Intermediate Algebra Tutorial 14


Intermediate Algebra
Answer/Discussion to Practice Problems
Tutorial 14: Graphing Linear Equations


WTAMU > Virtual Math Lab > Intermediate Algebra > Tutorial 14: Graphing Linear Equations


 

checkAnswer/Discussion to 1a

2x - 3y = -6


 
Step 1:  Find the x- and y- intercepts.

Let’s first find the x-intercept
What value are we going to use for y
You are correct if you said y = 0.


 
ad1a1

*Find x-int. by replacing y with 0
 

*Inverse of mult. by 2 is div. by 2

 


 
The x-intercept 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.


 
ad1a2

*Find y-int. by replacing x with 0
 

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

 


 
The y-intercept is (0, 2)

 
Step 2:  Find at least one more point.

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:


 
ad1a3

*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)
 
 
Step 3: Plot the intercepts and point(s) found in steps 1 and 2.

example 1


 
Step 4:  Draw the graph.

example 1a5

(return to problem 1a)


 


 

checkAnswer/Discussion to 1b

x = 3y


 
Step 1:  Find the x- and y- intercepts.

Let’s first find the x-intercept
What value are we going to use for y
You are correct if you said y = 0.


 
ad1b1

*Find x-int. by replacing y with 0

 
The x-intercept 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.


 
ad1b2

*Find y-int. by replacing x with 0
 

 


 
The y-intercept is (0, 0)

 
Step 2:  Find at least one more point.

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:


 
ad1b3

*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:


 
ad1b4

*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)
 
Step 3: Plot the intercepts and point(s) found in steps 1 and 2.

example 1b5


 
Step 4:  Draw the graph.

example 1b6

(return to problem 1b)


 


 

checkAnswer/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.

 
Step 1:  Find the x- and y- intercepts.
AND
Step 2:  Find at least one more point.

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 x-intercept is at (4, 0).

Do we have a y-intercept?  The answer is no.  Since x can never equal 0, then there will be no y-intercept 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 )
 
Step 3:  Plot the intercepts and point(s) found in steps 1 and 2.

example 2a1


 
 
Step 4:  Draw the graph.

ad2a2

(return to problem 2a)


 

checkAnswer/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.

 
Step 1:  Find the x- and y- intercepts.
AND
Step 2:  Find at least one more point.

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 y-intercept (where x = 0) is at (0, -5). 

Do we have a x-intercept? The answer is no.  Since y has to be -5, then it can never equal 0, which is the criteria of an x-intercept. 

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)
 
Step 3:  Plot the intercepts and point(s) found in steps 1 and 2.

ad2b1


 
 
Step 4:  Draw the graph.

ad2b2

(return to problem 2b)


 

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WTAMU > Virtual Math Lab >Intermediate Algebra >Tutorial 14: Graphing Linear Equations


Last revised on July 3, 2011 by Kim Seward.
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