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". 0:05:41.900,0:05:39.200 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.