3.5 solving systems of equations in three variables

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.

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 = 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.

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 = 213x + 2y – 5z = –302x – 5y + 2z = –6

Step 1

Rewrite the system as two smaller systems, each containing two of the three equations.

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

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) 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)

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 y – 10z = –48

11y – 23z = –93 –11y + 110z = 528

87z = 435 z = 5

y – 10(5) = –48 y – 50 = –48

y = 2

(–11)

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 = 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

Step 5

CHECK the solution in ALL 3 of the original equations.

Write the solution as an ordered triple.

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).

It is very helpful to neatly organize yourwork on your paper in the following manner.

(x, y, z)

Try this one.

x – 6y – 2z = –8–x + 5y + 3z = 23x – 2y – 4z = 18

(4, 3, –3)

Here’s another one to try.

–5x + 3y + z = –1510x + 2y + 8z = 1815x + 5y + 7z = 9

(1, –4, 2)