General Mathematics

Quarter 1 – Module 6:
Solving Rational Equations and
Inequalities

CO_Q1_General Mathematics SHS

Module 6
General Mathematics
Alternative Delivery Mode
Quarter 1 – Module 6: Solving Rational Equations and Inequalities
First Edition, 2021

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 General Mathematics
Quarter 1 – Module 6:
Solving Rational Equations and
Inequalities
Introductory Message
This Self-Learning Module (SLM) is prepared so that you, our dear learners,
can continue your studies and learn while at home. Activities, questions, directions,
exercises, and discussions are carefully stated for you to understand each lesson.

Each SLM is composed of different parts. Each part shall guide you step-by-
step as you discover and understand the lesson prepared for you.

Pre-tests are provided to measure your prior knowledge on lessons in each

SLM. This will tell you if you need to proceed on completing this module or if you
need to ask your facilitator or your teacher’s assistance for better understanding of
the lesson. At the end of each module, you need to answer the post-test to self-
check your learning. Answer keys are provided for each activity and test. We trust
that you will be honest in using these.

In addition to the material in the main text, Notes to the Teacher are also
provided to our facilitators and parents for strategies and reminders on how they can
best help you on your home-based learning.

Please use this module with care. Do not put unnecessary marks on any part
of this SLM. Use a separate sheet of paper in answering the exercises and tests. And
read the instructions carefully before performing each task.

If you have any questions in using this SLM or any difficulty in answering the
tasks in this module, do not hesitate to consult your teacher or facilitator.

Thank you.
 What I Need to Know

This module was designed and written for learners like you to determine a method
and set of steps for solving rational equations and inequalities. Learners like you can
also explore and develop new methods that you have synthesized and apply these
techniques for performing operations with rational expressions.

In this module, you will able to explain the appropriate methods in solving rational
equations and inequalities you used. You will also be able to check and explain
extraneous solutions.

After going through this module, you are expected to:

1. Apply appropriate methods in solving rational equations and inequalities.
2. Solve rational equations and inequalities using algebraic techniques for
simplifying and manipulating of expressions.
3. Determine whether the solutions found are acceptable for the problem by
checking the solutions.

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 What I Know

Choose the letter of the best answer. Write the chosen letter on a separate sheet of
paper.
1. Which of the following expressions is an equality between two expressions
containing one or more variables?
a. rational function
b. rational equation
c. rational inequality
d. irrational expression

2. What do you call a root obtained in the process of solving an equation which
looks correct but after analyzing it turns out as incorrect?
a. extraneous solution
b. rational expression
c. least common denominator
d. quotient

3. What do you call an inequality which involves one or more rational

expressions?
a. rational function
b. rational equation
c. rational inequality
d. irrational expression
4. What is the usual technique to a solve rational equation?
a. multiply both sides of the equation by its greatest common factor
b. multiply both sides of the equation by its least common denominator
c. multiply both sides of the equation by its inverse factor
d. multiply both sides of the equation by its greatest common
denominator

For items 5-9: Refer to the rational equation below.

+ =
𝑥 1 𝑥
3 4 2

5. What is the LCD of the denominator 3, 4 and 2?

a. 3
b. 6
c. 8
d. 12

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6. What property will be used if you multiply the LCD on both sides of the
equation?
a. Distributive Property
b. Associative Property
c. Commutative Property
d. Additive Property

7. What will be the new form of the equation after applying the property and

a. 4𝑥 + 3 = 6𝑥
simplifying?

b. 3𝑥 + 4 = 2𝑥
c. 6𝑥 + 4 = 3𝑥
d. 12𝑥 + 3 = 12𝑥

8. What will be the solution on the given rational equation?

a. 2
3
3
2
b.
c. 2
d. 3

9. How will you check if your solution is correct?

a. by eliminating the rational expressions.
b. by dividing both sides of the equation by LCD.
c. by applying Commutative Property.
d. by substituting the answer or solution in the original equation.

10. What will you if you obtained a solution that made the expression in the
equation undefined?
a. Accept even if it is untrue value.
b. Do not reject since it will satisfy the equation in a long run.
c. Discard the solution since it is unreal.
d. Continue the solution even if it will give undefined answer.

a. ≤
11. Which of the following is NOT an inequality sign?

b. √
c. ≥
d. <

12. Express the graph of solution set into interval notation.

a. {𝑥 | − 3 ≤ 𝑥 < 1}
b. {𝑥 | − 3 ≤ 𝑥 ≤ 1}
c. {𝑥 | 3 < 𝑥 ≤ 1}
d. {𝑥 | 3 ≤ 𝑥 < 1}

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13. Below are the steps in solving rational inequality EXCEPT
a. Put the inequality in general form.
b. Set the numerator and denominator equal to one and solve.
c. Plot the critical values on a number line, breaking the number line
into intervals and take a test number from each interval by
substituting into the original inequality.
d. Determine if the endpoints of the intervals in the solution should be
included in the intervals.

(𝑥+3)
≤ 1.
(𝑥−2)
14. Solve for the solutions of the rational inequality

a. [∞, 2)
b. (∞, 2]
c. (-∞, 2)
d. [-∞, 2)

15. How will you know that the critical points for item no. 14 will satisfy the
inequality?
a. If it makes a true statement, then the interval from which it came from
is not in the solution.
b. If it makes a false statement, then the interval from which it came
from is in the solution.
c. If it makes a true statement, then the interval from which it came from
is in the solution.
d. If it makes a false statement, then the interval from which it came
from is either in the solution or not.

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 Lesson
Solving Rational Equations
1 and Inequalities

In this lesson, you shall explore more about solving rational equations and inequalities
by carefully studying the step by step methods of solutions. You will first start from the
easiest procedures in solving this type of equation and as you progress, you will learn
more techniques and concepts that will help you to solve more complex problems
related to this topic. Exercises will range from the simplest problems to the most
complex.

At this point, students like you have already solved a variety of equations, including linear
and quadratic equations from the previous grade level. Rational equations and
inequalities follow the sequence of solving problems by combining the concepts used
in solving both linear and quadratic equations. Students will be assessed using both
formative and summative assessments along the way to best evaluate your progress.

What’s In

Let’s Review!
How do you solve algebraic expressions? What are the different properties you need to
apply to solve problems involving rational equations and inequalities?

For you to begin, you need to recall some properties and processes to simplify rational
expressions by answering the following problems below. Write your answer inside
the box.

𝑥−2
𝑥2−4
1. Simplify the given rational expression:

3𝑥+1 𝑥+1
∙2
2. Multiply the given rational expressions:
3𝑥2+
𝑥 −1
𝑥

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 3𝑥+4
3. Find the sum of given rational expressions with like denominators: 5𝑥−1
+
𝑥−8 𝑥−8

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 6 2
4. Find the difference of the given rational expressions with unlike denominators:

−
𝑥2 − 4 𝑥2 − 5𝑥 + 6

Let’s check if you have made it! You can also write your solution on the prepared box
to compare the techniques you apply.

1. To simplify the rational expression you can do the following steps.

𝑥−2
𝑥2 − 4
Steps in simplifying

𝑥−2
rational expression

(𝑥 − 2)(𝑥 + 2)
1. Factor the Write your previous
denominator of the solution here for
rational expression. comparison.
2. Cancel the common
factor.
1
𝑥+2
3. Write the simplified
rational expression.

2. To multiply rational expressions you can do the following steps.

Steps in multiplying 3𝑥+1 𝑥+1

rational expressions 𝑥2−1 . 3𝑥2+𝑥
1. Factor out all 3𝑥+1 𝑥+1 Write your previous
(𝑥+1)(𝑥−1) 𝑥(3𝑥+1)
. solution here for
possible common
comparison.
(3𝑥 + 1)(𝑥 + 1)
factors.

(𝑥 + 1)(𝑥 − 1)(𝑥)(3𝑥 + 1)
2. Multiply the
numerators and
denominators.
3. Cancel out all
common factors.

1
𝑥(𝑥 − 1)
4. Write the simplified
rational expression.

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3. To add and subtract rational expressions with like denominators you can do the
following steps.

5𝑥 − 1 3𝑥 + 4
+
𝑥−8 𝑥−8
Steps in addition or
subtraction of rational
expressions with like
denominators
1. Add or subtract the Write your previous

5𝑥 − 1 + 3𝑥 + 4
numerators of both solution here for
expressions and comparison.
keeping the 𝑥−8
common

5𝑥 + 3𝑥 + 4 − 1
denominator.

𝑥−8
2. Combine like terms

8𝑥 + 3
in the numerator.

𝑥−8
3. Write the simplified
rational expression.

4. To add and subtract rational expressions with unlike denominators you can do
the following steps.

6 2
Steps in adding or
− 2
𝑥 − 4 𝑥 − 5𝑥 + 6
subtracting rational
2
expressions with
unlike denominators

6 2
1. Factor the Write your previous
−
(𝑥 − 2)(𝑥 + 2) (𝑥 − 2)(𝑥 −
denominator of each solution here for

3)
fraction to help find comparison.
the LCD.

𝐿𝐶𝐷: (𝑥 − 2)(𝑥 + 2)(𝑥 − 3)

2. Find the least
common
denominator (LCD).
6(𝐿𝐶𝐷) 2(𝐿𝐶𝐷)
−
3. Multiply each
(𝑥 − 2)(𝑥 + 2) (𝑥 − 2)(𝑥 −
expression by

3)
its LCD

6(𝑥 − 3) − 2(𝑥 + 2)
4. Write the simplified
expression.
5. Let the simplified
6𝑥 − 18 − 2𝑥 − 4
(𝑥 − 2)(𝑥 + 2)(𝑥 − 3)
expression as the
numerator and the
LCD as the
denominator of the
new fraction

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 6. Combine like terms
and reduce the
4𝑥 − 22
rational expression

(𝑥 − 2)(𝑥 + 2)(𝑥 − 3)
if you can. In this
case, the rational
expression cannot
be simplified.

How was the activity? Did you answer all the reviewed items correctly? Great! If you
did, then you can now move forward on the next stage of this topic and I am confident
that it will be very easy for you to understand the lesson.

Notes to the Teacher

Please remind our students that learning mathematics is a linear
process wherein the math skills and knowledge from the previous
modules and grade level will be used throughout this topic. For example,
if the students have not mastered arithmetic properties and processes
then they will have difficulty with the current topic because it requires
all of these prerequisite skills. Therefore, it will be necessary to go back,
review previous topics and problem-solving skill before they can
continue. Inspire our students that learning is not always onward and
upward, sometimes we have to take a glimpse of the past before we can
move forward.

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 What’s New

Follow Me Activity
Solving Rational Equations and Inequalities

Before you proceed on the lesson proper try to answer the rational equation and
inequality using guided procedure. You can synthesize your own steps in solving the
problem. You can refer to previous activities if you are having difficulty processing
arithmetic properties. Hope you enjoy answering before you continue to the next part
of the discussion.
1. Solve example 2 of the rational equation by following the given steps.

𝑥−3 1 1 2 1 1
Example 1 Example 2
+ = − =
𝑥2 − 25 𝑥 + 5 (𝑥 − 5) 𝑥 −1 𝑥−1 2
2
Rational Equation

(𝑥 + 5)(𝑥 − 5)
1. Find the Least LCD:
Common Denominator

(𝑥 + 5)(𝑥 − 5)[
𝑥−3 1
+
(LCD).

= 2
2. Multiply both sides of
𝑥 −25 𝑥+5
the equation by its the
LCD.
1 ]
(𝑥−5)
(𝑥 − 3) + 1(𝑥 − 5) = 1(𝑥 +
5)
3. Apply the Distributive

𝑥−3+𝑥−5=𝑥+5
Property and then
simplify.

2𝑥 − 8 = 𝑥 + 5
simplify:

2𝑥 − 𝑥 = 8 + 5
𝑥 = 13

𝑥 = 13
4. Find all the possible
values of x.

𝑥−3 1 1
5. Check each value by Checking:
+ =
𝑥 − 25 𝑥 + 5 (𝑥 −
substituting into 2

5)
original equation and

13 − 3 1 1
reject any extraneous
+ =
13 − 25 13 + 5 (13 −
root/s 2

5)
10 1 1
+ =
169 − 25 18 8
10 1 1
+ =
144 18 8
10 + 8 1
=
144 8

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 1 1
=
8 8
✓
Note: No extraneous root

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2. Solve example 2 of rational inequality. You can refer to example 1 for the guided
steps.

3 3𝑥 + 1
Example 1 Example 2
≤ −1 ≥2
𝑥−2 𝑥−1
Rational Inequality

3
1. Put the rational inequality
+1≤0
𝑥−2
in general form.

𝑄𝑄(𝑥)> 0
(𝑥)
𝑅

by <, ≤ 𝑎𝑛𝑑 ≥
where > can be replaced

3 + 1(𝑥 − 2)
≤0
𝑥−2
2. Write the inequality into a
single rational expression

𝑥+1
on the left side. (You can
≤0
𝑥−2
refer to the review section
for solving unlike
denominators)

𝑥+1=0
3. Set the numerator and Numerator:

𝑥 = −1
denominator equal to zero
and solve. The values you

𝑥−2=0
get are called critical Denominator:
values.
𝑥=2
4. Plot the critical values on a
number line, breaking the
number line into intervals.
3
≤ −1
𝑥−2
5. Substitute critical values
to the inequality to

when 𝑥 = −1
determine if the endpoints
3
≤ −1
of the intervals in the
−1 − 2
3
solution should be
≤ −1
−3
included or not.

−1 ≤ −1 ✓
( 𝑥 = −1 is included in
the solution)

when 𝑥 = 2
3
≤ −1
2−2
3
≤ −1
0
𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 ≤ −1

( 𝑥 = 2 is not included
☓

in the solution)
5. Select test values in each

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 interval and substitute
those values into the

when 𝑥 =
inequality.

−2 3
Note:
≤ −1
−2 − 2
If the test value makes the
3
≤ −1 ☓ 𝑓𝑎𝑙𝑠𝑒
inequality true, then the

−4
entire interval is a solution
to the inequality.

when 𝑥 = 0
If the test value makes the
3
≤ −1
inequality false, then the

0−2
3
entire interval is not a

≤ −1 ✓ 𝑡𝑟𝑢𝑒
solution to the inequality.
−2

when 𝑥 = 3
3
≤ −1
3−2
3 ≤ −1 ☓ 𝑓𝑎𝑙𝑠𝑒

[−1,2)
6. Use interval notation or
set notation to write the
final answer.

How do you find the activity? Have you enjoyed it? Did you follow the steps
correctly? The activity tells you about solving rational equations and inequalities.
Yes, you read it right. You almost got it!

Let’s check if your answers are correct and which process did you find it difficult. I
hope you enjoyed answering by your own.

What is It

Rational equation is an equation containing at least one rational expression with a

polynomial in the numerator and denominator. It can be used to solve a variety of
problems that involve rates, times and work. Using rational expressions and
equations it can help us to answer questions about how to combine workers or
machines to complete a job on schedule.
Let us use the previous activity to discuss and deepen your knowledge and skills in
solving rational equation. The first thing to be in your mind in solving rational
equation is to eliminate all the fractions.

Let us solve

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 2 1 1
− =
𝑥2 − 1 𝑥−1 2

The LCD of the given fractions is 2(𝑥 − 1)(𝑥 + 1)

Step 1. You need to find the Least Common Denominator (LCD).

Step 2. You need to multiply LCD to both sides of the equation to eliminate the
fractions. You can also apply cross multiplication if and only if you have one
fraction equal to one fraction, that is, if the fractions are proportional. In this case
you cannot use the cross multiplication unless you simplify the left equation into a

2 1 1
single fraction.

2(𝑥 − 1)(𝑥 + 1) � 2 − 2= �
𝑥 −1 𝑥−1

Step 3. You simplify the resulting equation using the distributive property and then
combine all like terms.

2(2) − 2(𝑥 + 1) = (𝑥 − 1)(𝑥 + 1)

4 − 2𝑥 − 2 = 𝑥2 − 1
𝑥2 + 2𝑥 − 3 = 0
Step 4. You need to solve the simplified equation to find the value/s of x. In this

𝑥2 + 2𝑥 − 3 =
case, we need to get the equation equal to zero and solve by factoring.

0 (𝑥 + 3)(𝑥 − 1)
= 0
𝑥 + 3 = 0 𝑜𝑟 𝑥 − 1 = 0
𝑥 = −3 𝑜𝑟 𝑥 = 1
So possible solutions are -3 and 1.
Step 5. Finally, you can now check each solution by substituting in the original

2 1 1
equation and reject any extraneous root/s (which do not satisfy the equation).
− =
𝑥 −1 𝑥−1
2
2
When 𝑥 = −3
2 1 1
− =
(−3) − 1 (−3) − 1
2
2
2 1 1
+ =
8 4 2
1 1
=
2 2
✓
When 𝑥 =
1
2 1 1
− =
(1) − 1 (1) − 1
2
2
2 1 1
− =
0 0 2
1
0= ☓
2
In this case, 𝑥 = −3 is the only solution. That’s why it is always important to check
all solutions in the original equations. You may find that they yield untrue

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statements or produce undefined expressions.

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Rational inequality is an inequality which contains one or more rational
expressions. It can be used in engineering and production quality assurance as well
as in businesses to control inventory, plan production lines, produce pricing
models, and for shipping/warehousing goods and materials.
Solving an inequality is much like solving a rational equation except that there are
additional steps that focus on illustrating the solution set of an inequality on a
number line.
Let us use problem number 2 in the previous activity to discuss and deepen your

3𝑥 + 1
knowledge and skills in solving rational inequality.
≥2
𝑥−1

<, ≤ 𝑎𝑛𝑑 ≥.
Step 1. Put the rational inequality in the general form where > can be replaced by

𝑅(𝑥)
>0
𝑄𝑄(𝑥)
3𝑥 + 1
−2≥0
𝑥−1

3𝑥 + 1 − 2(𝑥 − 1)
Step 2. Write the inequality into a single rational expression on the left-hand side.
≥0
𝑥−1

𝑥+3
≥0
𝑥−1
Note: Remember that one side must always be zero and the other side is always a
single fraction, so simplify the fractions if there is more than one fraction.
Step 3. Set the numerator and denominator equal to zero and solve. The values you
get are called critical values.
𝑥+3=0
𝑥 = −3
Numerator:

𝑥−1=0
𝑥=1
Denominator:

Step 4. Plot the critical values on a number line, breaking the number line into
intervals.

Step 5. Substitute critical values to the inequality to determine if the endpoints of

the intervals in the solution should be included or not.

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 3𝑥 + 1
≥2
𝑥−1
when 𝑥 =
−3 3(−3) + 1
≥2
(−3) − 1
−8
≥2
−4
2 ≥ 2 ✓ ( 𝑥 = −3 is included in the solution)

when 𝑥 =
1 3(1) + 1
≥2
(1) − 1
4
≥2
0
𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 ≥ 2 ☓ ( 𝑥 = 1 is not included in the solution)
See the illustration below.

Step 6. Select test values in each interval and substitute those values into the
inequality.
3𝑥 + 1
≥2
𝑥−1

when 𝑥 =
−5 3(−5) + 1
≥2
(−5) − 1
−14
≥2
−6
7
𝑜𝑟 2.33 ≥ 2 ( 𝑥 = −5 TRUE)
3

when 𝑥 =
−1 3(−1) + 1
≥2
(−1) − 1

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 −
2 ≥2
−
1 ≥ 2 ( 𝑥 = −1 FALSE)
2
when 𝑥 =
3
≥2
3(3) +
1 ≥2
(3) −
1 5 ≥ 2 ( 𝑥 = 5 TRUE)
1
0
2
Note:

a. If the test value makes the inequality TRUE, then the entire interval is a solution
to the inequality.
b. If the test value makes the inequality FALSE, then the entire interval is not a
solution to the inequality.

Step 7. Use interval notation to write the final answer.

(−∞, −3] ∪ (1, ∞)

Let’s learn more!

Solve each rational equation and inequality.
4𝑥 + 1
1. −3 12
=
𝑥+1 𝑥 −1
2

2𝑥 − 8
2. ≥ 0
𝑥−2

Solution:

4𝑥 + 1 12
−3= 2
Rational Equation 𝑥+1 𝑥 −1

(𝑥 + 1)(𝑥 − 1)
1. Find the Least Common LCD:
Denominator (LCD).

4𝑥 + 1 12
]
2. Multiply both sides of the equation
(𝑥 + 1)(𝑥 − 1)[ −3= 2
𝑥+1 𝑥 −1
by its the LCD.

(𝑥 − 1)(4𝑥 + 1) − 3(𝑥 + 1)(𝑥 − 1)

= 12
3. Apply the Distributive Property and
then simplify.

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 simplify:
4𝑥2 − 3𝑥 − 1 − 3𝑥2 + 3 = 12
𝑥2 − 3𝑥 + 2 = 12
𝑥2 − 3𝑥 − 10 = 0
(𝑥 − 5)(𝑥 + 2) =
0
Factor

𝑥−5=0 𝑥=5
4. Find all the possible values of x. 𝑥+2=0 𝑥 = −2

4𝑥 + 1 12
5. Check each value by substituting Checking:
−3= 2
𝑥+1 𝑥 −1
into original equation and reject any
extraneous root/s

when 𝑥 = 5
4(5) + 1 12
−3= 2
5+1 5 −1
21 12
−3=
6 24
3 12
=
6 24
1 1
=
2 2
✓

when 𝑥 = −2
4(−2) + 1 12
−3=
(−2) + 1 (−2) − 1
2
−7 12
−3=
−1 3
4= 4 ✓

Note: No extraneous root

2𝑥 − 8
≥ 0
𝑥−2
Rational Inequality
1. Put the rational inequality in general
form. This inequality is already in general

𝑄𝑄(𝑥)> 0
(𝑥)
form. We are all set to go.
𝑅
where > can be replaced by <, ≤
𝑎𝑛𝑑 ≥
2. Write the inequality into a single
rational expression on the left side. This inequality is already in a single
(You can refer to the review section rational expression wherein 0 is on one

2𝑥 − 8
side.
≥ 0
for solving unlike denominators)

𝑥−2

19 CO_Q1_General mathematics SHS

Module 6
 2𝑥 − 8 = 0
3. Set the numerator and denominator Numerator:

2𝑥 = 8
equal to zero and solve. The values

𝑥=4
you get are called critical values.

𝑥−2=0
Denominator:

𝑥=2

4. Plot the critical values on a number

line, breaking the number line into
intervals.

2𝑥 − 8
≥ 0
𝑥−2
5. Substitute critical values to the
inequality to determine if the

when 𝑥 = 2
endpoints of the intervals in the
2(2) − 8
≥ 0
solution should be included or not.
2−2
−4
≥0
0
𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 ≥ 0 ☓
( 𝑥 = 2 is not included in the solution)

when 𝑥 = 4
2(4) − 8
≥ 0
4−2
0
≥0
2
0 ≥0 ✓
( 𝑥 = 4 is included in the solution)

20 CO_Q1_General mathematics SHS

Module 6
 6. Select test values in each interval
and substitute those values into the
inequality.
Note:
2𝑥 − 8
≥ 0
If the test value makes the inequality

𝑥−2
true, then the entire interval is a
solution to the inequality.
when 𝑥 = 1
2(1) − 8
If the test value makes the inequality

≥ 0
false, then the entire interval is not a
1−2
−6
solution to the inequality.

≥0
−1
6 ≥ 0 ✓ 𝑡𝑟𝑢𝑒

when 𝑥 = 3
2(3) − 8
≥ 0
3−2
−2
≥0
1
−2 ≥ 0 ☓ 𝑓𝑎𝑙𝑠𝑒

when 𝑥 = 5

2(5) − 8
≥ 0
5−2
2
≥ 0 ✓ 𝑡𝑟𝑢𝑒
3

7. Use interval notation or set notation (−∞, 2) ∪ [4, ∞)

to write the final answer.

What’s More

Activity 1.
Solve the following rational equations and inequalities using the guided procedure
on the table below.
1.

𝑥−2 1 1
+ =
𝑥 −4 𝑥+2 𝑥−2
2
Rational Equation

1. Find the Least Common

Denominator (LCD).

21 CO_Q1_General mathematics SHS

Module 6
2. Multiply both sides of the equation
by its the LCD.

3. Apply the Distributive Property and

then simplify.

𝑥=6
4. Find all the possible values of x.

5. Check each value by substituting

into the original equation and reject
any extraneous root/s

22 CO_Q1_General mathematics SHS

Module 6
2.

𝑥−6 2 1
+ =
𝑥2 − 4𝑥 − 12 𝑥 + 2 𝑥 − 6
Rational Equation

1. Find the Least Common

Denominator (LCD).

2. Multiply both sides of the equation

by its the LCD.

3. Apply the Distributive Property and

then simplify.

𝑥 = 10
4. Find all the possible values of x.

5. Check each value by substituting

into original equation and reject any
extraneous root/s

3.
2(𝑥 − 4)
< −4
Rational Inequality 𝑥

1. Put the rational inequality in general

form.

> 0
(𝑥)
𝑅
𝑄𝑄(𝑥)
where > can be replaced by <, ≤
𝑎𝑛𝑑 ≥
2. Write the inequality into a single
rational expression on the left side.
(You can refer to the review section
for solving unlike denominators)
3. Set the numerator and denominator
equal to zero and solve. The values
you get are called critical values.
4. Plot the critical values on a number
line, breaking the number line into
intervals.
5. Substitute critical values to the
inequality to determine if the
endpoints of the intervals in the
solution should be included or not.

23 CO_Q1_General mathematics SHS

Module 6
 6. Select test values in each interval
and substitute those values into the
inequality.
Note:
If the test value makes the inequality
true, then the entire interval is a
solution to the inequality.
If the test value makes the inequality
false, then the entire interval is not a

3
solution to the inequality.

( 0, )
4
7. Use interval notation or set notation
to write the final answer.
4.
𝑥2 + 𝑥 − 6
≤0
Rational Inequality 𝑥2 − 3𝑥 − 4

1. Put the rational inequality in general

form.

> 0
(𝑥)
𝑅
𝑄𝑄(𝑥)
where > can be replaced by <, ≤
𝑎𝑛𝑑 ≥
2. Write the inequality into a single
rational expression on the left side.
(You can refer to the review section
for solving unlike denominators)
3. Set the numerator and denominator
equal to zero and solve. The values
you get are called critical values.
4. Plot the critical values on a number
line, breaking the number line into
intervals.
5. Substitute critical values to the
inequality to determine if the
endpoints of the intervals in the
solution should be included or not.
6. Select test values in each interval
and substitute those values into the
inequality.
Note:
If the test value makes the inequality
true, then the entire interval is a
solution to the inequality.
If the test value makes the inequality
false, then the entire interval is not a
solution to the inequality.

[−3, −1 ) ∪ [ 2, 4 )
7. Use interval notation or set notation
to write the final answer.

24 CO_Q1_General mathematics SHS

Module 6
Activity 2
Solve each problem below and choose the letter that corresponds to the solution to
each problem. Place the correct answer in the corresponding lines.

What did the bible verses 1John 4:7-21 is all about?

1 2 3 4 5 6 7 8 9
𝑥+2 2𝑥−4
=
1.
2

7
3 −
1
=
2. A. -3
4𝑥 𝑥2 2𝑥2
2𝑥
=
C. -1 and 6
𝑥+
3.
1
+5
2
D. -5
2
𝑥

𝑥𝑥−3 = 𝑥−3
2
−1 8 E. (2, 11
2
4. ]

1 𝑥 4
+𝑥− =
5. G. 4
𝑥− 𝑥2−8𝑥+1
6 2 2
I. 3

< 4
5𝑥
𝑥−
6. L. (-4, 1)
1
2
−7=
𝑥
𝑥−2
7. N. -3 and 3
𝑥−
2
O. 2
𝑥2+𝑥−12
𝑥−1 ≤ 0
8.
V. (-∞, -4] ∪ (1, 3]
S. -1

𝑥+1
3𝑥−2 ≥5 Y. (-∞, -4) ∪ [1, 3)
9.

What I Have Learned

Complete the following statements by writing the correct word or words and
formulas.
1. A is an equation containing at least one rational
expression with a polynomial in the numerator and denominator.
2. To determine if the endpoints of the intervals in the solution should be included
or not you need to the critical values to the inequality.
3. In order to get the critical values you need to set and
equal to zero.

25 CO_Q1_General mathematics SHS

Module 6
4. The first step in solving rational inequality is to put the inequality in general form
where in one side must always be and the other side
is in a fraction.

26 CO_Q1_General mathematics SHS

Module 6
 5. If the test value makes the inequality , then the
entire interval is a solution to the inequality.

Topics for Discussion

1. Explain the similarities and differences in solving between rational equations
and inequalities?
2. Why do extraneous solutions sometimes occur and don’t work in the original
form of the equation?

What I Can Do

Read the situation carefully and solve the problem.

The new COVID-19 testing facility in Lucena City is operating with two laboratory
technicians. Technician A takes 2 hours to finish 50 samples of specimens from
CoVID-19 patients. Technician B takes 3 hours to finish 45 samples of specimens
from COVID-19 patients. Working together, how long should it take them to finish
150 samples of specimens from COVID-19 patients?

Hint:
Think about how many samples of specimens each technician can finish in one hour.
This is their testing rate.

27 CO_Q1_General mathematics SHS

Module 6
 Assessment

Choose the letter of the best answer. Write the chosen letter on a separate sheet of
paper.
1. It is an equation containing at least one fraction whose numerator and
denominator are polynomials.
a. rational function
b. rational equation
c. rational inequality
d. irrational equation

2. The usual technique to eliminate denominator in solving a rational equation

is to multiply both sides of the equation by its .
a. inverse factor
b. greatest common factor
c. least common denominator
d. greatest common denominator

3. An inequality which involves one or more rational expressions is called

.

a. rational function
b. rational equation
c. rational inequality
d. irrational equation

4. You can only use cross multiplication in solving rational equation if and only
if you have one fraction equal to one fraction, that is, if the fractions are
.
a. negative
b. positive
c. inequal
d. proportional

5. If the test value makes the inequality , then the

entire interval is not a solution to the inequality.
a. true
b. false
c. proportional
d. reciprocal

28 CO_Q1_General mathematics SHS

Module 6
 5 2 3
For items 6-7: Refer to the rational equation below.
+
2𝑥 − 𝑥−
=
4 𝑥+3 2

a. 𝑥 − 2
6. To solve the equation, we multiply both sides by

b. 𝑥 + 3
c. (𝑥 + 2)(𝑥 − 3)
d. (𝑥 − 2)(𝑥 + 3)

a. 11
7. Which of the following will be the solution to the given rational equation?
3
3

11
b.

−
11

3
c.

−
3

11
d.

For items 8-10: Refer to the rational inequality below.

𝑥+12
𝑥+2 ≤2

𝑥 = −2 𝑎𝑛𝑑 8
8. What are the critical values in the given rational inequality?

𝑥 = −2 𝑎𝑛𝑑 − 8
a.

𝑥 = −2 𝑎𝑛𝑑 12
b.

𝑥 = −2 𝑎𝑛𝑑 − 12
c.
d.

9. Which of the critical value or values is/are included as endpoints of the

a. −2
intervals?

b. 2
c. −8
d. 8

a. (−∞, −2) ∪ (8, ∞)

10. Which of the following is the solution in the given inequality?

b. (−∞, −2] ∪ [8, ∞)

c. (−∞, −2) ∪ [8, ∞)
d. (−∞, −2) ∪ [−8, ∞)

29 CO_Q1_General mathematics SHS

Module 6
For items 11-13, solve for the solutions of the following rational equations.
11. 2
𝑥+2 1 3
+ 𝑥−2 =𝑥
a. 6
b. −6
c. 8
d. −8
9
12. 8
+1=
𝑥2 𝑥
−1 𝑎𝑛𝑑 − 8
1 𝑎𝑛𝑑 8
a.

−1 𝑎𝑛𝑑 8
b.

1 𝑎𝑛𝑑 − 8
c.
d.

13. 1
− 16 = 0
𝑥2
a. ±1
b. ±2
1
c. ±
2
1
d. ±
4

For items 14-15, solve for the solutions of the following rational equations.

5
𝑥−3
14.
3
> 𝑥+1
(−∞, −7) ∪ (−1, ∞)
(−∞, −1) ∪ (3, ∞)
a.

(−7, −1) ∪ (3, ∞)

b.

(−7, −1] ∪ [3, ∞)

c.
d.

(𝑥−3)(𝑥+2)
𝑥−1 ≤ 0
(−∞, −2) ∪ (1,3]
15.

(−∞, −2] ∪ (1,3]

a.

(−∞, −2] ∪ [1,3)

b.

(−∞, −2) ∪ [1,3)

c.
d.

30 CO_Q1_General mathematics SHS

Module 6
 Additional Activities

Practice Worksheet: Solving Rational Equations and Inequalities

Solve each equation. Check extraneous solutions for rational equations. Write your
answer in interval notation for rational inequalities.

LEVEL 1

8 4 𝑥
1. 2.

= 2𝑥 + 3 =
𝑥+1 3 4

𝑥−4 𝑥+3
3. 4.

≤0 >0
𝑥+5 3𝑥 − 6

4 1 1
5. 6.
+ =9 ≤0
𝑥 3𝑥 𝑥 −4
2

31 CO_Q1_General mathematics SHS

Module 6
 LEVEL 2

20 20 4 2 1
7. 8.

− = =
𝑥 𝑥−2 𝑥 𝑥 −𝑥
2
𝑥−1

𝑥−9 𝑥 + 32
9. 10.

≥3 ≤6
3𝑥 + 2 𝑥+6

4 1 1 2 2
11. 12.
+ = 1+ <
𝑥 𝑥 2
5𝑥 2
𝑥+1 𝑥

32 CO_Q1_General mathematics SHS

Module 6
 LEVEL 3

3𝑥 12 2 1 3
13. 14.

= 2 +2 + =
𝑥+1 𝑥 −1 4𝑥 − 9
2
2𝑥 − 3 2𝑥 + 3

(𝑥 + 7)(𝑥 − 3)
15. 16.
(𝑥 + 1)(𝑥 − 2)
( >0
2

≥0
(𝑥 − 1)(𝑥 + 1) 𝑥 − 5)2

𝑥−3 𝑥2 + 3𝑥 − 18 12𝑥3 + 16𝑥2 − 3𝑥 − 4

17. 18.
+ 2𝑥 − 12 = < 0
2𝑥 + 10 2𝑥 + 10 8𝑥3 + 12𝑥2 + 10𝑥 + 15

33 CO_Q1_General mathematics SHS

Module 6
Module 6
29
CO_Q1_General mathematics SHS

Answer Key
References
Aunzo, Rodulfo, Flores Maricar, Gagani Ray Ferdinand M, and Quennie
Ypanto. 2016. General Mathematics Activity-based, Scaffolding of Student .
Quezon City: C&E Publishing, Inc.

2016. General Mathematics Learner’s Material . Meralco Avenue, Pasig City,

Philippines 1600: Lexicon Press Inc.

2016. General Mathematics Teacher’s Guide. Meralco Avenue, Pasig City,

Philippines 1600: Lexicon Press Inc.

Oronce, Orlando A. 2016. General Mathematics. Sampaloc, Manila: Rex Bookstore,

Inc.

30 CO_Q1_General mathematics SHS

Module 6
For inquiries or feedback, please write or call:

Department of Education - Bureau of Learning Resources (DepEd-

BLR) Ground Floor, Bonifacio Bldg., DepEd Complex

Meralco Avenue, Pasig City, Philippines 1600

Telefax: (632) 8634-1072; 8634-1054; 8631-

4985

Email Address: blr.lrqad@deped.gov.ph * blr.lrpd@deped.gov.ph