Rogationist College – Parañaque
Eriberta Lane, Don Jose Green Court, San Dionisio, Sucat, Parañaque City
Tel. No: (02) 404-7270 Email: director.rcp.edu@rcj.ph
MODULE 5: Linear Inequalities (Part 2)
OBJECTIVES / LEARNING OUTCOMES:
At the end of this module, the Rogationist students will be able to…
1. illustrates linear inequalities in two variables.
2. differentiates linear inequalities in two variables from linear
equations in two variables.
3. graphs linear inequalities in two variables.
Mentor: Mr. John Brian T. Pambago
Subject: Integrated Math (Grade 9)
Month: September (Week 5)
area above the line. The line is dashed as points on the line are not
true.
INTRODUCTION:
Did you know that you also use linear equations when you shop
online? When you use the option to view items within a specific
price range, you are asking the search engine to use a linear
inequality based on price. Essentially, you are saying, “Show me all
the items for sale between $50 and $100,” which can be written as
50 ≤ 𝑥 ≤ 100, where 𝑥 is price. In this section, you will apply
what you know about graphing linear equations to graphing linear To create a system of inequalities, you need to graph two or more
inequalities.
inequalities together. Let us use 𝑦 < 2𝑥 + 5 and 𝑦 > −𝑥 since
we have already graphed each of them.
CONTENT:
In the next section, we will see that points can be solutions to
Graph a System of Two Inequalities
systems of equations and inequalities. We will verify algebraically
Remember from the module on graphing that the graph of a single whether a point is a solution to a linear equation or inequality.
linear inequality splits the coordinate plane into two regions. On
one side lie all the solutions to the inequality. On the other side,
there are no solutions. Consider the graph of the inequality 𝑦 <
2𝑥 + 5.
The purple area shows where the solutions of the two inequalities
overlap. This area is the solution to the system of inequalities. Any
point within this purple region will be true for both 𝑦 > −𝑥 and 𝑦 <
The dashed line is 𝑦 = 2𝑥 + 5. Every ordered pair in the 2𝑥 + 5.
shaded area below the line is a solution to 𝑦 < 2𝑥 + 5, as all of
the points below the line will make the inequality true. If you doubt
that, try substituting the 𝑥 and 𝑦 coordinates of Points 𝐴 and 𝐵 into Determine Whether an Ordered Pair is a Solution to a System
the inequality; you will see that they work. So, the shaded area of Linear Inequalities
shows all of the solutions for this inequality.
On the earlier graph, you can see that the points 𝐵 and 𝑁 are
The boundary line divides the coordinate plane in half. In this case, solutions for the system because their coordinates will make both
it is shown as a dashed line as the points on the line do not satisfy inequalities true statements.
the inequality. If the inequality had been 𝑦 ≤ 2𝑥 + 5, then the
In contrast, points 𝑀 and 𝐴 both lie outside the solution region
boundary line would have been solid.
(purple). While point 𝑀 is a solution for the inequality 𝑦 > −𝑥 and
Now graph another inequality: 𝑦 > −𝑥. You can check a couple of point A is a solution for the inequality 𝑦 < 2𝑥 + 5, neither point is
points to determine which side of the boundary line to shade. a solution for the system. The following example shows how to test
Checking points 𝑀 and 𝑁 yield true statements. So, we shade the a point to see whether it is a solution to a system of inequalities.
1
Rogationist College – Parañaque
Eriberta Lane, Don Jose Green Court, San Dionisio, Sucat, Parañaque City
Tel. No: (02) 404-7270 Email: director.rcp.edu@rcj.ph
Mentor: Mr. John Brian T. Pambago
Subject: Integrated Math (Grade 9)
Month: September (Week 5)
EXAMPLE
(2,1) is not a solution for 3𝑥 + 𝑦 < 4.
Is the point (2,1) a solution of the system 𝑥 + 𝑦 > 1 and 2𝑥 +
Since (2,1) is not a solution of one of the inequalities, it is not a
𝑦 < 8?
solution of the system.
Check the point with each of the inequalities.
Substitute 2 for 𝑥 and 1 for 𝑦. Is the point a solution of both Here is a graph of the system above. Notice that (2,1)is not in the
inequalities?
purple area which is the overlapping area; it is a solution for one
inequality (the red region), but it is not a solution for the second
inequality (the blue region).
𝑥+𝑦 >1
2+1>1
3 > 1 𝑇𝑅𝑈𝐸
(2,1) is a solution for 𝑥 + 𝑦 > 1.
2𝑥 + 𝑦 < 8
2(2) + 1 < 8
4+1<8
5 < 8 𝑇𝑅𝑈𝐸
(2,1) is a solution for 2𝑥 + 𝑦 < 8.
Since (2,1) is a solution of each inequality, it is also a solution of
the system.
Here is a graph of the system in the example above. Notice
that (2,1) lies in the purple area which is the overlapping area for
the two inequalities.
As shown above, finding the solutions of a system of inequalities
can be done by graphing each inequality and identifying the region
they share. Below, you are given more examples that show the
entire process of defining the region of solutions on a graph for a
system of two linear inequalities. The general steps are outlined
below:
 Graph each inequality as a line and determine whether it
will be solid or dashed.
 Determine which side of each boundary line represents
solutions to the inequality by testing a point on each side.
 Shade the region that represents solutions for both
inequalities.
Systems with No Solutions
In the next example, we will show the solution to a system of two
inequalities whose boundary lines are parallel to each other. When
the graphs of a system of two linear equations are parallel to each
EXAMPLE
Is the point (2,1) a solution of the system 𝑥 + 𝑦 > 1 and 3𝑥 + other, we found that there was no solution to the system. We will
get a similar result for the following system of linear inequalities.
𝑦 < 4?
Check the point with each of the inequalities. EXAMPLES
Substitute 2 for 𝑥 and 1 for 𝑦. Is the point a solution of both Graph the system
inequalities?
𝑥+𝑦 >1
2+1>1
3 > 1 𝑇𝑅𝑈𝐸
(2,1) is a solution for 𝑥 + 𝑦 > 1.
3𝑥 + 𝑦 < 4
3(2) + 1 < 4
6+1<4
7 < 4 𝐹𝐴𝐿𝑆𝐸
𝑦 ≥ 2𝑥 + 1
𝑦 < 2𝑥 − 3
The boundary lines for this system are parallel to each other. Note
how they have the same slopes.
𝑦 = 2𝑥 + 1
𝑦 = 2𝑥 − 3
Plotting the boundary lines will give the graph below. Note that the
inequality 𝑦 < 2𝑥 − 3 requires that we draw a dashed line, while
the inequality 𝑦 ≥ 2𝑥 + 1 requires a solid line.
2
Rogationist College – Parañaque
Eriberta Lane, Don Jose Green Court, San Dionisio, Sucat, Parañaque City
Tel. No: (02) 404-7270 Email: director.rcp.edu@rcj.ph
Mentor: Mr. John Brian T. Pambago
Subject: Integrated Math (Grade 9)
Month: September (Week 5)
This is not true, so we know that we need to shade the other side
of the boundary line for the inequality 𝑦 < 2𝑥 − 3. The graph will
now look like this:
Now we need to shade the regions that represent the
inequalities. For the inequality 𝑦 ≥ 2𝑥 + 1, we can test a point on
either side of the line to see which region to shade. Test (0,0) to
make it easy.
Substitute (0,0) into 𝑦 ≥ 2𝑥 + 1
𝑦 ≥ 2𝑥 + 1
0 ≥ 2(0) + 1
0≥1
This is not true, so we know that we need to shade the other side
of the boundary line for the inequality 𝑦 ≥ 2𝑥 + 1. The graph will
now look like this:
This system of inequalities has no points in common so has no
solution.
Summary
 Solutions to systems of linear inequalities are entire
regions of points.
 You can verify whether a point is a solution to a system of
linear inequalities in the same way you verify whether a
point is a solution to a system of equations.
 Systems of inequalities can have no solutions when
boundary lines are parallel.
ASSESSMENT:
Answer the worksheet given along with this module.
REFERENCES:
 https://courses.lumenlearning.com/intermediatealgebra/c
hapter/read-graphing-using-ordered-pairs/
 https://courses.lumenlearning.com/intermediatealgebra/c
hapter/read-quadrants-on-the-coordinate-plane/
 https://courses.lumenlearning.com/intermediatealgebra/c
hapter/read-define-solutions-to-systems-of-linearinequalities/
Now shade the region that shows the solutions to the
inequality 𝑦 < 2𝑥 − 3. Again, we can pick (0,0) to test, because
it makes easy algebra.
Substitute (0,0) into 𝑦 < 2𝑥 − 3
𝑦 < 2𝑥 − 3
0 < 2(0) − 3
0 < −3
3