**Intermediate Algebra**

**Tutorial 19: Solving Systems of Linear Equations
**

**Learning Objectives**

After completing this tutorial, you should be able to:

- Know if an ordered pair is a solution to a system of linear equations in two variables or not.
- Solve a system of linear equations in two variables by graphing.
- Solve a system of linear equations in two variables by the substitution method.
- Solve a system of linear equations in two variables by the elimination method.

** Introduction**

In this tutorial we will be specifically looking at
systems that have
two equations and two unknowns. **Tutorial 20: Solving Systems of
Linear
Equations in Three Variables** will cover systems that have three
equations
and three unknowns. We will look at solving them three different
ways: graphing, substitution method and elimination method.
This will lead us into solving word problems with systems, which will
be
shown in **Tutorial 21: Systems of Linear Equations and Problem
Solving**.
That is where we get to answer the infamous question, when will we use
this? But first, we have to learn how to work with systems in
general.
That is why we use generic variables like* x* and *y *at this point. If
you know how to work it out in general, then when you have a specific
problem
that you are solving where the variables take on meaning like time or
money
(two things we don't ever seem to have enough of) you will be ready to
go.** ** So, let's go ahead and look at
systems in general to get us ready for the word problems that are ahead
of us.

** Tutorial**

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

In other words, it is where the two graphs intersect,
what they have
in common. So if an ordered pair is a solution to one equation,
but
not the other, then it is NOT a solution to the system.

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 equation. 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!

**The graph below illustrates a system of two equations
and two unknowns
that has one solution:**

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!

**The graph below illustrates a system of two equations
and two unknowns
that has no solution:**

If you get an infinite number of solutions for
your final answer, i**s
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!

**The graph below illustrates a system of two equations
and two unknowns
that has an infinite number of solutions:**

(3, -1) and (0, 2)

***True statement**

So far so good, (3, -1) is a solution to the
first equation *x* + *y* = 2.

**Now, let’s check (3, -1) in the second equation:**

***True statement**

Hey, we ended up with another true statement, which
means (3, -1) is
also a solution to the second equation *x* - *y* = 4.

**Here is the big question, is (3, -1) a solution to
the given system?????**

**Since it was a solution to BOTH equations in the system,
then it
is a solution to the overall system.**

**Now let’s put (0, 2) into the first equation:**

This is a true statement, so (0, 2) is a solution to
the first equation *x *+* y* = 2.

**Finally, let’s put (0,2) into the second equation:**

This time we got a false statement, you know what that
means.
(0, 2) is NOT a solution to the second equation *x* - *y *= 4.

**Here is the big question, is (0, 2) a solution to the
given system?????**

**Since it was not a solution to BOTH equations in the
system, then
it is not a solution to the overall system.**

There are three ways to solve systems of linear
equations in two variables:

Unless the directions tell you differently, you can use
any "legitimate"
way to graph the line. Our tutorials show three different
ways.
Feel free to review back over them if you need to: **Tutorial
12: Graphing Equations shows how to graph by plotting points**, **Tutorial
14: Graphing Linear Equations shows how to graph using intercepts**,
and **Tutorial 16:
Equations
of Lines shows how to graph using the ***y*-intercept
and slope.

You graph the second equation the same as any other
equation.
Refer to the first step if you need to review the different ways to
graph
a line.

The difference here is you will put it on the same coordinate system as the first. It is like having two graphing problems in one.

If the **two lines intersect at one place**, then
the **point of
intersection **is the solution to the system.

If the **two lines are parallel**, then they never
intersect, so
there is **no solution.**

If the** two lines lie on top of each other**, then
they are **the
same line** and you have an **infinite number of solutions.**.
In this case you can write down either equation as the solution to
indicate
they are the same line.

You can plug in the proposed solution into BOTH
equations. If
it makes BOTH 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.

*y*-intercept

**Find another
solution by letting x = 1.**

**Solutions:**

**Plotting the ordered pair solutions and drawing the
line:**

**y-intercept**

***Inverse of mult. by -1 is div.
by -1**

*** y-intercept**

**Find another
solution by letting x = 2.**

***Inverse of mult. by -1 is div
by -1**

**Solutions:**

**Plotting the ordered pair solutions and drawing the
line:**

We need to ask ourselves, is there any place that the
two lines intersect,
and if so, where?

**The answer is yes, they intersect at (2, 1).**

You will find that if you plug the ordered pair (2, 1)
into BOTH equations
of the original system, that this is a solution to BOTH of them.

**The solution to this system is (2, 1).**

*y*-intercept

*** y-intercept**

**Find another
solution by letting x = 1.**

**Solutions:**

**Plotting the ordered pair solutions and drawing the
line:**

***Inverse of mult. by -1 is div.
by -1**

*** x-intercept**

*y*-intercept

**Find another
solution by letting x = 1.**

**Solutions:**

**Plotting the ordered pair solutions and drawing the
line:**

We need to ask ourselves, is there any place that the
two lines intersect,
and if so, where?

**The answer is no, they do not intersect. We
have two parallel
lines.**

There are no ordered pairs to check.

**The answer is no solution.**

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.

It doesn't matter which equation you use or which
variable you choose
to solve for.

You want to make it as simple as possible. If one of the equations is already solved for one of the variables, that is a quick and easy way to go.

If you need to solve for a variable, then try to pick one that has a 1 as a coefficient. That way when you go to solve for it, you won't have to divide by a number and run the risk of having to work with a fraction (yuck!!).

This is why it is called the substitution method.
Make sure that
you substitute the expression into the OTHER equation, the one you
didn't
use in step 2.

This will give you one equation with one unknown.

Solve the equation set up in step 3 for the variable
that is left.

If you need a review on this, go to **Tutorial
7: Linear Equations in One Variable. **

**If your variable drops out and you have a FALSE
statement, that means
your answer is no solution. **

**If your variable drops 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 BOTH
equations. If
it makes BOTH 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.

Both of these equations are already simplified.
No work needs
to be done here.

Note how the second equation is already solved for *y*.
We can use that one for this step.

It does not matter which equation or which variable you choose to solve for. But it is to your advantage to keep it as simple as possible.

**Second equation solved for y:**

(when you plug in an expression like this, it is just like you plug in a number for your variable)

***Inverse of sub. 20 is add 20**

***Inverse of div. by -7 is mult.
by -7**

You will find that if you plug the ordered pair (-5,
-6) into BOTH
equations of the original system, that this is a solution to BOTH of
them.

**(-5, -6) is a solution to our system.**

Both of these equations are already simplified.
No work needs
to be done here.

This time, the problem was not so nice to us, we
will have to
do a little work to get one equation solved for one of our variables.

It does not matter which equation or which variable you choose to solve for. Just keep it simple.

Since the *x* in the first
equation has a
coefficient of 1, that would mean we would not have to divide by a
number
to solve for it and run the risk of having to work with fractions
(YUCK!!)
The easiest route here is to solve the first equation for *x*,
and we definitely want to take the easy route. You would not be
wrong
to either choose the other equation and/or solve for y, again you want
to keep it as simple as possible.

**Solving the first equation for x we get:**

***Solved for x**

(when you plug in an expression like this, it is just like you plug in a number for your variable)

***Variable dropped out AND false**

As mentioned above if your variable drops out and you
have a FALSE statement,
then there is no solution. ** **If we were to graph these two,
they would be parallel to each other.

Since we did not get a value for *y*,
there
is nothing to plug in here.

There are no ordered pairs to check.

**The answer is no solution.**

This methods is also known as the addition or the
elimination by addition
method.

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.

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

If neither variable drops out, then we are stuck with an equation with two unknowns which is unsolvable.

**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, we 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.

Add the two equations together.

The variable that has the opposite coefficients will drop out in this step and you will be left with one equation with one unknown.

Solve the equation found in step 3 for the variable
that is left.

If you need a review on this, go to **Tutorial
7: Linear Equations in One Variable.**

**If 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 the proposed solution into BOTH
equations. If it
makes BOTH 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.

This equation is full of those nasty fractions.
We can simplify
both equations by multiplying each separate one by it’s LCD, just like
you can do when you are working with one equation. As long as you
do the same thing to both sides of an equation, you keep the two sides
equal to each other.

**Multiplying each equation by it's respective LCD we
get:**

***Mult. by LCD of 6**

Again, you want to make this as simple as
possible. Note how
the coefficients on both *y*’s are
3. We
need to have opposites, so if one of them is 3 and the other is -3,
they
would cancel each other out when we go to add them.

If we added them together the way they are now, we would end up with one equation and two variables, nothing would drop out. And we would not be able to solve it.

**I propose that we multiply the second equation by -1,
this would
create a -3 in front of x and we will
have
our opposites.**

Note that we could just as easily multiply the first equation by -1 and not the second one. Either way will get the job done.

**Multiplying the second equation by -1 we get:**

*** y's
have opposite
coefficients**

You can choose any equation used in this problem to
plug in the found *x *value.

**I choose to plug in 11 for x into the
first simplified equation (found in step 1) to find y’s
value.**

***Inverse of add 55 is sub. 55**

***Inverse of mult. by 3 is div.
by 3**

You will find that if you plug the ordered pair (11,
-25/3) into BOTH
equations of the original system, that this is a solution to BOTH of
them.

**(11, -25/3) is a solution to our system.**

This problem is already simplified. However, the second
equation is
not written in the form Ax + By = C. In other words, we need to
write
it in this form so everything is lined up ready to go when we add the
two
equations together.

**Rewriting the second equation we get:**

***Everything lines up**

Note that if we multiply the first equation by 2, then
we will have
a -6*x* which is the opposite of the 6*x* found in the second equation.

**Multiplying the first equation by 2 we get:**

*** x'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.

There is no value 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 either equation to indicate that they are the same equation.

**Two ways to write the answer are {( x, y)| 3x - 2y = 1} OR {(x, y)
| 4y = 6x -
2}.**

** 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 Problem 1a:Solve the system by graphing.

Practice Problem 2a:Solve the system by the substitution method.

Practice Problem 3a:Solve the system by the elimination method.

** Need Extra Help on these Topics?**

**http://www.purplemath.com/modules/systlin1.htm **

This website helps you with the definition of and solving systems of linear equations.

**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 July 10, 2011 by Kim Seward.

All contents copyright (C) 2001 - 2011, WTAMU and Kim Seward. All rights reserved.