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Graphing Linear Equations
Section 1.2
Lehmann, Intermediate Algebra, 4ed Section 1.2
Consider the equation . Let’s find y when
So, when , which cab be represented by the ordered pair
Slide 2
Definition of Solution, Satisfy, and Solution Set
2 5y x= −3.x =
( )2 52 3 56 51
y xy= −
= −
= −=
Original Equation.
Substitute 3 for x.
Multiply before subtracting.
Subtract.
1y = 3x =( )3,1 .
D e f i n i t i o n o f S o l u t i o n , S a t i s f y , a n d S o l u t i o n S e t
Lehmann, Intermediate Algebra, 4ed Section 1.2
For an ordered pair , we write the value of the independent variable in the first (left) position and the value of the dependent variable in the second (right) position.
• The numbers a and b are called coordinates.
• For the x-coordinate is 3 and the y-coordinate is 1.
Slide 3
Definition of Solution, Satisfy, and Solution Set
( ),a b
( )3,1 ,
Definition
D e f i n i t i o n o f S o l u t i o n , S a t i s f y , a n d S o l u t i o n S e t
Lehmann, Intermediate Algebra, 4ed Section 1.2
The equation becomes a true statement when we substitute 3 for x-coordinate and 1 for y-coordinate.
Slide 4
Definition of Solution, Satisfy, and Solution Set
2 5y x= −
( )?
?
2 5
1 2 3 5
1 1true
y x= −
= −
=
Original Equation.
Substitute 3 for x and 1 for y.
D e f i n i t i o n o f S o l u t i o n , S a t i s f y , a n d S o l u t i o n S e t
Lehmann, Intermediate Algebra, 4ed Section 1.2
• An ordered pair is a solution of an equation in terms of x and y if the equation becomes a true statement when a is substituted for x and b is substituted for y.
• We say satisfies the equation.
• The solution set of the equation is the set of all solution of the equation.
Slide 5
Definition of Solution, Satisfy, and Solution Set
( ),a b
( ),a b
Definition
D e f i n i t i o n o f S o l u t i o n , S a t i s f y , a n d S o l u t i o n S e t
Lehmann, Intermediate Algebra, 4ed Section 1.2
Find five solutions to the equation and plot them in the coordinate system (on the right).
Slide 6
Graphing an Equation
2 1,y x= − +
Example
D e f i n i t i o n o f S o l u t i o n , S a t i s f y , a n d S o l u t i o n S e t
Lehmann, Intermediate Algebra, 4ed Section 1.2
We begin be arbitrarily choosing the values 0, 1, and 2 to substitute for x:
The ordered pairs and are also solutions.
Slide 7
Graphing an Equation
( )
( )
( )
( )
( )
( )
2 0 1 2 1 1 2 2 10 1 2 1 4 11 1 3
Solution: 0,1 Solution: 1, 1 Solution: 2, 3
y y y= − + = − + = − +
= + = − + = − += = − = −
− −
( )2,5− ( )1,3−
Solution
D e f i n i t i o n o f S o l u t i o n , S a t i s f y , a n d S o l u t i o n S e t
Lehmann, Intermediate Algebra, 4ed Section 1.2
• Create a table of solutions
x y -2 5 -1 3 0 1 1 -1 2 -3
Slide 8
Graphing an Equation
• Plot the solutions
• Points form a linear line.
Solution Continued
D e f i n i t i o n o f S o l u t i o n , S a t i s f y , a n d S o l u t i o n S e t
Lehmann, Intermediate Algebra, 4ed Section 1.2
• Every point on the line is a solution to the equation
Slide 9
Graphing an Equation
2 1y x= − +
D e f i n i t i o n o f S o l u t i o n , S a t i s f y , a n d S o l u t i o n S e t
Lehmann, Intermediate Algebra, 4ed Section 1.2
• The point lies on the line
• Should satisfy the equations
• Whereas is not on the line
• Thus should not satisfy the equation
Slide 10
Graphing an Equation
( )3, 5−
( )2,4
2 1y x= − +
D e f i n i t i o n o f S o l u t i o n , S a t i s f y , a n d S o l u t i o n S e t
Lehmann, Intermediate Algebra, 4ed Section 1.2 Slide 11
Graphing an Equation
( )?
?
2 1
4 2 2 1
4 3false
y x= − +
=− +
=−
Original Equation.
Substitute 2 for x and 4 for y.
• The is not a solution to the equation
( )2,4
D e f i n i t i o n o f S o l u t i o n , S a t i s f y , a n d S o l u t i o n S e t
Lehmann, Intermediate Algebra, 4ed Section 1.2
Use ZDecimal on a graphing calculator.
• To enter press (–) 2 X,T,ϴ,n + 1. The key – is used for subtraction, and the key . (–) is used for negative numbers as well as taking the opposite.
Slide 12
Graphing an Equation
2 1,y x= − +
Calculator
D e f i n i t i o n o f S o l u t i o n , S a t i s f y , a n d S o l u t i o n S e t
Lehmann, Intermediate Algebra, 4ed Section 1.2
The graph of an equation in two variables is the set of points that correspond to all solutions of the equation.
In the last example we found that the equation . is a line. Notice that the equation . is of the form (where and ).
Slide 13
Definition: Graph
2 1,y x= − +2 1,y x= − +
y mx b= +2m = − 1b =
Definition
D e f i n i t i o n o f S o l u t i o n , S a t i s f y , a n d S o l u t i o n S e t
Lehmann, Intermediate Algebra, 4ed
Equations of the form
Section 1.2
If an equation can be put into the form where m and b are constants, then the graph of the equation is a line.
Slide 14
Graphs of Linear Equations
y mx b= +
y mx b= +
What is m and b for the equations 3 2, 3 and 3?2
y x y x y= − = − =
Example
G r a p h s o f L i n e a r E q u a t i o n s
Lehmann, Intermediate Algebra, 4ed
is of the form : and
is of the form because we write the equation as (so and ). is of the form because we write the equation as (so and ). Sketch the graph of the equation
Graphing Linear Equations
3 22
y x= − y mx b= + 2b = −32
m =
2y x= − y mx b= +2 0y x= − + 0b =2m = −
3b =3y =
0 3y x= + 0m =y mx b= +
30 6 5 0.x y− + =
Definition
Example
G r a p h s o f L i n e a r E q u a t i o n s
Lehmann, Intermediate Algebra, 4ed Section 1.2
First we solve for y
Slide 16
Graphing Linear Equations
30 6 12 030 6 12
6 30 126 30 12
6 6 65 2
x yx y
y x
y x
y x
− + =− = −− = − −− −
= −
= − −
Original Equation.
Subtract 12 from both sides.
Subtract 30x from both sides.
Simplify.
Divide both sides by 6.
Solution
G r a p h s o f L i n e a r E q u a t i o n s
Lehmann, Intermediate Algebra, 4ed Section 1.2
• . is of the form • The graph of the equation is a line • Find 2 points of the line • Plot the two points • Sketch the line • Find a third point •Verify that the third point lies on the line
Slide 17
Graphing Linear Equations
5 2y x= − − y mx b= +
G r a p h s o f L i n e a r E q u a t i o n s
Solution Continued
Lehmann, Intermediate Algebra, 4ed Section 1.2 Slide 18
Graphing Linear Equations
Table of solutions
x y
( )( )( )
2 0 3 301 2 1 3 12 2 2 3 1
− = −
− = −
− =
Solution Continued
G r a p h s o f L i n e a r E q u a t i o n s
Lehmann, Intermediate Algebra, 4ed Section 1.2
• Enter for y1.
• Use Zstandard followed by Zsquare.
• The graph is correct assuming that y was isolated correctly.
Slide 19
Graphing Linear Equations
5 2x− −Graphing Calculator
G r a p h s o f L i n e a r E q u a t i o n s
Lehmann, Intermediate Algebra, 4ed Section 1.2
Sketch the graph of • Use the distributive property on the left-hand
side. • Collect like terms. • Isolate y.
Slide 20
Using the Distributive Law to Help Graph a Linear Equation
( )3 2 5 2 3 8 .y x x− = − −Example
Solution
G r a p h s o f L i n e a r E q u a t i o n s
Lehmann, Intermediate Algebra, 4ed Section 1.2 Slide 21
Using the Distributive Law to Help Graph a Linear Equation
( )3 2 5 2 3 86 15 6 3
6 15 15 6 3 156 6 126 6 6
2
y x xy x
y x
y x
y x
− = − −
− = − −− + = − − +
−= +
= − +
Original equation
Distributive property
Add 15 to both sides.
Divide both sides by 6.
Simplify.
Solution Continued
G r a p h s o f L i n e a r E q u a t i o n s
Lehmann, Intermediate Algebra, 4ed Section 1.2 Slide 22
Using the Distributive Law to Help Graph a Linear Equation
Table of solutions
x y
( )( )( )
0 2 201 1 2 12 2 2 0
− + =
− + =
− + =
Solution Continued
G r a p h s o f L i n e a r E q u a t i o n s
Lehmann, Intermediate Algebra, 4ed Section 1.2
Sketch a graph of • Avoid fraction values for y • Use even values for x
Slide 23
Graphing an Equation That Contains Fractions
1 1.2
y x= −Example
Solution
G r a p h s o f L i n e a r E q u a t i o n s
Lehmann, Intermediate Algebra, 4ed
Table of solutions x y 0 2 4
Section 1.2 Slide 24
Graphing an Equation That Contains Fractions
( )
( )
( )
1 0 1 121 2 1 021 4 1 12
− = −
− =
− =
Solution Continued
G r a p h s o f L i n e a r E q u a t i o n s
Lehmann, Intermediate Algebra, 4ed Section 1.2
Use Zdecimal to verify the solution.
Slide 25
Graphing an Equation That Contains Fractions
Graphing Calculator
G r a p h s o f L i n e a r E q u a t i o n s
Lehmann, Intermediate Algebra, 4ed Section 1.2
Sometimes we find intercepts to graph a line. • x-intercept is on the y-axis, so y = 0 • y-intercepts in on the x-axis, so x = 0
Slide 26
Property
• For an equation containing the variables x and y • x-intercept: Substitute y = 0 and solve for x • y-intercept: Substitute x = 0 and solve for y
Directions
F i n d i n g I n t e r c e p t s o f a G r a p h
Lehmann, Intermediate Algebra, 4ed Section 1.2
Use intercepts to sketch a graph of
Slide 27
Using Intercepts to Sketch a Graph
2 4.y x= − +
x-intercept: Set y = 0.
0 2 44 24 22 22
xx
x
x
= − +− = −− −=− −
= Simplify.
Substitute 0 for y.
Subtract both sides by 4.
Divide both sides by - 2.
Example
Solution
F i n d i n g I n t e r c e p t s o f a G r a p h
Lehmann, Intermediate Algebra, 4ed Section 1.2
y-intercept: Set x = 0.
So, the x-intercept is and y-intercept is
Slide 28
Using Intercepts to Sketch a Graph
( )2 0 44
yy= − +
=
Set x=0.
Simplify.
( )0,4 .( )2,0
Solution Continued
F i n d i n g I n t e r c e p t s o f a G r a p h
Lehmann, Intermediate Algebra, 4ed Section 1.2
Use ZStandard followed by ZSquare.
Use “zero” to verify the x-intercept.
Slide 29
Using Intercepts to Sketch a Graph
Use TRACE to verify the y-intercept.
Graphing Calculator
F i n d i n g I n t e r c e p t s o f a G r a p h
Lehmann, Intermediate Algebra, 4ed Section 1.2
Graph the equation of
Slide 30
Graphing a Vertical Line
3.x =
x y 3 5 3 3 3 1 3 -1 3 -3
Notice that the values of x must be 3, but y can have any value. Some solutions are listed to the left.
Example
Solution
F i n d i n g I n t e r c e p t s o f a G r a p h
Lehmann, Intermediate Algebra, 4ed Section 1.2
Graph the equation of
Slide 31
Graphing a Horizontal Line
5.y = −
x y –2 –5 –1 –5 0 –5 1 –5 2 –5
Notice that the values of y must be –5, but x can have any value. Some solutions are listed to the left.
Example
Solution
Ve r t i c a l a n d H o r i z o n t a l L i n e s
Lehmann, Intermediate Algebra, 4ed Section 1.2
Use ZStandard to verify the graph.
Slide 32
Graphing a Horizontal Line
Graphing Calculator
Ve r t i c a l a n d H o r i z o n t a l L i n e s
Lehmann, Intermediate Algebra, 4ed Section 1.2
If a and b are constants: • An equation that can be put
into the form . has a vertical line as its graph
• An equation that can be put into the form .has a horizontal line as its graph
Slide 33
Vertical and Horizontal Line Property
y b=
x a=
Property
Ve r t i c a l a n d H o r i z o n t a l L i n e s
Lehmann, Intermediate Algebra, 4ed Section 1.2
In an equation can be put into either form where m, a, and b are constants, then the graph of the equation is a line. We call such an equation a linear equation in two variables.
Slide 34
Vertical and Horizontal Line Property
ory mx b x a= + =
Property
Ve r t i c a l a n d H o r i z o n t a l L i n e s