Mathematics 5 Quarter 3 Module 14: Simple Equations Involving One or More Operations

Mathematics Quarter 3 – Module 14: Simple Equations Involving One or More Operations

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Mathematics– Grade 5 Alternative Delivery Mode Quarter 3 – Module 14: Simple Equations Involving One or More Operations First Edition, 2020

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Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio

Development Team of the Module

Writer: Eunilaine M. Serrato, Danilo S. Jadulco Editor: Jose J. Sagadal, Jr., Elizabeth Deligero Reviewer: Renato S. Cagomoc, Rolando Lacbo, Joshua Sherwin T. Lim Illustrator: Noel E. Sagayap Layout Artist: Joey Sustitudo, Joan T. Briz

Management Team: Ma. Gemma M. Ledesma Arnulfo R. Balane Rosemarie M. Guino Joy B. Bihag Ryan R. Tiu Sarah S. Cabaluna Thelma Cabadsan-Quitalig Elena S. De Luna Renato S. Cagomoc Noel E. Sagayap Geraldine P. Sumbise Joshua Sherwin T. Lim

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Mathematics

Quarter 3 – Module 14: Simple Equations Involving One or More Operations

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 with you in mind. It is here to help you master the use of different strategies to solve for the unknown in simple equations involving one or more operations on whole numbers and fractions. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. The lessons are arranged to follow the standard sequence of the course. But the order in which you read them can be changed to correspond with the textbook you are now using.

The module is divided into two lessons, namely:

• Lesson 1 – Simple Equations Involving One Operation
• Lesson 2 – Simple Equations Involving More Operations

After going through this module, you are expected to:

1. use different strategies to solve for the unknown in simple equations involving one or more operations on whole numbers and fractions; and
2. appreciate the use of different problem-solving strategies to make solving math problems easier.

What I Know

Good day, everyone! In this module, you will learn about solving equations. Let us find out first what you already know about it.

Directions: Choose the letter of the best answer. Write your answers on a separate sheet.

1. What do you call the symbols that do not have fixed values in an algebraic expression?

   • A. term
   • B. variable
   • C. constant
   • D. phrase

2. In the expression 12ab, what is the constant?

   • A. 12
   • B. a
   • C. b
   • D. ab

3. What is the value of 4x + 6, if x = 2?

   • A. 10
   • B. 12
   • C. 14
   • D. 16

4. Which is the verbal phrase of the algebraic expression 5y + 2?

   • A. five numbers increased by two
   • B. five times the sum of y and two
   • C. two more than five times a number y
   • D. two is more than five times a number y

5. Which is the algebraic translation of the phrase “a number x decreased by ten equals 5”?

   • A. x + 10 = 5
   • B. 10 – x = 5
   • C. 10x = 5
   • D. x – 10 = 5

6. What is the value of x that will satisfy the equation 3x = 15?

   • A. 5
   • B. 10
   • C. 12
   • D. 15

7. What is the solution of the equation y – 6 = 13?

   • A. 7
   • B. 10
   • C. 19
   • D. 21

8. Which equation has {3} as a solution set?

   • A. y + 1 = 3
   • B. 2y – 5 = 1
   • C. 4y = 7
   • D. 5y + 2 = 10

9. What is the value of a if 3a – 4 is equal to 2?

   • A. 2
   • B. 3
   • C. 4
   • D. 5

10. Which statement is TRUE?

    
    

• A. 5x = 20, if x = 5
• B. 8 + x = 21, if x = 16
• C. 3x – 6 = 9, if x = 5
• D. 2/3x = 10, if x = 5

CONGRATULATIONS! If you got a score of 9 or 10, you should not have any difficulty studying the lesson in this module.

If you got a score of 8 or below, you may need to study the lesson more carefully and do all the given activities.

Lesson 1: Simple Equations Involving One Operation

Arithmetic is concerned mainly with the study of numbers using different operations Algebra is concerned with the study of symbols or variables–and the operations or rules relating them. Variables represented by letters are often used as symbols for numbers.

The essence of algebra lies in representing quantities as symbols other than numerals. This is the advantage of applying algebra and not arithmetic alone in solving practical problems. These different symbols are grouped into expressions which in turn bring meaning to equations.

This module will enlighten us on the wonders and usefulness of algebra together with its importance in our lives.

Do the following activities to learn the different problem-solving strategies.

What’s In

Directions: Solve for the unknown. Match the mathematical statement in Column A with the answer in Column B that make the statement true. Write your answer on a separate sheet.

Example: _____ + 2 = 5 _____ + 2 – 2 = 5 -2 _____ = 3

Column A	Column B
_____ 1. _____ + 7 = 9	A. 4
_____ 2. 9 – _____ = 5	B. 12
_____ 3. 3 + 4 = 10 – _____	C. 2
_____ 4. _____ + 4 = 5 + 5	D. 10
_____ 5 _____ = 3 (4)	E. 3
	F. 6

What’s New

Simple Equations Involving One Operation

Consider the following:

2x 2b + 2d 2c + 5 = 25 3y – 2 a + 3 = 8 4b – 6 = 10 1/4 + x = 3/4 3/4 x + 1

Which are equations?

Can you solve the value of the variable to make each equation true?

The table below answers the first question.

Equations	Not Equations
a + 3 = 8	2x
4b – 6 = 10	2b + 2d
2c + 5 = 25	3y – 2
1/4 + x = 3/4	3/4 x + 1

What is It

An equation is a mathematical sentence that uses an equal sign “=”. The equal sign indicates that two mathematical expressions have the same value. Hence, the expression on the left side is equal to the expression on the right side of the equation.

Below are some examples of equations:

2 + 3 = 7 – 2 2 + 4 = 6 10 = 1 + 2 + 3 + 4 2/9 + p = 2
k + 3 = 3k – 5 2x – 5 = 7 2 (2) = 8 ÷ 2 3a = 18

An equation is a statement that two mathematical expressions are equal. Many equations contain variables.

A variable or unknown is represented by a symbol, usually a letter, that may take on different values.

A constant is a fixed value and does not change.

[chart: labeled diagram; Example: 2a, with an arrow to “variable” pointing at a and an arrow to “constant” pointing at 2]

Example: 2a variable constant

If an equation involves a variable, then a solution to the equation is a number that when substituted to the variable will make the equation true. The collection of all the solutions to an equation is called its solution set. The process of finding a solution is called solving an equation.

In solving an equation, you can try the following:

1. Write the equation
2. Group similar terms on one side.
3. Perform the indicated operations.
4. Simplify the answer.
5. Check.

Note that the truth of an equation does not change if we apply the following properties.

Properties of Equality

1. Addition Property of Equality:
   If a = b, then a + c = b + c, where a, b, c are whole numbers.

This property states if you add the same number to both sides of an equation, the two sides remain equal.

Example 1: Let’s try this numeric equation 5 + 7 = 12

5 + 7 + 2 = 12 + 2

14 = 14

Adding 2 to each side of the original equation results to 14. Notice that the resulting sides are still equal.

The addition property of equality states that if you add the same value to both sides of the equation, the resulting expressions will remain equal.

Example2: Solve x + 3 = 7 using models.

[chart: counter model of the equation x + 3 = 7, shown in three steps; Example2: Solve x + 3 = 7 using models.]

Step 1 (labels): x + 3 = 7. Caption: Model the equation.

Step 2 (labels): x + 3 − 3 = 7 − 3. Caption: Remove the same number of counters (stars) from each side.

Step 3 (labels): x = 4. Caption: The number of counters (stars) remaining on the right side that represents the value of x.

Check: Substitute x = 4 in the equation. x + 3 = 7 4 + 3 = 7 7 = 7

Subtraction has the same result as adding the opposite. Hence, the addition property of equality can be used for subtraction as well. So, the equality remains if you add or subtract the same value from both sides of an equation.

Example 3: Solve for f in the equation f – 2 = 7.

Solution: f – 2 = 7 f – 2 + 2 = 7 + 2 Add 2 to both sides of the equation f + 0 = 9 f = 9

To check the solution, substitute 9 for f in the equation.

Check: f – 2 = 7 9 – 2 = 7 7 = 7

Example 4: Solve for d in the equation d + 3 = 5.

Solution: d + 3 = 5 d + 3 – 3 = 5 – 3 Subtract 3 from both sides of the equation d = 2

To check the solution, substitute 2 for d in the equation.

Check: d + 3 = 5 2 + 3 = 5 5 = 5

Example 5: Solve for n in the equation 1/4 + n = 3/4

Solution: 1/4 + n = 3/4 1/4 – 1/4 + n = 3/4 – 1/4 Subtract 1/4 from both sides of the equation n = 2/4 or 1/2

To check the solution, substitute 2/4 for n in the equation.

Check: 1/4 + n = 3/4 1/4 + 2/4 = 3/4 3/4 = 3/4

2. Multiplication Property of Equality
   If a = b, then ac = bc, where a, b, c are whole numbers not equal to zero.

This property states that if you multiply the same nonzero number to both sides of an equation, the two sides remain equal.

Example 6: Let’s try the numeric equation 2 · 4 = 8. If you multiply both sides of this equation by 2, you will still get a true equation.

2 · 4 = 8 2 · 4 · 2 = 8 · 2 16 = 16

Example 7: Solve 2b = 10.

Solution: 2b/2 = 10/2 Divide both sides of the equation by 2 to get the value of b. Dividing by 2 is the same as multiplying by 1/2 which is the reciprocal of 2.

b = 5 Answer

To check the solution, substitute 5 for b in the equation.

Check: 2b = 10 2(5) = 10 10 = 10

Example 8: Solve 1/3 a = 2.

Solution: 1/3 a = 2 Multiply both sides of the equation by 3 (which is the reciprocal of 1/3) to get the value of a.

(1/3 a)3 = 2(3) a = 6

To check the solution, substitute 6 for a in the equation.

Check: 1/3 a = 2 1/3 (6) = 2 6/3 = 2 2 = 2

If you understand this topic, answer the activity below. If not, you can go back and review.

Direction: Solve for the unknown.

1. 3a + 6 = 33 2. 3/2 = c + ½ 3. b = 3/5 + 2/5

What’s More

For practice and understanding, do the activities below.

Activity 1: Looking Through Me

Directions: Solve each equation.

1. 4x – 5 = 11 3. ¾ x + 4 = 6 5. 3x = -9
2. 2 + 4x = 14 4. 20 – 5x = 30

Activity 2: Match, Match, Match

Directions: Inside the box are possible answers for the given equations. Match the letter of the correct solution to each equation.

A. -4 B. 4 C. 10 D. 3/8 E. 3

1. 5x – 1 = 14 3. 2x = ¾ 5. 2x – 9 = 11
2. -4x – 3 = 13 4. 3x + 3 = 15

Activity 3: Guess Me, Guess me

Directions: Solve for the value of the variable that will make each equation true. Write the letter of the correct answer and guess the word.

Answer	Item
	1. 2x + 5 = 25
	2. 5 a – 32 = 28
	3. 75 = 5c
	4. 1/5 + h = 3/5
	5. 20/y = 5
	6. 8d = 48
	7. b – 7 = 45
	8. 3b = 42

[transcriber’s note: the Answer column is blank in the printed table; the letter-and-value key below is printed as a single row of pairs and is shown here as two columns]

Letter	Value
N	14
T	4
I	6
O	52
Q	12
E	10
A	2/5
U	15

What I Have Learned

Remember the different terms in this module. Answer the activity below.

Directions: Identify what is being described by the statement. Write your answers on a separate sheet.

___________1. It is a sentence in mathematics that contains an equal sign. ___________2. It is a symbol or letter that may take different values. ___________3. It is a fixed value that does not change. ___________4. It is a number that makes an equation true. ___________5. It is a process of finding the solution of an equation.

What I Can Do

Directions: Solve the following problems using different strategies. Choose the letter of the correct answer.

1. Lyka’s grandmother left a pack of polvoron on the table. Althea ate 2 of them, his dad ate 4 of them, and they gave 11 of the polvorons to the kids playing outside. At the end of the day, only 4 polvorons were left in the plastic container. How many pieces of polvorons were there in the pack at the start?
   A. 6 B. 11 C. 17 D. 21

   
   

2. Athena has a weekly allowance of P 1,000.00 from her aunt. She plans to save some money for future use. On Tuesday, she deposited P12.00 in her bamboo bank. She deposited three times this amount on Wednesday and on Friday. How much money did she save in all?
   A. P 72.00 B. P 84.00 C. P916.00 D.P 928.00

   
   

3. Manilyn had P46.00 left after buying groceries. She bought 2 cans of sardines for P11.50 each and 1 kilo of salt for P30.00. How much money did she have at the start?
   A. P 57.50 B. P 76.00 C. P 87.50 D. P99.00

   
   

4. Albert ordered 3 pairs of sandals and a pair of black shoes. Each pair of sandals cost P140.00. He paid P840.00 in all. What was the cost of the pair of black shoes?
   A. P 140.00 B. P 280.00 C. P 420.00 D. P 840.00

   
   

5. Cecilia weighs 42.7 kg now. She gained 5.9 kg over the last two years. What was her weight two years ago?
   A. 36.8 kg B. 44.7kg C. 48.6kg D. 64.5 kg

   
   

Assessment

To check if you understand this module, answer the activity below. This covers all the lessons presented.

Multiple Choice

Directions: Choose the letter of the correct answer. Write your answers on a separate sheet.

1. After finishing buying vegetables at the market, Ruby wants to have P 35.00 left. She plans to buy slippers for P 25.00 and a coin purse for P15.00. How much money does she need?
   A. P 95.00 B. P 85.00 C. P 75.00 D.P 65.00

   
   

2. What is the value of x that will make the equation 6x – 9 = 27 true?
   A. 6 B. 7 C. 8 D. 9

   
   

3. Jasmine ordered 3 kilos of dalandan for P29.00 per kilo and 1 pack of kalamansi. The total cost was P 150.00. What was the cost of the kalamansi?
   A. P 103.00 B. P 93.00 C. P 73.00 D.P 63.00

   
   

4. What is the solution of 1/2 y = 30?
   A. 49 B. 50 C. 59 D. 60

   
   

5. Jeramie baked cupcakes. She put them on the table. After breakfast, Vanessa ate 4 cupcakes and got 7 for her playmates. During lunchtime, father and mother ate 13 cupcakes in all. There were 36 cupcakes left on the table. How many cupcakes did Jeramie bake?
   A. 30 B. 60 C. 90 D. 120

   
   

6. Three times a number decreased by four equals thirty-five. What is the number?
   A. 10 B. 11 C. 12 D. 13

   
   

7. What is the solution of 7y + 11 = 60?
   A. 7 B. 9 C. 11 D. 13

   
   

8. When Rizalyn rode on a jeepney, there were already passengers sitting. At the next stop, 3 people got on and 4 people got off. Three stops later, 6 people got on. All 14 people got off the terminal station. How many people were in the jeepney when Rizalyn got on the jeepney?
   A. 9 B. 8 C. 7 D. 6

   
   

9. Scarlet gave 10 stickers from her collection to Stephanie and Jaynard. Then she gave 14 stickers to Melissa and 6 stickers to Starla. She still had 275 stickers. How many stickers were in Scarlet’s collection, to begin with?
   A. 395 B. 315 C. 310 D. 30

   
   

10. Which equation is true if x = 4?
    A. 20 = 6x – 4 B. 4x – 1 = 7 C. 1 + 5x = 10 D. 3x = 7

    
    

Got a score of 8-10? CONGRATULATIONS! Job well done. See you in the next lesson. If you scored below 8, you may have to go over the lesson and do the exercises again.

Additional Activities

Directions: Translate each problem into an equation, then solve for the unknown.

1. A number decreased by 17 is 29. What is the number?

   
   

2. How many kilograms of mangoes can you buy with P 560.00 if one kilogram costs P 70.00?

   
   

3. Christopher and his four close friends went on a tour in Palawan. They decided to divide the expenses evenly. If each of them paid P 1,205.00, find their total expenses.

   
   

4. Twice a number decreased by 15 equals 9. Find the number.

   
   

5. After she lost 26 pounds in her diet, Melissa weighed 178 pounds. What was her original weight?

   
   

What I Know

Good day, everyone! Another lesson will be discussed. Answer first the activity below. Take note of the items unanswered and find the right answer as you go through this module.

Directions: Choose the letter of the correct answer. Write your answers on a separate sheet.

1. Which is the correct translation of “Twice the number x increased by seven is equal to thrice the number x decreased by five.”?
   • A. 2x – 5 = 3 (x + 7)
   • B. 2 (x – 5) = 3x + 7
   • C. 2x + 7 = 3x – 5
   • D. 2 (x – 5) = 3 (x + 7)
2. Refer to number 1. Twice the number x increased by seven is equal to thrice the number x decreased by five. What is the number?
   • A. 2
   • B. 5
   • C. 7
   • D. 12
3. What is the value of x that will make the equation 5x = 3x + 32 true?
   • A. 4
   • B. 8
   • C. 12
   • D. 16
4. Which equation has a solution set of {2}?
   • A. 5b = 45
   • B. 6m – 1 = 9
   • C. 4x + 3 = 11
   • D. 3y + 5 = 0
5. What is the solution of 6y = 2y + 36?
   • A. 2
   • B. 4
   • C. 7
   • D. 9
6. Which of the following is true if a = 5?
   • A. 10a = 6a + 20
   • B. 6 (a+4) = 5a + 25
   • C. 7a + 5 = 4a + 8
   • D. a = 2a – 10
7. One of the following equations has {0} as the solution set. Which one is it?
   • A. 3y = 2y + 1
   • B. 2 (y + 4) = y – 4
   • C. 5y + 8 = 3y + 8
   • D. 4y + 15 = y
8. What is the solution of the equation 3 (x + 5) = x + 21?
   • A. 1
   • B. 2
   • C. 3
   • D. 4
9. Which equation is true when the given value of x is substituted?
   • A. 2x = x + 1, if x = 1
   • B. 2x – 1 = x + 2, if x = 0
   • C. 2 (x+1) = x + 2, if x = 2
   • D. 2x + 1 = x, if x = 1
10. Which is the verbal translation of 3x + 4 = 2x – 5?
    • A. Three times a number and four is less than five and twice the number.
    • B. The sum of thrice a number and four is equal to the difference of twice the number and five.
    • C. Thrice the sum of a number and four is twice the difference of the number of five.
    • D. Three times four is equal to two times the opposite of five.

Lesson 2: Simple Equations Involving More Operations

In the previous lesson, you were taught how to solve for the unknown in simple equations. However, some equations have more than one variable term such as in 3x – 6 = 2x + 10. In this module, you will learn how to use different strategies to solve equations involving more operations.

This module will continue to enlighten us on the importance of algebra. Do the following activities to learn different problem-solving strategies.

What’s In

Simple Equations Involving More Operations

Consider the following:

5x + 1 = 3x + 7 y – 1 = 7 3m – 1 = 4 3(x + 1) = 2x + 8 b + 5 = 8 – 3 a = 8 – 5a 2/5 + 2x = 3/5 – x 5b = 20

Which equations has more than one term with variables?

Can you find the solution of each equation?

CONGRATULATIONS! If you got a score of 9 or 10, you should not have any difficulty studying the lesson in this module.

If you got a score of 8 or below, you may need to study the lesson more carefully and do all the given activities.

What’s New

To answer the first question above, we group the given equations into two.

Equations with More than One Term with Variables	Equations with Only One Term with Variable
5x + 1 = 3x + 7	b + 5 = 8 – 3
a = 8 – 5a	y – 1 = 7
2/5 + 2x = 3/5 – x	5b = 20
3(x + 1) = 2x + 8	3m – 1 = 4

To find the solution of each equation involving more terms with variables, the following discussion will be helpful.

What is It

Solving an equation means to find the value of the variable which will make the equation true. A basic technique in solving equations is to place all terms with variables on one side of the equation.

We must combine the terms with variables into a single term. We then find the value of the variable that will make the equation true.

There are terms in the equation which are called “like terms” because they have the same variable raised to the same exponent. For example, 5y and 6y are like terms, but 5y and 6m are not like terms. Like terms can be combined by adding their coefficients and copying the same variable. Thus:

5y + 6y = 11y .

To solve an equation, the following properties are helpful.:

Addition Property of Equality:

• Add the same number to both sides of an equation.
• Subtract the same number from both sides of an equation.

Multiplication Property of Equality:

• Multiply the same nonzero number to both sides of an equation.
• Divide both sides of an equation by the same nonzero number.

Consider the given examples:

Example 1: Solve 3x + 7 = 2x + 12.

Solution: 3x + 7 = 2x + 12 Write the equation. 3x + 7 – 7 = 2x + 12 – 7 Subtract 7 from both sides of the equation. 3x = 2x + 5 7 – 7 = 0, and 12 – 7 = 5 3x – 2x = 2x – 2x + 5 Subtract 2x from both sides of the equation. 3x – 2x = x, and 2x – 2x = 0 x = 5 Answer

Example 2: Solve 5a + 3 = 3a + 4.

Solution: 5a + 3 = 3a + 4 Write the equation. 5a + 3 – 3 = 3a + 4 – 3 Subtract 3 from both sides of the equation. 5a = 3a + 1 3 – 3 = 0, and 4 – 3 = 1 5a – 3a = 3a – 3a + 1 Subtract 3a from both sides of the equation. 5a – 3a = 2a, and 3a – 3a = 0 2a/2 = 1/2 Divide both sides by 2 to find the value of a. a = 1/2 Answer

Example 3: Find the solution of 3 (y – 2) = 2y.

Solution: 3 (y – 2) = 2y Write the equation. 3y – 6 = 2y Simplify by multiplying 3 (y – 2) 3y – 6 + 6 = 2y + 6 Add 6 to both sides of the equation. 3y = 2y +6 6 – 6 = 0 3y – 2y = 2y – 2y +6 Subtract 2y from both sides of the equation. 3y – 2y = y, and 2y – 2y = 0 y = 6 Answer

Application and Problem Solving

Example 4: Four times a number decreased by 5, is thrice the number increased by 10. What is the number?

Understand

a. What is asked?

• The number

b. What are the given facts?

• Four times a number decreased by 5, is thrice the number increased by 10.

Plan

Translate the sentence into an equation.

Four times a number decreased by 5, is thrice the number increased by 10.

Let x be the number.

4x – 5 = 3x + 10

Solve

Solution:

4x – 5 = 3x + 10 Equation formed 4x – 3x – 5 = 3x – 3x + 10 Subtract 3x from both sides of the equation x – 5 = 10 Simplify x – 5 + 5 = 10 + 5 Add 5 to both sides of the equation x = 15 Answer

Answer

Therefore, the number is 15.

Check

Substitute x = 15 in the equation.

4x – 5 = 3x + 10 Original equation 4 (15) – 5 = 3 (15) + 10 Substitute x = 15. 60 – 5 = 45 + 10 Simplify. 55 = 55 True statement

Therefore, x = 15 is the solution to the equation 4x – 5 = 3x + 10.

Activity

Directions: Solve the following problems. Show your solutions.

1. 5x – 6 = 4(x – 3)
2. 4(x – 3) = 2 + 3(x – 1)
3. Don Emilio, split his money amounting to P19,250,000.00 among his 3 children and a grandchild. Each child gets the same amount and the grandchild gets one half as much as each child gets. How much does each person receive?

In summary, in solving an equation, follow the steps below:

1. Write the equation.
2. Group similar terms on one side.
3. Perform the indicated operations.
4. Simplify the answer.
5. Check.

The problem-solving steps are stated below.

1. Understand
2. Plan
3. Solve
4. Answer
5. Check

This is the UPSAC method.

Did you understand the lesson? Well, if you do, work on the next activities. If not, review the previous lessons and activities.

What’s More

Activity 1: Match Me

Directions: Match the equation in Column A with its solution in Column B.

Column A	Column B
1. 3x = x + 8	a. 25
2. 3y + 2 = 5y – 4	b. 10
3. 5 (x – 2) = 4x	c. 4
4. 16 – 5y = 9y + 2	d. 3
5. 4/5 x – 5 = 3/5 x	e. 1

Activity 2: Know Me

Directions: Translate each equation into verbal sentences. Solve each equation.

1. 3x + 6 = 2x – 3
2. 2x = 5x – 6
3. 5 + 4x = 9 – 3x

Activity 3: Compare Me

Directions: Compare the solutions of equations in each number. Use >, <, or = in the circle.

1. 2x + 5 = 3(x – 2) ( ) 3x – 1 = 2(x + 5)
2. 3x + 4 = 6x – 2 ( ) 4x + 3 = 2x + 6
3. 2 + 3x = x – 6 ( ) 3 + 2x = x – 1
4. 2(3x – 1) = 5x ( ) 4x = 4 + 2(x + 3)
5. 2x + 1 = 3(x + 1) ( ) 4x + 1 = 3(x + 1)

[transcriber’s note: the circles between the two equations in each item are empty in the source]

If you get a score of 10 or more in these 3 activities, very good. To enhance more your abilities in solving word problems, try to answer the next activity.

What I Have Learned

Directions: Solve the problem below step by step using the UPSAC method.

Problem: The regular price of a bracelet was reduced by P20.00. Emily paid P140.00 for the bracelet. What was the regular price?

What I Can Do

Activity: Message Hunting

Directions: Find the value of the variable that will make each equation true. Match each letter with the correct answer in the code below to answer the question “What is your idea about Math in your life”?

1. 3x + 2 = x – 4, H
2. y + 4 = 5y – 8, E
3. 2b – 1 = b + 4, F
4. 2 + 7a = 4a – 4, M
5. 5y = 3y – 8, A
6. -4 + 3x = -2x + 6, I
7. 9a + 1 = 8a – 4, T
8. 2(x – 4) = 3(x – 3), L
9. 2x + 4 = 3(x – 1), S

CODE:

---

• 2 – 4 -5 -3 2 7 1 2 5 3

Assessment

Multiple Choice: Choose the letter of the correct answer. Write your answers on a separate answer sheet.

1. Which is true when the given value of x is substituted in the equation?

   • a. 3x + 5 = 2x + 8, if x = 2
   • b. 5x = x + 12, if x = 5
   • c. 2x = 10 + 7x, if x = -2
   • d. 6x – 2 = 2x + 14, if x = 4

2. Which equation is true if y = 1?

   • a. 3y = 2y + 8
   • b. 4y + 5 = 3y + 6
   • c. 1 – 2y = 3y – 9
   • d. y = 3y – 1

3. Which equation has {2} as a solution set?

   • a. 5x + 2 = 3x + 6
   • b. 3x = x + 2
   • c. 6 (x + 1) = 2x + 10
   • d. 8x + 5 = 4x – 11

4. What is the solution of 86 + 9a = 5a + 100?

   • a. 1
   • b. 2
   • c. 3
   • d. 4

5. What is the solution of 13y = 5y + 104?

   • a. 8
   • b. 11
   • c. 13
   • d. 15

6. Which equation is FALSE if a = 2?

   • a. 6a + 1 = 4a + 5
   • b. 7a = 2a + 10
   • c. 5 (a-3) = 2a + 3
   • d. 9a – 10 = 3a + 2

7. What is the value of x that will make the equation 6 (x – 5) = 3x + 12 TRUE?

   • a. 11 `
   • b. 12
   • c. 13
   • d. 14

8. Which is the mathematical translation of “The difference of five times a number and eight is four times the number increased by twenty.”?

   • a. 5x – 8 = 4x + 20
   • b. 5 (x + 8) = 4 (x + 20)
   • c. -5x + 8 = 4x + 20
   • d. 8 – 5x = 4 (x + 20)

9. What is the solution of 11a = 6a + 35?

   • a. 4
   • b. 5
   • c. 6
   • d. 7

10. What is the value of x that will satisfy the equation ½x + 3 = x?

    • a. 6
    • b. 8
    • c. 10
    • d. 12

Got a score of 8-10? CONGRATULATIONS! Job well done. See you in the next lesson. If you scored below 8, you may have to go over the lesson and do the exercises again.

Additional Activities

A. Directions: Translate each problem into an equation, then solve for the unknown.

Find the number of calories in a lemon and mango if:

a. A mango has 40 calories more than that in a lemon. b. Fifteen lemons have as many calories as 5 mangoes.

B. Directions: Solve the following problems.

1. The sum of thrice a number and 4 is the same as twice the number increased by 15. What is the number?

   • a. 11
   • b. 14
   • c. 17
   • d. 19

2. What is the solution of 15a – 49 = 8a?

   • a. 6
   • b. 7
   • c. 8
   • d. 9

3. A videoke machine rental has two plans. Under Plan A, a videoke machine is rented for P 300.00 plus P 15.00 for each additional hour. Under Plan B, a videoke machine is rented for P 350.00 plus P 10.00 for each additional hour. What number of hours would result in the same cost?

   • a. 5
   • b. 10
   • c. 15
   • d. 20

4. What is the value of y that will satisfy 5 (2 – y) = 3y?

   • a. 2/5
   • b. 4/5
   • c. 5/4
   • d. 5/2

5. What is the value of x that will make the equation 6x – 7 = 3x + 11 TRUE?

   • a. 3
   • b. 4
   • c. 5
   • d. 6

Answer Key – Lesson 1

What I Know

1. B
2. A
3. C
4. C
5. D
6. A
7. C
8. B
9. A
10. C

What’s In

1. c
2. a
3. e
4. f
5. b

What is It

1. a = 9
2. c = 1
3. b = 1

What’s More

Looking Through Me

1. X = 4
2. X = 3
3. X = 8/3
4. X = -2
5. X = -3

Match, Match, Match

1. E
2. A
3. D
4. B
5. C

Guess Me, Guess Me

1. E
2. Q
3. U
4. A
5. T
6. I
7. O
8. N

What I Have Learned

1. Equation
2. Variable
3. Constant
4. Solution of an Equation
5. Solving an Equation

What I Can Do

1. D
2. B
3. D
4. C
5. A

Additional Activities

1. 46
2. 8
3. P 6,025.00
4. 12
5. 204

Assessment

1. C
2. A
3. D
4. D
5. B
6. D
7. A
8. A
9. D
10. A

Lesson 2

Additional Activities

Equation:

15a=5(a+40)

• There are 20 calories in a lemon
• There are 60 calories in a mango

1. A
2. B
3. B
4. C
5. D

Assessment

1. D
2. B
3. A
4. B
5. C
6. C
7. D
8. A
9. B
10. A

What I Can Do

[transcriber’s note: this box is printed upside down and its letter-number pairs are read as displayed; the signs and order are uncertain]

-2 M -4 A -5 T -3 H 2 I 7 S 1 L 2 I 5 F 3 E

What is It?

1. 5x – 6 = 4(x – 3)
   5x – 6 = 4x – 12 + 6
   5x + (–4x) = 4x + (–4x) – 12 + 6
   x = –6

   
   

2. 4(x – 3) = 2 + 3(x – 1)
   4x – 12 = 2 + 3x – 3
   4x + (–3x) – 12 + 12 = –1 + 12 [?]
   x = 11

   
   

3. Let x = represent each child’s part
   ½x = the grandchild’s part

   
   

   x + x + x + ½x = 19,250,000 ; ½ = 0.5
   3.5x = 19,250,000 ; x = 19,250,000 / 3.5
   x = 5,500,000

   
   

   Answer:
   Each child will receive P5,500,000.00 and the grandchild will receive P2,750,000.00.

   
   

What I Know

1. C
2. D
3. D
4. D
5. C
6. A
7. C
8. C
9. A
10. B

What’s More

Know Me Match Me

[transcriber’s note: this matching layout is printed upside down; the letters, numbers, and phrases below are in the order read from the page, and some line breaks are uncertain]

1. C – Thrice a number added by six is twice that number minus three ; x = -9
2. D
3. B – Twice a number is five times a number less six ; x = 2
4. E
5. A – Five added by four times a number is equal to nine minus thrice a number. ; x = 4/7

Compare Me

1. =
2. 
   
3. =
4. <
5. =

What I Have Learned

Solutions:

❖ Understand

a. What is asked?

• What is the original price of the bracelet?

b. What are the given facts?

• The bracelet is reduced by P20.00
• The bracelet amounted to P140.00

❖ Plan

• Translate the phrase “the bracelet is reduced by P20.00 and the bracelet amounted to P140.00”

❖ Solve

Let x represent the original price of the bracelet.

x – 20 = 140

Equation

x – 20 = 140 x – 20 + 20 = 140 + 20 x = 160

❖ Answer

The original price of the bracelet is P160.00.

❖ Check

Substitute x = 160 in the equation

x – 20 = 140 160 – 20 = 140 140 = 140

Therefore, P160 is the original price of the bracelet.

References

Lumbre, Angelina P. and Ursua, Alvin C.. 2016. 21st Century Mathletes 5 Textbook. Quezon City: Vibal Group, Inc.

Project EASE (Effective and Alternative Secondary Education) Module. Pasig City.

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