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Lesson-2 1

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28 views23 pages

Lesson-2 1

Uploaded by

simoncabahug07
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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GRAPHS OF LINEAR

EQUATIONS IN TWO
VARIABLES
LEARNING TARGETS:

❑ I can illustrate a linear equation in two


variables;
❑ I can graph a linear equation in two variables;
and,
❑ I can solve solutions of linear equations in two
variables.
LINEAR EQUATIONS
IN TWO VARIABLES
What is a linear equation in two variables?
DEFINITION:
✓ It is an equation that can be written as 𝑨𝒙 + 𝑩𝒚 = 𝑪.
✓ The degree of the x- and y-terms in the equation is 1.

EXAMPLES: NON-EXAMPLES:
𝟑𝒙 + 𝒚 = 𝟐 𝟑𝒙𝟐 − 𝒚 = 𝟐
𝒙−𝒚=𝟎 𝟏
=𝟐
𝒙 = 𝟐𝒚 + 𝟏 𝒙𝒚
𝒚=𝒙−𝟏 𝒙+𝟒=𝟐
LINEAR EQUATIONS
IN TWO VARIABLES
ILLUSTRATIVE EXAMPLES:
Determine whether each equation is a LINEAR EQUATION IN TWO VARIABLES or
NOT. Write LETV if it is, otherwise NLETV. If LETV, identify the values of A, B, and C.
𝟏) 𝒚 = −𝟐𝒙 + 𝟒
Solution:
𝒚 = −𝟐𝒙 + 𝟒 By Addition Property of Equality, write the equation in standard
+𝟐𝒙 +𝟐𝒙
form.
𝟐𝒙 + 𝒚 = 𝟒 The equation is already in standard form.

Thus, LETV; 𝑨 = 𝟐 𝑩 = 𝟏 𝑪 = 𝟒

𝟐) 𝒙 + 𝒚 = 𝟒 The equation is already written in standard form.

Thus, LETV; 𝑨 = 𝟏 𝑩 = 𝟏 𝑪 = 𝟒
LINEAR EQUATIONS
IN TWO VARIABLES
SOLUTIONS OF A
LINEAR
EQUATION IN
TWO VARIABLES
SOLUTIONS OF A LINEAR EQUATION IN TWO VARIABLES
DEFINITION: TRUE: FALSE:
A solution of a linear equation in two variables are ordered 3=3 3 ≠ -3
pairs [(x,y)] that can make an equation true.
4=4 6≠5
ILLUSTRATIVE EXAMPLE 1:
Determine if the given ordered pair is a solution of 𝟐𝒙 − 𝒚 = 𝟓.
(𝟑, 𝟏) (−𝟐, −𝟏) (𝟐, −𝟏)
𝟐𝒙 − 𝒚 = 𝟓 𝟐𝒙 − 𝒚 = 𝟓 𝟐𝒙 − 𝒚 = 𝟓
𝟐 3 − 1=𝟓 𝟐 −𝟐 − −𝟏 = 𝟓 𝟐 𝟐 − −𝟏 = 𝟓
𝟔 −𝟏 = 𝟓 −𝟒 + 𝟏 = 𝟓 𝟒 +𝟏 = 𝟓
𝟓=𝟓 −𝟑 = 𝟓 𝟓=𝟓
Hence, (𝟑, 𝟏) is a Hence, (−𝟐, −𝟏) is Hence, (𝟐, −𝟏) is a
SOLUTION. NOT A SOLUTION. SOLUTION.
LINEAR EQUATIONS
IN TWO VARIABLES
SOLUTIONS OF A LINEAR EQUATION IN TWO VARIABLES
DEFINITION: TRUE: FALSE:
A solution of a linear equation in two variables are ordered 3=3 3 ≠ -3
pairs [(x, y)] that can make an equation true.
4=4 6≠5
ILLUSTRATIVE EXAMPLE 2:
Determine if the given ordered pair is a solution of 𝟑𝒚 = 𝒙 − 𝟏.
(𝟏, 𝟏) (𝟒, 𝟏) (−𝟓, 𝟐)
𝟑𝒚 = 𝒙 − 𝟏 𝟑𝒚 = 𝒙 − 𝟏 𝟑𝒚 = 𝒙 − 𝟏
𝟑 1 = 𝟏 −𝟏 𝟑 1 = 𝟒 −𝟏 𝟑 2 = −𝟓 −𝟏
𝟑 = 𝟏 −𝟏 𝟑 = 𝟒 −𝟏 𝟔 = −𝟓 −𝟏
𝟑=𝟎 𝟑=𝟑 𝟔 = −𝟔
Hence, (𝟏, 𝟏) is NOT Hence, (𝟒, 𝟏) is a Hence, (−𝟓, 𝟐) is
A SOLUTION. SOLUTION. NOT A SOLUTION.
LINEAR EQUATIONS
IN TWO VARIABLES
SOLUTIONS OF A LINEAR EQUATION IN TWO VARIABLES

PROCEDURE:
To find a solution of a linear equation in two variables,
select a particular value of “x” and substitute it to the
“x” of the given equation, then solve the resulting
equation for “y”.
It is also possible to select a particular value of “y” and
substitute it to the “y” of the given equation, then
solve the resulting equation for “x”.

LINEAR EQUATIONS
IN TWO VARIABLES
SOLUTIONS OF A LINEAR EQUATION IN TWO VARIABLES
ILLUSTRATIVE EXAMPLE 1: Solve for the three solutions of 𝒙 − 𝟑𝒚 = 𝟒.
Solution:
We have to select 3 ordered pairs that make 𝒙 − 𝟑𝒚 = 𝟒 true.
Let 𝒙 = 𝟐 Let 𝒙 = −𝟐 Let 𝒙 = 𝟓
𝒙 − 𝟑𝒚 = 𝟒 𝒙 − 𝟑𝒚 = 𝟒 𝒙 − 𝟑𝒚 = 𝟒
x y
𝟐 − 𝟑𝒚 = 𝟒 −𝟐 − 𝟑𝒚 = 𝟒 𝟓 − 𝟑𝒚 = 𝟒
𝟐 -2/3 −𝟐 −𝟐 +𝟐 +𝟐 −𝟓 −𝟓
−𝟑𝒚 = 𝟐 −𝟑𝒚 = 𝟔 −𝟑𝒚 = −𝟏
−𝟐 -2 −𝟑 −𝟑 −𝟑 −𝟑 −𝟑 −𝟑
𝟐 𝒚 = −𝟐 𝟏
𝟓 1/3 𝒚=− 𝟏 𝒚=
𝟐 𝟑 𝟓, 𝟑
𝟐, − −𝟐, −𝟐 𝟑
𝟑
LINEAR EQUATIONS
IN TWO VARIABLES
SOLUTIONS OF A LINEAR EQUATION IN TWO VARIABLES
ILLUSTRATIVE EXAMPLE 2: Solve for the three solutions of 𝟒𝒙 + 𝒚 = 𝟏𝟐.
Solution:
We have to select 3 ordered pairs that make 𝟒𝒙 + 𝒚 = 𝟏𝟐 true.
Let 𝒙 = 𝟎 Let 𝒙 = 𝟏 Let 𝒙 = 𝟐
𝟒𝒙 + 𝒚 = 𝟏𝟐 𝟒𝒙 + 𝒚 = 𝟏𝟐 𝟒𝒙 + 𝒚 = 𝟏𝟐
x y
𝟒 𝟎 + 𝒚 = 𝟏𝟐 𝟒 𝟏 + 𝒚 = 𝟏𝟐 𝟒 𝟐 + 𝒚 = 𝟏𝟐
𝟎 12 𝟎 +𝒚 = 𝟏𝟐 𝟒 + 𝒚 = 𝟏𝟐 𝟖 + 𝒚 = 𝟏𝟐
−𝟒 −𝟒 −𝟖 −𝟖
𝟏 8 𝒚 = 𝟏𝟐
𝒚=𝟖 𝒚=𝟒
𝟐 4
𝟎, 𝟏𝟐 𝟏, 𝟖 𝟐, 𝟒
LINEAR EQUATIONS
IN TWO VARIABLES
GRAPHING A
LINEAR
EQUATION
GRAPHING A LINEAR EQUATION IN TWO VARIABLES

PROCEDURE: (by plotting points)


Solve for at least two solutions (ordered
pairs/points) of the given equation.
Plot the points on the Cartesian plane.
Draw a straight line that passes through the
points.
LINEAR EQUATIONS
IN TWO VARIABLES
ILLUSTRATIVE EXAMPLE 1: Graph 𝒙 + 𝒚 = 𝟐.
Solution: 𝒚
Solve for at least two solutions of 𝒙 + 𝒚 = 𝟐.
𝟔

x y 𝒙=𝟎 𝒙=𝟏 𝟓

𝟎 𝟐 𝒙+𝒚=𝟐 𝒙+𝒚=𝟐 𝟑

𝟏 𝟏 𝟎 +𝒚=𝟐 𝟏 +𝒚=𝟐
𝟐

𝟏
−𝟏 −𝟏
𝒚=𝟐 𝒙
𝒚=𝟏 −𝟖 −𝟕 −𝟔 −𝟓 −𝟒 −𝟑 −𝟐 −𝟏 𝟏
−𝟏
𝟐 𝟑 𝟒 𝟓 𝟔 𝟕 𝟖

𝟎, 𝟐 −𝟐 𝒙+𝒚=𝟐
𝟏, 𝟏
−𝟑

−𝟒

−𝟓

LINEAR EQUATIONS −𝟔
IN TWO VARIABLES
ILLUSTRATIVE EXAMPLE 2: Graph 𝟑𝒙 − 𝒚 = 𝟏.
𝒚
Solution:
𝟔
Solve for at least two solutions of 𝟑𝒙 − 𝒚 = 𝟏.
𝟓

x y 𝒙=𝟎 𝒙 = −𝟏 𝟒

𝟎 𝟑𝒙 − 𝒚 = 𝟏
−𝟏 𝟑𝒙 − 𝒚 = 𝟏 𝟐
𝟑𝒙 − 𝒚 = 𝟏
−𝟏 −𝟒 𝟑 𝟎 − 𝒚 = 𝟏 𝟑 −𝟏 − 𝒚 = 𝟏 𝟏

𝟎−𝒚=𝟏 −𝟑 − 𝒚 = 𝟏 𝒙
+𝟑 +𝟑 −𝟖 −𝟕 −𝟔 −𝟓 −𝟒 −𝟑 −𝟐 −𝟏 𝟏 𝟐 𝟑 𝟒 𝟓 𝟔 𝟕 𝟖

−𝒚 = 𝟏 −𝒚 = 𝟒
−𝟏

−𝟏 −𝟏 −𝟐
−𝟏 −𝟏 −𝟑
𝒚 = −𝟏 𝒚 = −𝟒 −𝟒

𝟎, −𝟏 −𝟏, −𝟒 −𝟓

−𝟔

LINEAR EQUATIONS
IN TWO VARIABLES
ILLUSTRATIVE EXAMPLE 3: Graph 𝟑𝒚 = 𝒙 − 𝟏.
Solution: 𝒚
Solve for at least two solutions of 𝟑𝒚 = 𝒙 − 𝟏.
𝟔

x y Let 𝒙 = 𝟏 Let 𝒙 = 𝟒 𝟓

𝟏 𝟎 𝟑𝒚 = 𝒙 − 𝟏 𝟑𝒚 = 𝒙 − 𝟏 𝟑

𝟒 −𝟒 𝟑𝒚 = 𝟏 − 𝟏 𝟑𝒚 = 𝟒 − 𝟏 𝟐 𝟑𝒚 = 𝒙 − 𝟏
𝟑𝒚 = 𝟎 𝟑𝒚 = 𝟑
𝟏

𝒙
𝟑𝒚 = 𝟎 𝟑 𝟑 −𝟖 −𝟕 −𝟔 −𝟓 −𝟒 −𝟑 −𝟐 −𝟏 𝟏 𝟐 𝟑 𝟒 𝟓 𝟔 𝟕 𝟖
−𝟏
𝟑 𝟑
−𝟐
𝒚=𝟎 𝒚=𝟏
−𝟑

𝟏, 𝟎 𝟒, 𝟏 −𝟒

−𝟓

LINEAR EQUATIONS −𝟔
IN TWO VARIABLES
GRAPHING using the INTERCEPTS
𝒚

x-intercepts 𝒚 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕 𝟔

It is defined as the point of the 𝟓

intersection of a line and the x-axis. 𝟒 𝟎, 𝒃


𝟑

y-intercepts 𝟐

It is defined as the point of the 𝒂, 𝟎 𝟏

intersection of a line and the y-axis. 𝒙


−𝟖 −𝟕 −𝟔 −𝟓 −𝟒 −𝟑 −𝟐 −𝟏 𝟏 𝟐 𝟑 𝟒 𝟓 𝟔 𝟕 𝟖
−𝟏

PROCEDURE: −𝟐

• To solve for the x-intercept, let 𝒚 = 𝟎, 𝒙 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕 −𝟑

then solve for 𝒙. −𝟒

• To solve for the y-intercept, let 𝒙 = 𝟎, −𝟓

then solve for 𝒚. −𝟔

LINEAR EQUATIONS
IN TWO VARIABLES
ILLUSTRATIVE EXAMPLE 1: Graph 𝒙 + 𝒚 = 𝟏.
Solution: 𝒚
Solve for at least two solutions of 𝒙 + 𝒚 = 𝟏.
𝟔

x y Let 𝒙 = 𝟎 Let 𝒚 = 𝟎 𝟓

𝟎 𝟏 𝒙+𝒚=𝟏 𝒙+𝒚=𝟏 𝒙+𝒚=𝟏 𝟑

𝟏 𝟎 𝟎 +𝒚=𝟏 𝒙+ 𝟎 =𝟏 𝟐
𝒙 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕
𝟏

𝟎+𝒚=𝟏 𝒙+𝟎 =𝟏 𝒙
−𝟖 −𝟕 −𝟔 −𝟓 −𝟒 −𝟑 −𝟐 −𝟏 𝟏 𝟐 𝟑 𝟒 𝟓 𝟔 𝟕 𝟖
𝒚=𝟏 𝒙=𝟏 −𝟏

−𝟐
𝒚 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕
−𝟑
𝟎, 𝟏 𝟏, 𝟎 −𝟒

𝒚 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕 𝒙 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕 −𝟓

LINEAR EQUATIONS −𝟔
IN TWO VARIABLES
ILLUSTRATIVE EXAMPLE 2: Graph 𝟑𝒙 + 𝟓𝒚 = 𝟏𝟓.
Solution: 𝒚
Solve for at least two solutions of 𝟑𝒙 + 𝟓𝒚 = 𝟏𝟓.
𝟔
𝒚 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕
x y Let 𝒙 = 𝟎 Let 𝒚 = 𝟎 𝟓

𝟎 𝟑 𝟑𝒙 + 𝟓𝒚 = 𝟏𝟓 𝟑𝒙 + 𝟓𝒚 = 𝟏𝟓 𝟑
𝟑𝒙 + 𝟓𝒚 = 𝟏𝟓 𝒙 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕
𝟏 𝟎 𝟑 𝟎 +𝟓𝒚 = 𝟏𝟓 𝟑𝒙 + 𝟓 𝟎 = 𝟏𝟓 𝟐

𝟏
𝟎 + 𝟓𝒚 = 𝟏𝟓 𝟑𝒙 + 𝟎 = 𝟏𝟓 𝒙
𝟓𝒚 = 𝟏𝟓 𝟑𝒙 = 𝟏𝟓 −𝟖 −𝟕 −𝟔 −𝟓 −𝟒 −𝟑 −𝟐 −𝟏 𝟏
−𝟏
𝟐 𝟑 𝟒 𝟓 𝟔 𝟕 𝟖

𝟓 𝟓 𝟑 𝟑
−𝟐
𝒚=𝟑 𝒙=𝟓 −𝟑

𝟎, 𝟑 𝟓, 𝟎 −𝟒

𝒚 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕 𝒙 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕 −𝟓

LINEAR EQUATIONS −𝟔
IN TWO VARIABLES
GRAPHING with the SLOPE-INTERCEPT FORM

PROCEDURE:
Change the given equation into slope-intercept form (𝒚 = 𝒎𝒙 + 𝒃).

SLOPE (m) INTERCEPT (b):


It is defined as the ratio of It is defined as the intersection
change in y (vertical rise) to the of the line and any of the axes.
change in x (horizontal run).

𝒓𝒊𝒔𝒆
The intercept is the starting point while the slope is the movement through .
𝒓𝒖𝒏

LINEAR EQUATIONS
IN TWO VARIABLES
ILLUSTRATIVE EXAMPLE 1: Graph 𝟑𝒙 − 𝒚 = 𝟏.
𝒚
Solution:
Write 𝟑𝒙 − 𝒚 = 𝟏 in slope intercept form 𝟔

(y = mx + b). 𝟓

𝟒
𝟑𝒙 − 𝒚 = 𝟏 𝟑
−𝟑 −𝟑𝒙 𝟑𝒙 − 𝒚 = 𝟏
𝟐
−𝒚 = −𝟑𝒙 + 𝟏
𝟏
−𝟏 −𝟏
𝟑 𝒙
𝒚 = 𝟑𝒙 − 𝟏 𝒎=
−𝟖 −𝟕 −𝟔 −𝟓 −𝟒 −𝟑 −𝟐 −𝟏 𝟏
−𝟏
𝟐 𝟑 𝟒 𝟓 𝟔 𝟕 𝟖

𝟏 −𝟐

−𝟑
𝒃 = (𝟎, −𝟏)
−𝟒

−𝟓

−𝟔
LINEAR EQUATIONS
IN TWO VARIABLES
ILLUSTRATIVE EXAMPLE 2: Graph 𝟐𝒚 = 𝟑𝒙 − 𝟔.
𝒚
Solution:
Write 𝟑𝒙 − 𝒚 = 𝟏 in slope intercept form 𝟔

(y = mx + b). 𝟓

𝟐𝒚 = 𝟑𝒙 − 𝟔 𝟑
𝟐 𝟐 𝟐

𝟑 𝟐𝒚 = 𝟑𝒙 − 𝟔
𝒚= 𝒙 −𝟑
𝟏

𝟐 𝟑 𝒙
−𝟖 −𝟕 −𝟔 −𝟓 −𝟒 −𝟑 −𝟐 −𝟏 𝟏 𝟐 𝟑 𝟒 𝟓 𝟔 𝟕 𝟖
𝒎= −𝟏
𝟐 −𝟐

−𝟑
𝒃 = (𝟎, −𝟑)
−𝟒

−𝟓

−𝟔
LINEAR EQUATIONS
IN TWO VARIABLES
ILLUSTRATIVE EXAMPLE 3: Graph 𝟐𝒙 + 𝒚 = −𝟑.
𝒚
Solution:
Write 𝟑𝒙 − 𝒚 = 𝟏 in slope intercept form 𝟔

(y = mx + b). 𝟓

𝟒
𝟐𝒙 + 𝒚 = −𝟑 𝟑
−𝟐 −𝟐𝒙
𝟐
𝒚 = −𝟐𝒙 − 𝟑 𝟐𝒙 + 𝒚 = −𝟑
𝟏

𝟐 𝒙
𝒚 = −𝟐𝒙 − 𝟏 𝒎 = − −𝟖 −𝟕 −𝟔 −𝟓 −𝟒 −𝟑 −𝟐 −𝟏 𝟏
−𝟏
𝟐 𝟑 𝟒 𝟓 𝟔 𝟕 𝟖

𝟏 −𝟐

−𝟑
𝒃 = (𝟎, −𝟏)
−𝟒

−𝟓

−𝟔
LINEAR EQUATIONS
IN TWO VARIABLES
LEARNING TARGETS:

❑ I can illustrate a linear equation in two


variables;
❑ I can graph a linear equation in two variables;
and,
❑ I can solve solutions of linear equations in two
variables.
LINEAR EQUATIONS
IN TWO VARIABLES

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