Intermediate Algebra
Answer/Discussion to Practice Problems
Tutorial 12: Graphing Equations
Answer/Discussion
to 1a
A(3, 1), B(2, 1/2), C(2, 2), and D(0,1)
A(3, 1) lies in quadrant I.
B(2, 1/2) lies in quadrant III.
C(2, 2) lies in quadrant IV.
D(0, 1) lies on the y axis.
(return to problem
1a) 
Answer/Discussion
to 2a
y = 4 x  10 (0, 10), (1, 14), (1, 14) 
Let's start with the ordered pair (0, 10).
Which number is the x value and which one
is the y value? If you said x = 0 and y = 10, you are correct!
Let's plug (0, 10) into the equation and see what we get: 

*Plug in 0 for x and 10 for y

This is a TRUE statement, so (0, 10) is a solution to the equation y = 4x  10.
Now, let's take a look at (1, 14).
Which number is the x value and which one
is the y value? If you said x = 1 and y = 14, you are right!
Let's plug (1, 14) into the equation and see what we get: 

*Plug in 1 for x and 14 for y

This is a FALSE statement, so (1, 14) is NOT a solution to the
equation y = 4x  10.
Now, let's take a look at (1, 14).
Which number is the x value and which one
is the y value? If you said x = 1 and y = 14, you are right!
Let's plug (1, 14) into the equation and see what we get: 

*Plug in 1 for x and 14 for y

Answer/Discussion
to 3a
y = 2 x  1 
If we subtract 2x from both sides, then
we can write the given equation as 2x + y = 1.
Since we can write it in the standard form, Ax + By = C, then we have a linear equation.
This means that we will have a line when we go to graph this. 
The three x values I'm going to use are
1, 0, and 1. (Note that you can pick ANY three x values that
you want. You do not have to use the values that I picked.) You
want to keep it as simple as possible. The following is the chart
I ended up with after plugging in the values I mentioned for x.
x

y = 2x 
1

(x, y)

1

y = 2(1)  1 = 3

(1, 3)

0

y = 2(0)  1 = 1

(0, 1)

1

y = 2(1)  1 = 1

(1, 1)


Answer/Discussion
to 3b

If we add x squared to both sides
we would end up with .
Is this a linear equation? Note how we have an x squared
as opposed to x to the one power.
It looks like we cannot write it in the form Ax + By = C, because the x has to be to the one power, not squared. So this is not a linear
equation.
However, we can still graph it. 
The seven x values that I'm going to use are 3, 2, 1, 0, 1, 2, and
3. (Note that you can pick ANY x values
that you want. You do not have to use the values that I picked.) You
want to keep it as simple as possible. The following is the chart
I ended up with after plugging in the values I mentioned for x.
x


(x, y)

3

y = (3)^2 = 9

(3, 9)

2

y = (2)^2 = 4

(2, 4)

1

y = (1)^2 = 1

(1, 1)

0

y = (0)^2 = 0

(0, 0)

1

y = (1)^2 = 1

(1, 1)

2

y = (2)^2 = 4

(2, 4)

3

y = (3)^2 = 9

(3, 9)


Answer/Discussion
to 3c

Do you think this equation is linear or not? It is a tricky problem,
because both the x and y variables are to the one power. However, x is inside the absolute value sign and we can't just take it out of there.
In other words, we can't write it in the form Ax +
By = C. This means that this equation
is not a linear equation. 
The seven x values that I'm going to use are 3, 2, 1, 0, 1, 2, and
3. (Note that you can pick ANY x values
that you want. You do not have to use the values that I picked.) You
want to keep it as simple as possible. The following is the chart
I ended up with after plugging in the values I mentioned for x.
x


(x, y)

3

y = 3  1 = 2

(3, 2)

2

y = 2  1 = 1

(2, 1)

1

y = 1  1 = 0

(1, 0)

0

y = 0  1 = 1

(0, 1)

1

y = 1  1 = 0

(1, 0)

2

y = 2  1 = 1

(2, 1)

3

y = 3  1 = 2

(3, 2)


Last revised on July 3, 2011 by Kim Seward.
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