Solving 3x3 Systems of Linear Equations

Instructor: Mia Primas

Mia has taught math and science and has a Master's Degree in Secondary Teaching.

Have you ever solved an equation with three variables in it? In this lesson, you will learn how to solve systems of equations that have three equations and three variables.

What Is a 3x3 System of Linear Equations?

To define a 3x3 system of linear equations we need to understand what each part of the term means. Linear equations form straight lines when they are graphed. They have a degree of one, meaning that the variables have an exponent no greater than one. A system of equations has two or more equations that are solved simultaneously. When a system of equations is 3x3, it has three equations and three variables. The goal of solving a system of equations is to find a value for each of the variables that satisfies all of the equations. In a 3x3 system of linear equations, we need to find a value for each of the three variables that makes each equation true.

Solving with the Substitution Method

In the example, we see how an expression from one equation can be substituted for a variable in another equation. The goal is to have an equation with one variable that we can solve for. Once we find the value of one variable, we can use it to solve for the others.


Example of a 3x3 system of linear equations
system1

The first equation tells us that x is equal to the expression 2y + 1. We can substitute this expression for x in the other equations. The second equation becomes z = -3(2y + 1) = -6y - 3. The expression -6y - 3 can be substituted for z in the third equation. After substituting the expressions for x and z, the third equation becomes (2y + 1) + 3y - (-6y - 3) = 4.


Applying the substitution method
system2

Now that we have one equation with just one variable, we can solve for the variable y.


Solving for the variable y
system3

Since y = 0, we can substitute it into the other equations and solve for the other variables. For the first equation, we get x = 2(0) + 1 = 1. Substituting one for x in the second equation gives us z = -3(1) = -3. We have now found the values for all three variables.

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