# Algebra 44 – Solving Systems of Equations in Three Variables

Hello. I’m Professor Von Schmohawk
and welcome to Why U. In the last lecture, we saw that just as systems
of two linear equations in two variables have a single unique solution when the two
lines in the system intersect at a single point systems of three linear equations
in three variables have a unique solution when the three planes intersect
at a single point. Likewise, just as systems in two variables
can be solved to find this unique solution using the substitution and elimination methods these same techniques can be used
to solve systems in three variables. There are also powerful methods
for solving systems of linear equations that we will learn in future lectures
which involve the use of matrices. But before we see how to solve a system
in three variables let’s briefly review
how the elimination method works when solving a system in two variables. As we saw in the chapter
“Solving Systems of Equations by Elimination” when using the elimination method
to solve a system in two variables we add a multiple of one equation to the other choosing a multiple which will eliminate
one of the variables. For instance, in this example
we could chose to eliminate the variable y by multiplying the top equation by negative one and then adding the equations. This gives us an equation
with only one variable, x which can then be solved
to find the value of x. Once we know the value of one variable that value can then be substituted back
into either of the original equations to find the value of the other variable. Since in this example
the top equation is the simplest we will use it to find the value of y. Substituting negative one for x
in the top equation we can solve for y by adding one to both sides giving us “y=2”. So the values of the variables
which are the solution for this system are “x=-1” and “y=2”. This solution can also be represented
as the ordered pair (-1,2). Now let’s see how elimination and substitution can be used to solve systems
of three equations in three variables. To solve a system of three equations
in three variables we start by selecting two of the equations and use the elimination method
to eliminate one of the three variables. For instance, let’s use the first two equations
and eliminate the variable y. In this example we can eliminate y
without multiplying either equation by a constant. Adding the equations eliminates y giving us an equation in only two variables
x and z. Now if we can create a second equation
in these same two variables we will have a system
of two equations in two variables which can then be solved to find the values
of x and z. In order to create a system of two equations
in two variables with a unique solution the second equation must be different
from the one we just created so we must use a different pair of equations
than we used to produce the first equation. This time, instead of adding
the first and second equations we could add either the
second and third equations or the first and third equations so let’s use the first and third equations. We want to once again eliminate y
so that we end up with another equation in x and z. We can do this if we first multiply
the top equation by three before adding the equations. This gives us a second equation in x and z. We now have a system of two equations
in two variables which can be solved to find the values
of x and z. Using the elimination method we will first multiply the top equation by eight
so that we can eliminate the x term. Now when the equations are added we get “11z=44” or dividing by eleven, “z=4”. Now that we know the value of z we can substitute that value back
into either of the two equations we just added to find the value of x. We’ll use the first equation
since it has the smallest coefficients. Setting the value of z in that equation to four we get “-x + 8=5”. And subtracting eight from both sides gives us “-x=-3” or “x=3”. Now that we know the values of x and z these values can be substituted back
into any of the three original equations to find the value of the remaining variable, y. We’ll use the first equation
since it’s the simplest. Setting the values of x and z
to three and four we get “y-1=1” or adding one to both sides “y=2”. So the solution for this system is
x=3, y=2, and z=4. This solution can also be represented as
the ordered triple (3,2,4). If we substitute these values for x, y, and z
into the three original equations and complete the arithmetic we will find that these values for x, y, and z
are a common solution to all three equations. This confirms that our answer is correct. Even though we have not graphed
the three planes representing the three linear equations
in this system we know that all three planes
intersect at the point whose x, y, and z coordinates
are the ordered triple (3,2,4). We can see from this example, that solving
a system of three equations in three variables involves more steps than solving a system
of two equations in two variables. However, the basic techniques are the same. We start by choosing two of the three equations
and eliminate a variable giving us an equation in two variables. We then repeat this process
with a different pair of equations giving us a second equation
in the same two variables. We can then use substitution or elimination to solve this system of two equations
in two variables to find the values of those two variables. And once we know the values
of two of the variables we can substitute those values back
into any of the original equations to find the value of the third variable. In this lecture, we have seen how to solve
a system of three equations in three variables which has a single unique solution. However, we have not said how we know
that this solution is unique. Perhaps these three planes intersect in such a way
that there are an infinite number of solutions and this solution is only one of them. In the next lecture
we will see how to determine if a system of three linear equations
in three variables has a single unique solution an infinite number of solutions or no solutions.

### 4 thoughts on “Algebra 44 – Solving Systems of Equations in Three Variables”

• April 18, 2015 at 11:46 pm

Thank you.

• April 20, 2015 at 12:37 pm

Really well done. Love these series of lectures.