# Linear Equations With Three Variables

These are my notes on linear equations with three variables.

Systems of linear equations can have any number of variables. When writing

systems of linear equations in three variables, we usually use x,y, and z.

\(2x-3y+4z=4\)

This is an example of a system of linear equations in three variables. The

solution to this system can be written as (3,2,1). This system of linear

equations has exactly one solution. In general, systems of linear equations can

have zero, one, or infinitely many solutions.**Example 1**

Determine whether (-1,-3,2) or (1,-10,-13) is the solution to the system of

linear equations.

\(x-4y+2z=15\)

\(4x-y+z=1\)

\(6x-2y-3z=-6\)

Solution: First, substitute x=-1, y=-3, and z=2 in the system of linear

equations and then substitute x=1, y=-10, and z=-13.

\((-1)-4(-3)+2(2)=15\)

\(4(-1)-(-3)+2=1\)

\(6(-1)-2(-3)-3(2)=-6\)

The first combination (-1,-3,2) seems to work out as true statements. **Solving With Elimination and Substitution**

We can solve systems of linear equations in three variables by hand. The

following procedure uses substitution and elimination and assumes that the

variables are x,y, and z.

Step 1: Eliminate one variable, such as x, from two of the equations.

Step 2: Apply the techniques in the previous two sections to solve the two

resulting equations in two variables from Step 1. If x is eliminated, then solve

these equations to find y and z. Note: if there are no solutions for y and z,

then the given system also has no solutions. If there are infinitely many

solutions for y and z, then write y in terms of z and proceed to step 3.

Step 3: Substitute the values for y and z in one of the given equations to find

x. The solution is (x,y,z).**Example 2**

Solve the following system.

\(x-y+2z=6\)

\(2x+y-2z=-3\)

\(-x-2y+3z=7\)

Solution:

Step 1-We begin by eliminating the variable x from the second and third

equations. To eliminate x from the second equation we multiple the first

equation by -2 and then add it to the second equation. To eliminate x from the

third equation, we add the first and third equations.

\(-2x+2y-4z=12\) - First equation times -2

\(2x+y-2z=-3\) - Second equation

\(3y-6z=-15\) - Equations added together

Now, we add the first and third equations together.

\(x-y+2z=6\) - First equation

\(-x-2y+3z=7\) - Third equation

\(-3y+5z=13\) - Equations added together

Step 2 - Take the two resulting equations from Step 1 and eliminate either

variable. Here we add the two equations to eliminate the variable y.

\(3y-6z=-15\)

\(-3y+5z=13\)

\(-z = -2\)

\(z = 2\)

Now we can use substitution to find the value of y. We let z=2 in either

equation used in Step 2 to find y:

\(3y-6z=-15\)

\(3y-6(2)=-15\)

\(3y-12=-15\)

\(3y=-3\)

\(y = -1\)

Step 3 - Finally, we substitute y=-1 and z=2 in any of the given equations to

find x.

\(x-y+2z=6\)

\(x-(-1)+2(2)=6\)

\(x+1+4=6\)

\(x = 1\)

The solution is (1,-1,2)**Systems With No Solutions**

Regardless of the number of variables, a system of linear equations can have

zero, one, or infinitely many solutions. Some systems of linear equations have

infinitely many solutions. In this case, we say the system is consistent, but

the equations are dependent.