Chapter 7
Graphing Linear Equations
7.1
-The Cartesian Coordinate System
-Linear Equations in two variables
-Plot points in the Cartesian Coord.System
-Determine whether an ordered pair is a
solution of an equation.
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7.1 Cartesian Coordinate System
This is the Cartesian (also referred to as the
Rectangular) Coordinate System
Y-axis is the vertical axis
X-axis is the horizontal axis
Origin-the point where the two
axes intersect; has coordinates
of (0,0)
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Notice that the quadrants are numbered in a counter-clockwise direction with
Roman Numerals.
All the points in a particular quadrant have the same sign orientation. The right
hand picture below shows the signs of all the points in each quadrant.
Ordered pairs correspond to points in each quadrant as well as on each axis.
(3,5) is an ordered pair. 3 is the x-coordinate and 5 is the y-coordinate.
(x,y)
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To plot points on the Cartesian Coordinate System:
1) Start at the origin
2) Move right or left according to the x-coordinate
rt=(+) and left=(-); Keep your pencil here.
3) From this new location, move up or down according to the ycoordinate; up= (+) and down = (-)
4) Place a dot and label when necessary.
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7.1 Linear Equations in two
Variables
A linear equation in two variables is an
equation that can be written in the form
ax+by=c where a,b, and c are real
numbers.
4x + 3y = 12
x = 5 - 3y
2x = 5y
y=7
When we graph equations of this form, we
get LINES.
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The previous examples are all linear equations,
but only one of them is in STANDARD FORM:
ax + by =c
4x + 3y = 12
x = 5 - 3y
2x = 5y
y=7
Because these equations have two variables, they
will have two solutions. – an x and a y given as
an ordered pair. (x,y)
A solution is something that makes the equation
true.
One way to find solutions is to take a possible
solution and plug it in to see if it makes a true
statement.
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Consider the point (1,2) as (x,y)
4x + 3y = 12
4(1) + 3(2) = 12
4 + 6 = 10
(1,2) is not a solution of this equation.
Consider the point (3,0)
4x + 3y = 12
4(3) + 3(0) = 12
12 + 0 = 12
(3,0) is a solution of this equation.
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While equations with one variable (like we
solved in chapter 2) have only one
solution, equations in two variables have
an infinite number of solutions.
Can you find another solution for the
equation 4x + 3y = 12?
Consider the point (0,4)
4x + 3y = 12
4(0) + 3(4) = 12
0 + 12 = 12
(0,4) is a solution for the equation.
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We can find more solutions by:
1)Choosing an x-value and
2) Solving for y
(or vice versa)
4x + 3y = 12 Now choose an x-value (1)
4(1) + 3y = 12 by substituting 1 for x
4 + 3y = 12
Now solve for y
3y = 8
2
y = 8/3 or 2 3
2
So if x=1 then y= 2 3 so the ordered pair
2
(1,2 3 ) is a solution of the equation.
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If we want to graph an equation, we need at least two ordered
pairs, but preferably three. This will show us where our line
will be. The line represents all the points that are solutions for
this particular equation.
(0,4), (3,0), (1,2 23 )
When points fall in a
line they are said to
be collinear.
These points are
collinear so we could
draw the line.
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7.2 Graphing Linear Equations
1)
2)
3)
4)
5)
Choose a value for one of the variables
Substitute that value into the equation
Solve for the other variable
This gives you an ordered pair
Repeat the above steps until you have 3
points.
6) Plot the points and see if they are
collinear.
7) Draw the line through the points.
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3x + 2y = 6
Find point #1
Choose x = 0
3(0) + 2y = 6
2y = 6
y=3
(0,3)
1) Choose a value for one
of the variables
2) Substitute that value
into the equation
3) Solve for the other
variable
4) This gives you an
ordered pair
5) Repeat the above steps
until you have 3 points.
6) Plot the points and see
if they are collinear.
7) Draw the line through
the points.
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3x + 2y = 6
Find point #2
Choose x = 1
3(1) + 2y = 6
3 + 2y = 6
2y = 3
y=1½
(1, 1 ½ )
1) Choose a value for one
of the variables
2) Substitute that value
into the equation
3) Solve for the other
variable
4) This gives you an
ordered pair
5) Repeat the above steps
until you have 3 points.
6) Plot the points and see
if they are collinear.
7) Draw the line through
the points.
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3x + 2y = 6
Find point #3
Choose x = -1
3(-1) + 2y = 6
-3 + 2y = 6
2y = 9
y=4½
(-1, 4 ½ )
1) Choose a value for one
of the variables
2) Substitute that value
into the equation
3) Solve for the other
variable
4) This gives you an
ordered pair
5) Repeat the above steps
until you have 3 points.
6) Plot the points and see
if they are collinear.
7) Draw the line through
the points.
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3x + 2y = 6
Plot the points and
then graph the line
(0,3)
(1, 1 ½ )
(-1, 4 ½ )
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Graph using intercepts
X-intercept is where the line crosses the xaxis. An x-intercept always has the form
(x,0)
Y-intercept is where the line crosses the yaxis. A y-intercept always has the form
(0,y)
The book presents this as a totally different
method, however, it is just picking a
certain x or y value and solving for the
other variable. Use this method when you
see the problem lends itself to it.
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Graph using intercepts
4x + 5y = 20
4(0) + 5y = 20
0 + 5y = 20
5y = 20
Y=4
4x + 5y = 20
4x + 5(0) = 20
4x = 20
x=5
Choose x=0 and solve
for y
So the point is (0,4)
Choose y = 0 and solve
for x
So the point is (5,0)
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Graph 4x + 5y = 20
Plot the two intercepts
(0,4)
(5,0)
A third point could be
graphed as a check
point. It is not
required
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Horizontal and Vertical Lines
These are special cases. In these equations,
only one variable appears. The equation is
saying that no matter what the other
variable is, this given variable remains
constant.
These equations look like
x=3
or y = -5
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Horizontal and Vertical Lines
Let’s take
x=3
That tells me that x=3 no matter what y is.
Y can be equal to anything and x will still be
equal to three.
X
Y
3
-1
3
0
3
5
3
-5
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Horizontal and Vertical Lines
Let’s take
y = -5
That tells me that y=-5 no matter what x is.
x can be equal to anything and y will still be
equal to -5
X
-3
1
0
6
Y
-5
-5
-5
-5
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7.3 Slope of a Line
-represented by the letter m
-a number that describes the steepness of a line
-read from left to right on a graph (uphill lines are
positive; downhill lines are negative)
-a ratio of vertical change to horizontal change
between any two points on a line
-referred to also as rise over run.
m
change in y
change in x
y2 – y 1
x2 – x 1
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Find the slope using a graph
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3
Count up and over from one
point on the line to any
2nd point on the line.
Start at (-4,1); Go up three
units and right five units
and land at (1,4)
Vert change
3
horiz change
5
This won’t reduce so the
slope is 3/5
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Find the slope using the formula
Using the points (-9,1) and (7,4) for the
same line
And the formula
y2 – y1
x2 – x1
4–1
7- -9
3
16
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Find the slope using the formula
y2 – y1
x2 – x1
(5,7) and (1,7)
m
7-7
5-1
0
4
0
what kind of line?
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Find the slope using the formula
y2 – y1
x2 – x1
(1,-2) and (1,3)
m
3 - -2
1–1
5
0
undefined
What kind of line is this?
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Uphill lines have
positive slope
Downhill lines have
negative slope
Read the lines from left
to right on the graph.
+
slope
- slope
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Parallel and Perpendicular lines
Parallel lines have
equal slopes
Perpendicular lines
have slopes that are
negative reciprocals
of each other
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7.4 Slope Intercept Form
Slope Intercept Form:
y = mx + b
where m is the slope
and b is the yintercept
To put an equation in
slope-intercept form,
solve for y
2x + 3y = 6
3y = -2x + 6
y = 32x + 2
Now you can pull out
the m and b values
m= -2/3 and b = 2
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Graph using y=mx+b
Now using this info:
m= -2/3 and b = 2
We can quickly and
easily graph this line
y = 32 x + 2
b= 2 means the line
cross the y-axis at 2
m= -2/3 means it goes
down 2 and right 3
Down 2
Right 3
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y = (½) x - 3
To graph a line using
the m and b
1) Place a dot at the bvalue on y-axis (b=-3)
2) From this point move
according to the slope
and place a 2nd dot
(up 1 and right 2)
3) Draw line
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Slope Intercept Form
We can also use this form and its resulting
information to answer questions about the
line –
such as is it parallel to another line?
Or perpendicular to another line?
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If we know certain things about a line, we
can write the equation of the line in slopeintercept form. We need to know the slope
and the intercept values.
Sometimes they will give you slope. If they
don’t, do they give you two points so that
you can find the slope using the formula?
If they do not give us an intercept value, we
cannot approximate it using the graph.
They must tell us what the b-value is.
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7.4 Point Slope Form
The point slope form is not as user-friendly
as the slope intercept form, but it can be
used in instances where y=mx + b cannot
be used. The point slope form requires
only a point and the slope
The point slope form: y - y1 = m(x – x1)
where m is the slope and (x1, y1) is any
point on the line.
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Given (-1,3) and (-2,7), write the equation of the
line passing through the two points.
Because they do not give us the slope, we must
use the formula to find it:
m
7–3
4
-4
-2 - -1
-1
Now choose either point to be (x1, y1)
y - y1 = m(x – x1)
y – 7 = -4(x - -2)
y – 7 = -4(x + 2)
y – 7 = -4x -8
y = -4x -1 in slope-int or 4x + y = -1 in standard
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“Easiest Case” #1
m is known
b is known
Go straight to
Y=mx + b
Substitute in m and b
and you are done!
m = -1/2
b=3
y = mx + b
y = (-1/2) x + 3
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“Easyish Case” #2
m is unknown
b is known but maybe
hidden
Find the slope using the
formula
Go straight to
Y=mx + b
Substitute in m and b
and you are done!
through
(10,3) and (0,-2)
m = 3- -2 = 5 = 1
10-0 10 2
y = mx + b
y = (1/2) x + -2
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“Harder Case” #3
m is known
b is unknown
a point is given
use this form:
y – y1 = m (x – x1)
rearrange into this form:
y = mx + b
m = -3 through (2,5)
y – y1 = m (x – x1)
y – 5 = -3 (x – 2)
y – 5 = -3x + 6
y = -3x + 11
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“Hardest Case” #4
m and b are unknown
two points are given
use this form:
y – y1 = m (x – x1)
then rearrange to this:
y = mx + b
through (7,4) and (6,3)
m = 4-3 = 1
7-6 1
y – y1 = m (x – x1)
y – 4 = 1(x – 7)
y–4=x–7
y=x-3
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7.5 Graph Linear Inequalities
1) 3x + 2y > 5
3x + 2y = 5
2) 2y = -3x + 5
y =  32 x + 52
1) Replace > with = for
now
2) Graph the line (I like
using y=mx+b
method)
3) Choose a point
clearly on either side
of the line; (0,0) is a
good one if the line
doesn’t pass through
this point
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4) 3x + 2y > 5
3(0) + 2(0) > 5
5) 0 > 5 is False
Shade the opposite side
from (0,0)
4)Subsitute the check
point into the
ORIGINAL inequality
to see if it is T or F
5) If it is T, shade the
same side of the line.
If it is F, shade the
opposite side of the
line.
6) Check your line
>,< are dotted lines
>,< are solid lines
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