system of linear inequalities into variables

Mathemati
cs 8
Quarter 2 Week 1

Most Essential Learning Competencies

differentiates linear inequalities in two
variables from
linear equations in two variables.
 Illustrates and graphs linear inequalities
in two variables.
 solves problems involving linear
inequalities in two variables.

A linear
inequality
is almost the same as a linear equation,
except the equals sign is replaced with an
inequality symbol. It can be written in
different forms such as:

Linear Inequality
“ax + by is less than c”
“ax + by is greater than c”
“ax + by is less than or
equal to c”
“ax + by is greater than
or equal to c”
ax + by ≤ c ax + by ≥ c
ax + by < c
ax + by > c

Differentiate Linear
equations and Linear
Inequality

Example of Linear
EQUATION
1. 2x + 5y = 10
As you can see it forms a
line

Example of Linear
EQUATION
2. 3x -4y =12
Again this example forms a
line

Example of Linear
INEQUALITY
1. 3x – y 9
≤
It has a shaded region and
a solid line as a boundary.

Example of Linear
INEQUALITY
2. 2x - 3y > 60
It has a shaded region and
a solid line as a boundary.

Graphs Linear
Inequalities in Two
Variables

The Graph of a Linear Inequality is the set of
all points that are solutions to the inequality.
Step 2: Solve for the boundary line of
the graph using x and y intercept
method.
Let y = 0 Let x = 0
x + y = 4 x + y = 4
x + 0 = 4 0 + y = 4
x = 4 y = 4
(4 ,0) (0,4)
Points (4,0) and (0,4) will be the
boundaries of the graph.
2
Step 1: Change the inequality
x + y < 4
into an equation x + y = 4.
1
Example 1:
x + y < 4

The Graph of a Linear Inequality

Step 4: Decide which part of the plane
will be shaded by performing test point
(0,0) to the inequality.
Substitute x = 0 and y= 0
to x + y < 4
x + y < 4
0 + 0 < 4
0 < 4 - TRUE
If proven TRUE, the half plane that
contains the origin will be shaded and if
it is FALSE, the origin must not be
included in shading.
4
Step 3: Determine what type
of line (solid or broken) will
be used. For symbols > or <
use broken lines. Therefore.
the two points on the boundary
line (4,0) and (0,4) will be
holes.
3

Step 5: Since 0 < 4 is TRUE,
the graph will be:
The origin (0,0) is included.
The shaded region contains all
the possible solutions to the
inequality x + y < 4.
5

Step 2: Solve for the boundary line of
the
graph using x and y intercept method.
Let y = 0 Let x = 0
8x – 6y = 24 8x – 6y = 24
8x – 6(0) = 24 8(0) - 6 y = 24
8x – 0 = 24 0 - 6y = 24
8x = 24 -6y = 24
x = 3 y = -4
(3,0) (0,-4)
Points (3,0) and (0,-4) will be the
boundaries of the graph.
2
Step 1: Change the inequality
8x – 6y ≥ 24 into equation
8x – 6y = 24.
1
Example 2:
8x – 6y 24
≥

Step 4: Decide which part of the plane
will be shaded by performing test point
(0,0) to the inequality.
Substitute x = 0 and y= 0
to 8x – 6y ≥ 24
8(0) – 6(0) ≥ 24
0 - 0 ≥ 24
0 ≥ 24 – FALSE
If proven TRUE, the half-plane that
contains the origin will be
shaded and if it is FALSE, the origin
must not be included in
shading.
4
Step 3: Determine what type
of line (solid or broken) will
be
used. For symbols ≥ or ≤ use
a solid line. Therefore the
two points on the boundary
line (4,0) and (0,4) will be part
of the line.
3

Step 5: Since 0 ≥ 24 is FALSE, the
graph will
be:
The origin (0,0) is not
included.
The shaded region contains all the
possible solutions for the given
inequality
5

A Picture
Always
Reinforces the
Concept
Images reveal large
amounts of data, so
remember: use an
image instead of
long texts

THANK
YOU!
Excited to be
BTS!
Back To School!

system of linear inequalities into variables

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