Intermediate Algebra
Answer/Discussion to Practice Problems
Tutorial 17: Graphing Linear Inequalities
Answer/Discussion
to 1a

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 3 is div. by 3
*xintercept 
xintercept is (0, 0)
Since the xintercept came out to be (0,
0), then it stands to reason that when we put in 0 for x to find the yintercept, we will get (0,
0).
Let's move on and plug in 1 for x to get a second solution: 
(1, 3) is another solution on the boundary line.
Plug in 1 for x to get a third solution: 
(1, 3) is another solution on the boundary line.
Solutions:
x

y

(x, y)

0

0

(0, 0)

1

3

(1, 3)

1

3

(1, 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 (1, 1). 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 (1, 1) into the original problem and see what happens: 

*Replacing x and y with 1
*True statement

Since our test point (1, 1) 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.
(return
to problem 1a) 
Answer/Discussion
to 1b

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.
(return
to problem 1b) 
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 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.

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
*False Statement

Since our test point (0, 0) made our inequality FALSE, this
means it is not 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.

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.
(return
to problem 1c) 
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 2 is div. by 2
*xintercept 

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

(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: 

*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 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.

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 4 is div. by 4
*yintercept 
yintercept is (0, 1).
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 4 is div. by 4

(1, 1/2) is another solution on the boundary line.
Solutions:
x

y

(x, y)

2

0

(2, 0)

0

1

(0, 1)

1

1/2

(1, 1/2)

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 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 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.
(return
to problem 1d) 
Last revised on July 5, 2011 by Kim Seward.
All contents copyright (C) 2001  2011, WTAMU and Kim Seward.
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