Intermediate Algebra
Tutorial 17: Graphing Linear Inequalities

WTAMU > Virtual Math Lab > Intermediate Algebra
Learning Objectives
After completing this tutorial, you should be able to:
 Graph linear inequalities in two variables.
 Graph the union of two linear inequalities in two variables.
 Graph the intersection of two linear inequalities in two variables.

Introduction
In this tutorial we will be looking at linear
inequalities in two variables.
It will start out exactly the same as graphing linear equations and
then
we get to color in the region of the coordinate system that correlates
with the inequality. Some of these problems may get a little
long,
especially the ones that involve two inequalities. Don't let that
discourage you, you can do it. Hang in there, a lot of the steps are concepts from the past, things you
should already have seen and done before. I will put in links to
the material that you need to remember from the past, in case you need
a review. A lot of times math works that way, use what you
know to learn the new concept. Let's see what you can do with
these
inequalities.

Tutorial
Linear Inequalities in Two
Variables
A linear inequality in two variables
is any expression
that can be put in the form
where a, b, and c are constants

Note that the above definition
can be applied
to any of the following four inequalities: <, >, <, or >.
The solution set and graph for a linear inequality is
a region of
the rectangular coordinate system. Recall that the graph of a
linear equation is a straight line. The inequality sign extends
this
to being on one side of the line or the other on the graph. 
Graphing a Linear
Inequality

Step 1: Graph the boundary
line. 
When you draw the boundary line, you need to have a way
to indicate
if the line is included or not in the final answer.
Solid boundary
line: < or >
If the problem includes where it is equal, then you will have a solid
boundary line. In other words, if you have < or > , you will have a solid line for your boundary line.
This shows the boundary line for x +
y < 6:
(note that this does not show the inequality part)
Dashed boundary
line: < or
>
If the problem does not include where it is equal, then you will use
a dashed boundary line. In other words, if you have < or >,
you
will have a dashed line for your boundary line.
This shows the boundary line for x +
y < 6:
(note that this does not show the inequality part)
In either case, you still graph the line the
same. You just
have to decide if you are needing a solid line or a dashed line.
The boundary line separates the rectangular coordinate
system into two
parts. One of those parts will make the inequality true and be
it’s
solution. 
Step 2: Plug in a test point
that is not on
the boundary line. 
Pick a test point on either side of the boundary line
and plug it into
the original problem. This will help determine which side of the
boundary line is the solution. 
Step 3: Shade in the answer to
the inequality. 
If you get a true statement when you plug in the
test point in step 2, then you have found a solution. Shade the region
that
the test point is in.
If you get a false statement when you plug in the
test point in step 2, then you don’t have a solution. Shade in the
region
that is on the other side of the test point.
It doesn’t matter what you use for the test point as
long as it is not
on the
boundary line. You want to keep it as simple as possible. 
Example
1: Graph the inequality . 
I’m going to use the intercepts to help me graph the
boundary line.
Again, you can use any method that you want, unless the directions say
otherwise.
When I’m working with only the
boundary line,
I will put an equal sign between the two sides to emphasize that we are
working on the boundary line. That doesn’t mean that I
changed the problem. When we put it all together in the end, I will put
the inequality back in.
What value is y on
the xintercept?
If you said 0, you are correct.
If you need a review on xintercepts,
go to Tutorial 14: Graphing Linear Equations. 

*Replace y with
0
*xintercept


*Replace x with
0
*yintercept

yintercept is (0, 2).
Plug in 1 for x to get a
third solution: 

*Replace x with 1
*Inverse of add 1 is sub. 1

(1, 1) is another solution on the boundary line.
Solutions:
x

y

(x, y)

2

0

(2, 0)

0

2

(0, 2)

1

1

(1, 1)

Since the original problem has a >, this means it
DOES NOT include the
boundary line.
So are we going to draw a solid or a dashed line for
this problem?
It looks like it will have to be a dashed line.
Putting it all together, we get the following boundary
line for this problem:

Note how the boundary line separates it into two
parts.
An easy test point would be (0, 0). Note
that it is a point
that is not on the boundary line. In fact, it is located below the
boundary
line.
Let’s put (0, 0) into the original problem and see
what happens: 

*Replacing x and y with 0
*False statement

Since our test point (0, 0) made our inequality
FALSE, this
means it is not a solution.
Since it has to be on one side or the other of the
boundary line, and
it is not below it, then our solution would lie above the boundary
line.
This means we will shade in the part that is above it.
Note that the gray lines indicate where you would
shade your
final answer.

Example
2: Graph the inequality . 
I’m going to use the intercepts to help me graph the
boundary line.
Again, you can use any method that you want, unless the directions say
otherwise.
When I’m working with only the
boundary line,
I will put an equal sign between the two sides to emphasize that we are
working on the boundary line. That doesn’t mean that I
changed the problem. When we put it all together in the end, I will put
the inequality back in.
What value is y on
the xintercept?
If you said 0, you are correct.
If you need a review on xintercepts,
go to Tutorial 14: Graphing Linear Equations. 

*Replace y with
0
*Inverse of mult. by 2 is div.
by 2
*xintercept 

*Replace x with
0
*Inverse of mult. by 3 is div.
by 3
*yintercept 
yintercept is (0, 2).
Plug in 1 for x to get a
third solution: 

*Replace x with
1
*Inverse of add 2 is sub. 2
*Inverse of mult. by 3 is div.
by 3

(1, 4/3) is another solution on the boundary line.
Solutions:
x

y

(x, y)

3

0

(3, 0)

0

2

(0, 2)

1

4/3

(1, 4/3)

Since the original problem has a <, this
means it DOES
include the boundary line.
So are we going to draw a solid or a dashed line for
this problem?
It looks like it will have to be a solid line.
Putting it all together, we get the following boundary
line for this problem:

Note how the boundary line separates it into two
parts.
An easy test point would be (0, 0). Note
that it is a point
that is not on the boundary line. In fact, it is located above the
boundary
line.
Let’s put (0, 0) into the original problem and see
what happens: 

*Replace x and y with
0
*True statement

Since our test point (0, 0) made our inequality TRUE,
this means
it is a solution.
Our solution would lie above the boundary line.
This means
we will shade in the part that is above it.
Note that the gray lines indicate where you would
shade your
final answer.

Example
3: Graph the inequality . 
Since the original problem has a <, this means it
DOES NOT include
the boundary line.
So are we going to draw a solid or a dashed line for
this problem?
It looks like it will have to be a dashed line.
Putting it all together, we get the following boundary
line for this problem:

Note how the boundary line separates it into two
parts.
An easy test point would be (0, 0). Note
that it is a point
that is not on the boundary line. In fact, it is located to the left of
the boundary line.
Let’s put (0, 0) into the original problem and see
what happens: 

*Replace x with 0
*True Statement 
Since our test point (0, 0) made our inequality TRUE,
this means
it is a solution.
Our solution would lie to the left of the boundary
line.
This means we will shade in the part that is to the left of it
Note that the gray lines indicate where you would
shade your
final answer.

Graphing the Union
of Two Linear Inequalities

The key word for union is “or”.
This means we will put both graphs on the same
coordinate system and
shade in the area that includes anything that was shaded in one
graph
or the other or both. Basically you are putting the two
inequalities
onto one graph, kind of like having a "+" sign in between the two
inequalities..
Note that when you do graph
each individual
inequality you follow the exact same steps as shown in the example
above.
The only difference is that you have to do it twice in one problem. 
Example
4: Graph the union
OR . 
I’m going to use the intercepts to help me graph the
boundary line.
Again, you can use any method that you want, unless the directions say
otherwise.
When I’m working with only the
boundary line,
I will put an equal sign between the two sides to emphasize that we are
working on the boundary line. That doesn’t mean that I
changed the problem. When we put it all together in the end, I will put
the inequality back in.
What value is y on
the xintercept?
If you said 0, you are correct.
If you need a review on xintercepts,
go to Tutorial 14: Graphing Linear Equations. 

*Replace x with 0
*Inverse of mult. by 3 is div.
by 3
*xintercept 

*Replace x with
0
*Inverse of mult. by 5 is div.
by 5
*yintercept 
yintercept is (0, 3).
Plug in 1 for x to get a
third solution: 

*Replace x with
1
*Inverse of add 3 is sub. 3
*Inverse of mult. by 5 is div.
by 5

(1, 12/5) is another solution on the boundary line.
Solutions:
x

y

(x, y)

5

0

(5, 0)

0

3

(0, 3)

1

12/5

(1, 12/5)

Since the original problem has a >, this
means it DOES include
the boundary line.
So are we going to draw a solid or a dashed line for
this problem?
It looks like it will have to be a solid line.
Putting it all together, we get the following boundary
line for this problem:

Note how the boundary line separates it into two
parts.
An easy test point would be (0, 0). Note
that it is a point
that is not on the boundary line. In fact, it is located below the
boundary
line.
Let’s put (0, 0) into the original problem and see
what happens: 

*Replace x and y with 0
*False statement

Since our test point (0, 0) made our inequality
FALSE, this
means it is not a solution.
Since (0, 0) was below the line and we need to be on
the other side,
our solution lies above the boundary line. This means we will
shade in the part that is above it.
Note that the gray lines indicate where you would
shade your
final answer.

I’m going to use the intercepts to help me graph the
boundary line.
Again, you can use any method that you want, unless the directions say
otherwise.
When I’m working with only the
boundary line,
I will put an equal sign between the two sides to emphasize that we are
working on the boundary line. That doesn’t mean that I
changed the problem. When we put it all together in the end, I will put
the inequality back in.
What value is y on
the xintercept?
If you said 0, you are correct.
If you need a review on xintercepts,
go to Tutorial 14: Graphing Linear Equations. 

*Replace y with 0
*xintercept


*Replace x with
0
*Inverse of mult. by 1 is div.
by 1
*yintercept 
yintercept is (0, 5).
Plug in 1 for x to get a
third solution: 

*Replace x with 1
*Inverse of add 1 is sub. 1
*Inverse of mult. by 1 is div.
by 1

(1, 4) is another solution on the boundary line.
Solutions:
x

y

(x, y)

5

0

(5, 0)

0

5

(0, 5)

1

4

(1, 4)

Since the original problem has a <, this
means it DOES include
the boundary line.
So are we going to draw a solid or a dashed line for
this problem?
It looks like it will have to be a solid line.
Putting it all together, we get the following boundary
line for this problem:

Note how the boundary line separates it into two
parts.
An easy test point would be (0, 0). Note
that it is a point
that is not on the boundary line. In fact, it is located above the
boundary
line.
Let’s put (0, 0) into the original problem and see
what happens: 

*Replace x and y with
0
*True statement

Since our test point (0, 0) made our inequality TRUE,
this means
it is a solution.
Our solution lies above the boundary line.
This means we
will shade in the part that is above it.
Note that the gray lines indicate where you would
shade your
final answer.

The union of the two inequalities is the area on the
graph that was
shaded in either the first inequality OR the second one OR both.
This is what we get when we union these two
inequalities:
Note that the gray lines indicate where you would
shade your
final answer.

Graphing the Intersection
of Two Linear Inequalities

The key word for intersection is “and”.
This means in the end, we only want to shade in the area
that satisfies
BOTH linear inequalities. We will graph each one separately and
then
assess what they have in common, this will be our final answer.
Note that when you do graph
each individual
inequality you follow the exact same steps as shown in the example
above.
The only difference is that you have to do it twice in one problem. 
Example
5: Graph the intersection
AND . 
Since the original problem has a >, this
means it DOES
include the boundary line.
So are we going to draw a solid or a dashed line for
this problem?
It looks like it will have to be a solid line.
Putting it all together, we get the following boundary
line for this problem:

Note how the boundary line separates it into two
parts.
An easy test point would be (0, 0). Note
that it is a point
that is not on the boundary line. In fact, it is located to the right
of
the boundary line.
Let’s put (0, 0) into the original problem and see
what happens: 

*Replace x with
0
*True Statement

Since our test point (0, 0) made our inequality TRUE,
this means
it is a solution.
Our solution would lie to the right of the boundary
line.
This means we will shade in the part that is to the right of it
Note that the gray lines indicate where you would
shade your
final answer.

Since the original problem has a <, this means it
DOES NOT include
the boundary line.
So are we going to draw a solid or a dashed line for
this problem?
It looks like it will have to be a dashed line.
Putting it all together, we get the following boundary
line for this problem:

Note how the boundary line separates it into two
parts.
An easy test point would be (0, 0). Note
that it is a point
that is not on the boundary line. In fact, it is located below the
boundary
line.
Let’s put (0, 0) into the original problem and see
what happens: 

*Replace y with
0
*True statement

Since our test point (0, 0) made our inequality TRUE,
this means
it is a solution.
Our solution would lie below the boundary line.
This means
we will shade in the part that is below it
Note that the gray lines indicate where you would
shade your
final answer.

The intersection of the two inequalities is the area on
the graph that
was shaded in BOTH the first inequality AND the second
inequality.
It is the region where they overlap
This is what we get when we intersect these two
inequalities:
Note that the gray lines indicate where you would
shade your
final answer.

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  1d: Graph each inequality.
Need Extra Help on these Topics?
Last revised on July 5, 2011 by Kim Seward.
All contents copyright (C) 2001  2011, WTAMU and Kim Seward. All rights reserved.
