**College Algebra**

**Tutorial 50: **** Solving Systems of
**

**Learning Objectives**

After completing this tutorial, you should be able to:

- Solve a system of linear equations in three variables by the elimination method.

**Introduction**

In this tutorial we will be specifically looking at
systems that have
three linear equations and three unknowns. In **Tutorial
49: Solving Systems of Linear Equations in Two Variables** we
covered
systems that have two linear equations and two unknowns. We will
only look at solving them using the elimination method. Don't get
overwhelmed by the length of some of these problems. Just keep in
mind that a lot of the steps are just like the ones from the
elimination
method of two equations and two unknowns that were covered in Tutorial
49: Solving Systems of Linear Equations in Two Variables, just more of
them.

** Tutorial**

**System of Linear Equations**

In this tutorial, we will be looking at systems that have three linear equations and three unknowns.

**Tutorial 49:
Solving a System
of Linear Equations in Two Variables** looked at three ways to
solve
linear equations in two variables.

In other words, it is what they all three have in common. So if an ordered triple is a solution to one equation, but not another, then it is NOT a solution to the system.

Note that the linear equations in two variables found in Tutorial 49: Solving a System of Linear Equations in Two Variables graphed to be a line on a two dimensional Cartesian coordinate system. If you were to graph (which you won't be asked to do here) a linear equation in three variables you would end up with a figure of a plane in a three dimensional coordinate system. An example of what a plane would look like is a floor or a desk top.

Recall the following from

**Tutorial 49: Solving
a System
of Linear Equations in Two Variables:**

A **consistent system** is a system that has **at
least one solution.**

An** inconsistent system** is a system that **has
no solution**.

The equations of a system are **dependent** if ALL
the solutions
of one equation are also solutions of the other two equations. In
other words, they end up being **the same line**.

The equations of a system are **independent** if
they **do not share
ALL solutions**. They can have one point in common, just not
all
of them.

If you do get one solution for your final answer,** is
this system consistent or inconsistent?**

If you said consistent, give yourself a pat on the back!

If you do get one solution for your final answer, **would
the equations be dependent or independent?**

If you said independent, you are correct!

If you get no solution for your final answer, **is
this system consistent or inconsistent?**

If you said inconsistent, you are right!

If you get no solution for your final answer, **would
the equations be dependent or independent?**

If you said independent, you are correct!

If you get an infinite number of solutions for
your final answer, **is
this system consistent or inconsistent?**

If you said consistent you are right!

If you get an infinite number of solutions for
your final answer, **would
the equations be dependent or independent?**

If you said dependent you are correct!

Note that there is more than one way that you can solve
this type of
system. Elimination (or addition) method is one of the more
common
ways of doing it. So I choose to show it this way.

If you have another way of doing it, by all means, do it your way, then you can check your final answers with mine. No matter which way you choose to do it, if you are doing it correctly, the answer is going have to be the same.

This would involve things like removing ( ) and
removing fractions.

To remove ( ): just use the distributive property.

To remove fractions: since fractions are another way to write division, and the inverse of divide is to multiply, you remove fractions by multiplying both sides by the LCD of all of your fractions.

This works in the same manner as eliminating a variable
with two linear
equations and two variables as shown in **Tutorial
49: Solving a System of Linear Equations in Two Variables.**

**At this point, you are only
working with two
of your equations. In the next step you will incorporate the
third
equation into the mix. **

Looking ahead, **you will be adding these two
equations together**.
In that process you need to make sure that one of the variables drops
out,
leaving one equation and two unknowns. The only way you can
guarantee
that is if you are **adding opposites**. The sum of opposites
is 0.

**It doesn't matter which variable you choose to drop
out.**
You want to keep it as simple as possible. If a variable already
has opposite coefficients than go right to adding the two equations
together.
If they don't, you need to multiply one or both equations by a number
that
will create opposite coefficients in one of your variables. You
can
think of it like a LCD. Think about what number the original
coefficients
both go into and multiply each separate equation accordingly.
Make
sure that one variable is positive and the other is negative before you
add.

For example, if you had a 2*x* in one equation
and a 3*x* in another equation, you could
multiply
the first equation by 3 and get 6*x* and
the
second equation by -2 to get a -6*x*.
So
when you go to add these two together they will drop out.

Basically, you are going to do another elimination
step, eliminating
the same variable we did in step 2, just with a different pair of
equations.
For example, if you used equations 1 and 3 in step 2, then you
can
use either 1 and 2 OR 2 and 3 in this step. As long as you are
using
a different combination of equations you are ok. This will get
that
third equation into the mix. We need to do this to give us two
equations
to go with our two unknowns that are left after the first elimination.

After steps 2 and 3, there will be two equations and
two unknowns which
is exactly what was shown how to solve in **Tutorial
49: Solving a System of Linear Equations in Two variables.**
You can use any method you want to solve it. **Generally,
I find it easiest to use the elimination method, if that is what I used
in steps 2 and 3.** It is less confusing that way.

When you solve this system that has two equations and two variables, you will have the values for two of your variables.

**Remember that** i**f both variables drop out and
you have a FALSE
statement, that means your answer is no solution. **

**If both variables drop out and you have a TRUE
statement, that means
your answer is infinite solutions, which would be the equation of the
line.**

You can plug in the proposed solution into ALL THREE
equations.
If it makes ALL THREE equations true then you have your solution to the
system.

If it makes at least one of them false, you need to go back and redo the problem.

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 *y* to eliminate. I need
to do
this with ANY pair of equations.

**Let’s first eliminate y using the first
and second equations. ** This process is identically to how we
approached
it with the systems found in

If I multiply 3 times the first equation, then the *y* terms will be opposites of each other and ultimately drop out.

**Multiplying 3 times the first equation and then
adding that to the
second equation we get:**

*** y's
have opposite
coefficients**

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. We
could
use the second and third, as long as we have not used both of the same
ones used in step 2 above.

It looks like we will have to multiply the first
equation by 2, to get
opposites on *y*.

**Multiplying the first equation by 2 and then adding
that to equation
(3) we get:**

*** y's
have opposite
coefficients**

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:**

Now **I’m going to choose ***z* to eliminate.
We can either multiply the first equation by -1 or the second, either
way
will create opposites in front of the *z* terms.

**I’m going to go ahead and multiply equation (5) by -1
and then add
the equations together:**

*** z's
have opposite
coefficients**

*** x is 3/4**

***Plug in 3/4 for x**

***Solve for z**

*** z is 1/2**

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 (1) to plug in our 2 for x and
1 for z that we found:**

***Solve for y**

*** y is -2**

You will find that if you plug the ordered triple (3/4,
-2, 1/2) into
ALL THREE equations of the original system, this is a solution to ALL
THREE
of them.

**(3/4, -2, 1/2) is a solution to our system.**

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 the *z*'s
already have opposite
coefficients, so all we have to do is add these two equations
together.

**Adding equations (2) and (3) together we get:**

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:**

Now **I’m going to choose ***x* to eliminate.
We can either multiply the first equation by -1 or the second, either
way
will create opposites in front of the *z* terms.

**I’m going to go ahead and multiply equation (1) by -1
and then add
the equations together:**

*** x's
and y's
have opposite coefficients**

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 any of 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 + y = 9} OR {(x, y, z)| y + z = 7} OR {(x, y, z)| x - z = 2}.**

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 to get rid of a few parenthesis
and line everything
up.

***Rewrite (1) and (3) in the
form A x + By + Cz =
D**

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. ** This process is identically to how we
approached
it with the systems found in

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:**

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. We
could
use the second and third, as long as we have not used both of the same
ones used in step 2 above.

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:**

*** x's, y's,
and z's have opposite coefficients**

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.**

** Practice Problems**

These are practice problems to help bring you to the next level.
It will allow you to check and see if you have an understanding of these
types of problems. **Math works just like anything
else, if you want to get good at it, then you need to practice it.
Even the best athletes and musicians had help along the way and lots of
practice, practice, practice, to get good at their sport or instrument.**
In fact there is no such thing as too much practice.

To get the most out of these, **you should work the problem out on
your own and then check your answer by clicking on the link for the answer/discussion
for that problem**. At the link you will find the answer
as well as any steps that went into finding that answer.

Practice Problems 1a - 1c:Solve each system.

Need Extra Help on these Topics?

The following is a webpage that can assist you in the topics that were covered on this page.

http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut20_systhree.htm

This webpage helps you with solving linear systems with three unknowns.

Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions.

Last revised on April 25, 2011 by Kim Seward.

All contents copyright (C) 2002 - 2011, WTAMU and Kim Seward. All rights reserved.