solve the system for the two remaining variables. 11y – 23z = –93 (–11) y – 10z = –48 11y – 23z = –93 –11y + 110z = 528 87z = 435 z=5 y – 10(5) = –48 y – 50 = –48 y=2 Step 4 Substitute the value of the variables from the system of two equations in one of the ORIGINAL equations with three variables. 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 Step 5 CHECK the solution in ALL 3 of the original equations. Write the solution as an ordered triple. x – 3y + 6z = 21 P –3 – 3(2) + 6(5) = 21 3x + 2y – 5z = –30 P 3(–3) + 2(2) – 5(5) = –30 2x – 5y + 2z = –6 2(–3) – 5(2) + 2(5) = –6 P The solution is (–3, 2, 5). It is very helpful to neatly organize your work on your paper in the following manner. (x, y, z) Try this one. x – 6y – 2z = –8 –x + 5y + 3z = 2 3x – 2y – 4z = 18 (4, 3, –3) Here’s another one to try. –5x + 3y + z = –15 10x + 2y + 8z = 18 15x + 5y + 7z = 9 (1, –4, 2)