Intermediate Algebra
Answer/Discussion to Practice Problems
Tutorial 14: Graphing Linear Equations
Answer/Discussion
to 1a
2x  3y = 6 
Let’s first find the xintercept.
What value are we going to use for y?
You are correct if you said y = 0. 

*Find xint.
by
replacing y with 0
*Inverse of mult. by 2 is div.
by 2

The xintercept is
(3, 0).
Next we will find the y
intercept.
What value are we going to plug in for x?
If you said x = 0, you are right. 

*Find yint.
by
replacing x with 0
*Inverse of mult. by 3 is div.
by 3

The yintercept is
(0, 2) 
We can plug in any x value
we want as long
as we get the right corresponding y value and
the function exists there.
Let’s put in an easy number x =
1: 

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

So the ordered pair (1, 8/3) is another solution to our function.
Note that we could have plugged in any value for x: 5,
10, 25, ...,
but it is best to keep it as simple as possible.
The solutions that we found are:
x

y

(x, y)

3

0

(3, 0)

0

2

(0, 2)

1

8/3

(1, 8/3)


Answer/Discussion
to 1b
x = 3y 
Let’s first find the xintercept.
What value are we going to use for y?
You are correct if you said y = 0. 

*Find xint.
by
replacing y with 0

The xintercept is
(0, 0).
Next we will find the y
intercept.
What value are we going to plug in for x?
If you said, x = 0 you are right. 

*Find yint.
by
replacing x with 0

The yintercept is
(0, 0) 
Since we really have found only one point this time,
we better find
two additional solutions so we have a total of three points.
We can plug in any x value
we want as long
as we get the right corresponding y value and
the function exists there.
Let’s put in an easy number x =
1: 

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

So the ordered pair (1, 1/3) is another solution to our function.
Let’s put in another easy number x = 1: 

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

So the ordered pair (1, 1/3) is another solution to our function.
The solutions that we found are:
x

y

(x, y)

0

0

(0, 0)

1

1/3

(1, 1/3)

1

1/3

(1, 1/3)


Answer/Discussion
to 2a
x = 4 
This is in the form x = c.
So, what type of line are we going to end up with?
Vertical. 
Since this is a special type of line, I thought I would
talk about steps
1 and 2 together.
It does not matter what y is, as long as x is 4.
Note that the xintercept
is at (4, 0).
Do we have a yintercept? The answer is no. Since x can
never
equal 0, then there will be no yintercept for this equation.
Some points that would be solutions are (4, 0), (4,
1), and (4, 2).
Again, I could have picked an infinite number of
solutions.
The solutions that we found are:
x

y

(x, y)

4

0

(4, 0)

4

1

(4, 1)

4

2

(4, 2 )


Answer/Discussion
to 2b
y + 5 = 0 
If you subtract 5 from both sides, you will have y = 5. It looks like it fits the form y = c.
With that in mind, what kind of line are we going to end up with?
Horizontal. 
Since this is a special type of line, I thought I would
talk about steps
1 and 2 together.
It doesn’t matter what x is, y is always 5. So for our solutions we just need three ordered
pairs
such that y = 5.
Note that the yintercept
(where x =
0) is at (0, 5).
Do we have a xintercept? The answer is no.
Since y has to be 5, then it can never equal 0, which is the criteria of an xintercept.
So some points that we can use are (0, 5), (1, 5)
and (2, 5).
These are all ordered pairs that fit the criteria of y having to be 5.
Of course, we could have used other solutions, there are
an infinite
number of them.
The solutions that we found are:
x

y

(x, y)

0

5

(0, 5)

1

5

(1, 5)

1

5

(1, 5)


Last revised on July 3, 2011 by Kim Seward.
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