College Algebra
Answer/Discussion to Practice Problems
Tutorial 50: Solving Systems of
Linear Equations in Three Variables
Answer/Discussion
to 1a

Note that the numbers in ( ) are equation
numbers. They will be used throughout the problems for reference
purposes.
Basically, we are going to do the same thing we did with the systems
of two equations, just more of it. In other words, we will have to
do the elimination twice, to get down to just one variable since we are
starting with three variables this time. 
No simplification needed here. Let's go on to the next step. 
Let's start by picking our first variable to eliminate. I'm
going to pick z to eliminate. I need
to do this with ANY pair of equations.
Note how equation (1) already has z eliminated.
We can use this as our first equation with z eliminated
Using equation (1) as one of our equations where z is eliminated: 
Now we can't just stop here for of two reasons. First, we would
be stuck because we have one equation and two unknowns. Second, when
we solve a system it has to be a solution of ALL equations involved and
we have not incorporated the third equation yet. Let's do that now. 
We are still going after eliminating z,
this time we want to use the second and the third equations.
It looks like we will have to multiply the second equation by 2, to
get opposites on z.
Multiplying the second equation by 2 and then adding that to equation
(3) we get: 

*Mult. eq. (2) by 2
*z's have opposite
coefficients
*z's dropped out

Putting the two equations that we have found together, we now have
a system of two equations and two unknowns, which we can solve just like
the ones shown in Tutorial 49: Solving a System of Linear Equations in
Two variables. You can use either elimination or substitution.
I'm going to go ahead and stick with the elimination method to complete
this.
Let's first put those equations together: 

*Put equations found in steps 2 and 3
together into one system 
Now I'm going to choose y to eliminate.
It looks like they already have opposite coefficients, so we just need
to add the two equations together.
I'm going to go ahead and add the equations together: 

*y's have opposite
coefficients
*y's dropped out

If we go back one step to the system that had two equations and
two variables and put in 3 for x in what is
labeled equation 1, we would get: 

*Eq. (1)
*Plug in 3 for x
*Solve for y
*y is 1 
Now we need to go back to the original system and pick any equation
to plug in the two known variables and solve for our last variable.
I choose equation (2) to plug in our 3 for x that
we found: 

*Eq. (2)
*Plug in 3 for x
*Solve for z
*z is 2 
You will find that if you plug the ordered triple (3, 1, 2) into ALL
THREE equations of the original system, this is a solution to ALL THREE
of them.
(3, 1, 2) is a solution to our system. 
Answer/Discussion
to 1b

Note that the numbers in ( ) are equation
numbers. They will be used throughout the problems for reference
purposes.
Basically, we are going to do the same thing we did with the systems
of two equations, just more of it. In other words, we will have to
do the elimination twice, to get down to just one variable since we are
starting with three variables this time. 
No simplification needed here. Let's go on to the next step. 
Let's start by picking our first variable to eliminate. I'm going
to choose z to eliminate. I need to do
this with ANY pair of equations.
Let's first eliminate z using the first
and second equations.
If I multiply 1 times the second equation, then the z terms will be opposites of each other and ultimately drop out.
Multiplying 1 times the second equation and then adding that to
the first equation we get: 

*Mult. eq. (2) by 1
*z's have opposite
coefficients
*z's dropped out

Now we can't just stop here for of two reasons. First, we would
be stuck because we have one equation and two unknowns. Second, when
we solve a system it has to be a solution of ALL equations involved and
we have not incorporated the third equation yet. Let's do that now. 
We are still going after eliminating z,
this time we want to use the first and the third equations.
It looks like we will have to multiply the first equation by 2, to get
opposites on z.
Multiplying the first equation by 2 and then adding that to equation
(3) we get: 

*Mult. eq. (1) by 2
*z's have opposite
coefficients
*z's dropped out

Wait a minute, where did our variables go????
As mentioned above, if the variable drops out AND we have a FALSE statement,
then we have no solutions. 
Since we did not get a value for our variables, there is nothing to
solve here. 
Since we did not get a value for our variables, there is nothing to
plug in here. 
There is no value to plug in here.
The answer is no solution. 
Answer/Discussion
to 1c

Note that the numbers in ( ) are equation
numbers. They will be used throughout the problems for reference
purposes.
Basically, we are going to do the same thing we did with the systems
of two equations, just more of it. In other words, we will have to
do the elimination twice, to get down to just one variable since we are
starting with three variables this time. 
It looks like we have some fractions to get rid of.
Multiplying equation (3) by its LCD we get: 

*Multiply eq. (3) by LCD of 2

Let's start by picking our first variable to eliminate. I'm going
to choose y to eliminate. I need to do
this with ANY pair of equations.
Let's first eliminate y using the first
and second equations.
It looks like the y's already have opposite
coefficients so all we need to do is add the two equations and the y's
will drop out.
Adding equations (1) and (2) together we get: 

*y's have opposite
coefficients
*y's dropped out

Now we can't just stop here for of two reasons. First, we would
be stuck because we have one equation and two unknowns. Second, when
we solve a system it has to be a solution of ALL equations involved and
we have not incorporated the third equation yet. Let's do that now. 
We are still going after eliminating y,
this time we want to use the first and the third equations.
It looks like we will have to multiply the third equation by 2, to get
opposites on y.
Multiplying the third equation by 2 and then adding that to equation
(1) we get: 

*Mult. eq. (3) by 2
*y's have opposite
coefficients
*y's dropped out

Putting the two equations that we have found together, we now have
a system of two equations and two unknowns, which we can solve just like
the ones shown in Tutorial 49: Solving a System of Linear Equations in
Two variables. You can use either elimination or substitution.
I'm going to go ahead and stick with the elimination method to complete
this.
Let's first put those equations together: 

*Put equations found in steps 2 and 3
together into one system 
Now I'm going to choose x to eliminate.
It looks like we will have to multiply equation (5) by 1, to get opposites
on x.
Multiplying equation (5) by 1 and then adding that to equation (4)
we get: 

*Mult. eq. (5) by 1
*x's and z's
have opposite coefficients
*x's and z's
dropped out

Wait a minute, where did our variables go????
As mentioned above, if the variable drops out AND we have a TRUE statement,
then when have an infinite number of solutions. They end up being the same
line. 
Since we did not get a value for our variables, there is nothing to
plug in here. 
There is no value to plug in here.
When they end up being the same equation, you have an infinite number
of solutions. You can write up your answer by writing out any of the three
equations to indicate that they are the same equation.
Three ways to write the answer are {(x, y, z) x + 2y + 4z=
3} OR {(x, y, z)
4x  2y  6z = 2} OR {(x, y, z) x  1/2y  3/2z = 1/2}. 
Last revised on April 25, 2011 by Kim Seward.
All contents copyright (C) 2002  2011, WTAMU and Kim Seward. All rights reserved.

