3.5 solving systems of equations in three variables
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Transcript of 3.5 solving systems of equations in three variables
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3.5 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.
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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.
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Is (–3, 2, 4) a solution of this system?
3x + 2y + 4z = 112x – y + 3z = 45x – 3y + 5z = –1
3(–3) + 2(2) + 4(4) = 112(–3) – 2 + 3(4) = 45(–3) – 3(2) + 5(4) = –1
Yes, it is a solution to the system because it is a solution to all 3
equations.
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Methods Used to Solve Systems in 3 Variables
1. Substitution
2. Elimination
3. Cramer’s Rule
4. Gauss-Jordan Method
….. And others
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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.
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This lesson will focus on the
Elimination Method.
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Use elimination to solve the following system of equations.
x – 3y + 6z = 213x + 2y – 5z = –302x – 5y + 2z = –6
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Step 1
Rewrite the system as two smaller systems, each containing two of the three equations.
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x – 3y + 6z = 213x + 2y – 5z = –302x – 5y + 2z = –6
x – 3y + 6z = 21 x – 3y + 6z = 213x + 2y – 5z = –30 2x – 5y + 2z = –6
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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.
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(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)
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Step 3
Write the resulting equations in two variables together as a system of equations.
Solve the system for the two remaining variables.
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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)
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Step 4
Substitute the value of the variables from the system of two equations in one of the ORIGINAL equations with three variables.
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x – 3y + 6z = 213x + 2y – 5z = –302x – 5y + 2z = –6
I choose the first equation.
x – 3(2) + 6(5) = 21x – 6 + 30 = 21 x + 24 = 21
x = –3
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Step 5
CHECK the solution in ALL 3 of the original equations.
Write the solution as an ordered triple.
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x – 3y + 6z = 213x + 2y – 5z = –302x – 5y + 2z = –6
–3 – 3(2) + 6(5) = 213(–3) + 2(2) – 5(5) = –302(–3) – 5(2) + 2(5) = –6
The solution is (–3, 2, 5).
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It is very helpful to neatly organize yourwork on your paper in the following manner.
(x, y, z)
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Try this one.
x – 6y – 2z = –8–x + 5y + 3z = 23x – 2y – 4z = 18
(4, 3, –3)
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Here’s another one to try.
–5x + 3y + z = –1510x + 2y + 8z = 1815x + 5y + 7z = 9
(1, –4, 2)