Systemsof3 equations

Solving Systems of Three Linear Equations in Three

VariablesThe 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)