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Bittinger
Pre-Algebra, 6th edition
Section 6.4 – Graphing Linear Equations
Tom Atwater
Draft 1
Example from Text:
Exercises from Text: 7, 17, 37, 51, 61, 65
Student: No
Video
Audio
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Section 6.4
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Graphing Linear Equations
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INTRODUCTION:
Hi, my name is Tom. Today we are going
to learn how to graph linear equations.
READ Objectives.
Objectives
Find solutions of equations in two
variables.
Graph linear equations in two
variables.
Graph equations for horizontal or
vertical lines.
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SAY: To solve an equation with two
variables, we will first replace one
variable with some number choice and
then solve the resulting equation. Let’s
work two examples.
READ Exercise 7
For the equation, use the indicated value
to find an ordered pair that is a solution.
x + 3y = 1; let x = 10
MATH STEPS:
x + 3y = 1; let x = 10
Substitute 10 for x and solve for y
10 + 3y = 1
3y = –9
y=–3
The ordered pair (10, –3) is a solution of
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Bittinger
Pre-Algebra, 6th edition
Section 6.4 – Graphing Linear Equations
Tom Atwater
Draft 1
x + 3y = 1.
READ Exercise 17
For the equation, complete the given
ordered pair.
2x + 3y = 30; (0, ); (, 0)
MATH STEPS:
2x + 3y = 30; (0, )
Substitute 0 for x and solve for y
2(0) + 3y = 30
3y = 30
y = 10
The ordered pair is (0, 10) is a solution of
2x + 3y = 30. Since, 2(0) + 3(10) = 30, the
solution checks.
2x + 3y = 30; ( , 0)
Substitute 0 for y and solve for x
2x + 3(0) = 30
2x = 30
x = 15
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The ordered pair is (15, 0) is a solution of
2x + 3y = 30. Since, 2(15) + 3(0) = 30, the
solution checks.
SAY: In summary, here are the steps to
finding a solution of an equation with two
variables.
READ To find a solution of an equation
with two variables
To find a solution of an equation with
two variables
1. Choose a replacement for one variable.
2. Solve for the other variable.
3. Write the solution as an ordered pair.
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SAY: Here’s an example.
READ Exercise 37
For the equation, find three solutions.
Answers may vary.
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Bittinger
Pre-Algebra, 6th edition
Section 6.4 – Graphing Linear Equations
Tom Atwater
Draft 1
3x + 2y = 12
MATH STEPS:
3x + 2y = 12
Substitute 1 for x and solve for y
3(1) + 2y = 12
2y = 9
y = 9/2
The ordered pair is (1, 9/2) is a solution of
3x + 2y = 12.
In general, it is a good idea to use a
replacement value of zero for both x and
y. Let’s do that now.
Substitute 0 for x and solve for y
3(0) + 2y = 12
2y = 12
y=6
The ordered pair is (0, 6) is a solution of
3x + 2y = 12.
Substitute 0 for y and solve for x
3x + 2(0) = 12
3x = 12
x=4
The ordered pair is (4, 0) is a solution of
3x + 2y = 12.
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SAY: Next we consider graphing linear
equations. Equations like those considered
in Examples 1–3 are in the form
WRITE ON THE BOARD
Ax + By = C
All equations that can be written this way
are said to be linear because the solutions
of each equation, when graphed, form a
straight line. An equation Ax + By = C is
called the standard form of a linear
equation. When the line representing the
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Bittinger
Pre-Algebra, 6th edition
Section 6.4 – Graphing Linear Equations
Tom Atwater
Draft 1
solutions is drawn, we say that we have
graphed the equation.
Since solutions of Ax + By = C are
written in the form
WRITE ON THE BOARD
(x, y),
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we label the horizontal axis as the x-axis
and the vertical axis as the y-axis.
SAY: Let’s work two examples.
READ Exercise 51
Graph the equation y = 2x – 1.
MATH STEPS:
First, we find some ordered pairs that are
solutions. To find three ordered pairs, we
can choose any three values for x and then
calculate the corresponding values for y.
We will organize these values in a table.
One good choice is 0, and then a negative
value and a positive value.
x y=2x–1 (x, y)
0 –1
(0, –1)
-2 –5
(–2, –5)
1 1
(1, 1)
Plot these points and we draw the line, or
graph, with a ruler and label it y = 2x – 1.
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READ Exercise 57
Graph the equation y = –x + 4.
MATH STEPS:
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Bittinger
Pre-Algebra, 6th edition
Section 6.4 – Graphing Linear Equations
Tom Atwater
Draft 1
Make a table.
x y=–x+4 (x, y)
0 4
(0, 4)
-3 7
(–3, 7)
2 2
(2, 2)
Plot these points and we draw the line, or
graph, with a ruler and label it y = –x + 4.
FULL SCREEN PRESENTER
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SAY: Any equation in the form Ax + By
= C is linear, provided A and B are
not both zero. If A is 0 and B is nonzero,
there is no x-term and the graph is a
horizontal line. If B is 0 and A is nonzero,
there is no y-term and the graph is a
vertical line. Let’s work an example of
each.
READ Exercise 61
Graph y = 2.
MATH STEPS:
We regard y = 2 as 0 • x + y = 2. No
matter what number we choose for x, we
find that y must be 2 if the equation is to
be solved. We can make a table to see
this.
x y=2 (x, y)
0 2
(0, 2)
-3 2
(–3, 2)
2 2
(2, 2)
When we plot these points and connect
them, we obtain a horizontal line. Any
ordered pair of the form (x, 2) is a
solution, so the line is 2 units above the yaxis.
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Bittinger
Pre-Algebra, 6th edition
Section 6.4 – Graphing Linear Equations
Tom Atwater
Draft 1
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READ Exercise 65
Graph x = –3.
MATH STEPS:
No matter what number we choose for y,
we find that x must be –3 if the equation
is to be solved.
Therefore, any ordered pair of the form
(–3, y) is a solution, so the line is 3 units
to the left of the x-axis.
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SAY: Let’s summarize what we just
discovered.
READ HORIZONTAL AND VERTICAL
LINES
HORIZONTAL AND VERTICAL
LINES
The graph of y = b is a horizontal line.
The graph of x = a is a vertical line.
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CONCLUSION:
Today we learned how to graph linear
equations. Go back and review this
section before beginning the exercise set.
If you're having any trouble with the
material see your instructor immediately.
Good Luck.
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