3rd quarter math g7 las (week 1 - 2).pdf

8/14/2019 3rd Quarter Math G7 LAS (Week 1 - 2).pdf

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3rdQuarter Mathematics 7 LAS 1| P a g e

ROMAN CATHOLIC BISHOP OF NOVALICHESEDUCATIONAL SYSTEM (RCBN-ES, Inc.)

Good Shepherd Cathedral, Regalado Ave., Fairview Park, Quezon City

Telephone 430-7819

MATHEMATICS 7S.Y. 20132014

Name: ____________________________________________________________ Date:___________________________

Year & Section: ____________________________________________________ Quarter: 3rd QuarterActivity Title: Introduction to Equations Activity No.: 01Value/s: TruthfulnessReference/s: Oronce, O.A., Mendoza, M.O., (2013) e-Math 7, Manila, Rex Bookstore pp. 224 - 228Designer:Miss Mary Jane A. Palattao RCBN-ES School:OLLCS

I. LEARNING TARGETSAt the end of the lesson the students should be able to:

1.Differentiate between mathematical expressions and mathematical equations2.Identify conditional, inconsistent and identity equations3. Identify if the given number is a solution of the given equation

II. LEARNING EXPERIENCESA. Explore Activities

Pair Activity:(Ask students the following questions) Do not copy!!!

Choose two numbers for the left part of the scale and two numbers for the right part of the scale so that the two sums are

equal.a. 22, 29, 46, 53b. 32, 34, 53, 55

B. Concept Notes

Equation is a statement that shows that the two numbers or expressions are equal.

In the equation x + 9 = 15, the expression x + 9 is called the left-hand sideand 15 is called the right-hand side. To

solve an equation means to find all of the solutions to the equation. Any value of the variable that makes the equation a true

statement is a root or solutionto the equation.

Equations that havethe same solution areequivalentequations.Examples:

x + 2 = 4 and x = 2

Equations that is satisfied by everynumber for which both sides aredefined are identity equations.Examples:3a = 2a + a and 2( a + 1 ) = 2a + 2

Equation that has nosolution is inconsistentequation.Examples:4b = 4b1 and x + 3 = x

Equation that has at leastone solution but not anidentity is conditionalequation.Examples:

5y + 3 = 13Examples:Is the given number a solution of the equation?

a. x4 = 12 ; 16 b. a + 5 = 15; 20 c. 2x + 3 = 15 ; 4 d. 93

n ; 27

You are given a bag of marbles, all of which are of the same shape and appearance. All marbles are also of the same weight

except for one of which is heavier than the rest. You are to use a balance to make exactly two weighings to pick out the heavier

marble from the rest of the marbles. Describe clearly how you can do it and explain it to your partner if the bag contains exactly 6

marbles.

An equation can be represented with an equality sign =.Two algebraic expressions that are equal form an algebraic equation.These are some examples of equations:

x5 = 7 2x + 7 = 26 232

5x

x

K:_______S:_______U:_______

P:_______

SCORE


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Solution:a. Substitute 16 forxin the equation and then solve.

x4 = 12

164?

1212 = 12

Yes, 16 is a solution of x 4 = 12.

b. Substitute 20 for ain the equation and then solve.a + 5 = 15

20 + 5?

1525 15

No, 20 is not a solution.

C. Firm-Up Activity (Knowledge)A. Fill in the blanks to make the statement true.

1. An ___________________ is a mathematical statement indicating that two expressions are equal.2. Any number that makes an equation true when substituted for its variable is called a ___________________ or

___________________ to the equation.3. Two equations are ___________________ when they have the same solution.4. To ___________________ the solution of an equation, you substitute the value of the variable in the original

equation and see of the result is a ___________________ statement.B. Choose the correct meaning. Write only the letter of your answer.

5. inconsistent equationa. an equation with two variables c. an equation that has no solutionb. an equation that is not correct d. an equation is sometimes consistent

6. identitya. an equation that is identical c. an equation that is wrongb. an equation that has 1 as its solution d. an equation that is satisfied by every real number for which

both sides are defined.7. conditional equation

a. an equation that is correct c. an equation that is true or falsean equation that you are not sure how d. an equation that is satisfied by at least one real number but isto solve not an identity.

C. Identify each equation as a conditional, inconsistent or identity equation.1. 15 = 11x 2. b - 2 = 2 - b 3. a + 5 = a 4. 2(x + 3) = 2x + 6 5. n5 = 11

C. Deepen Activity (Knowledge)A. Which of the following numbers had shown in the parentheses a solution for the given equation?

1. a + 7 = 19 (-12, 4, 12, 26) 2. 3(x + 5) = 3x + 15 ( -1, 2, 5, 9) 3. 3b5 = 10 (-5, 3, 0, 5)

D. Synthesis (Skills)The five symbols in the box represent the numbers 0, 1, 2, -1, and 2. Examine the equation and write the numberrepresented by each symbol. (Answers only)

E. ReflectionTo find the solution in a given equation we substituted the given and check if this will make the both sides of equation

equal. In real life how do you determine if you able to find the solution to your problem? Cite an example.

c. Substitute 4 forxin the equation and then solve.2x + 3 = 15

2(4) + 3?

15

8 + 3?

1511 15

No, 4 is not a solution.

d. Substitute 27 for nin the equation and then solve.

99

93

27

93

n

?

yes, 27 is a solution.


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Telephone 430-7819


Name: ____________________________________________________________ Date:___________________________

Year & Section: ____________________________________________________ Quarter: 3rd QuarterActivity Title: Introduction to Equations Activity No.: 02Value/s: TruthfulnessReference/s: Oronce, O.A., Mendoza, M.O., (2013) e-Math 7, Manila, Rex Bookstore pp. 224 - 228Designer:Miss Mary Jane A. Palattao RCBN-ES School:OLLCS

I. LEARNING TARGETAt the end of the lesson the students should be able to:

1. Translate English sentences to mathematical equations and vice versa.

II. LEARNING EXPERIENCES

A. Concept NotesTranslating sentences into algebraic equations is just the same on how you translated phrases into algebraic

expressions. However, the verb is in the sentence is translated as = in the equation.

Examples:Write a mathematical equation for each sentence.

a. A number increased by 4 is 20. c. Four times a number increased by 3 is 15.b. Twenty diminished by a number is 6. d. The quotient of a number and 5 is 7.

Answer:a. c.

b. d.

B.FirmUpActivity (Knowledge)Match the given sentence in Column A with the correct mathematical equation in Column B. Write the letter of youranswer. (Answers only)

Column A Column B1.8 more than a number is 28. a. x28 = 82.A number subtracted from 28 is 8. b. x + 28 = 83.28 less than a number is 8. c. x + 8 = 284.The product of a number and 8 is 28. d. 28x = 85.The sum of a number and 28 is 8. e. 8x = 28C. Deepen Activity (Skills)

Write an algebraic equation for each.1. The quotient of 15 and a number is 5. 4. Twice a number increased by 4 is 16.2. One-third of a number is 30. 5. The difference between 9 times a number and 16 is 74.3. The product of a number and 15 is 3.

D. Synthesis (Skills)

S is the number of students in a school. T is the number of teachers. There are 20 times as many students asteachers. Write down an equation using S and T.Is the answer 20S = T or is it 20T = S? Give your reason.

A number increased by 4 is 20

x + 4 = 25

x + 4 = 20

Twenty diminished by a number is 6

20 - n = 620 - n = 6

The quotient of a number and 5 is 7

= 7

K:_______S:_______U:_______

P:_______

SCORE

Four times a number increased by 3 is

4 x + 3 =

4x + 3 = 15

15

15


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Telephone 430-7819


Name: ____________________________________________________________ Date:___________________________Year & Section: ____________________________________________________ Quarter: 3rd Quarter

Activity Title: Solving Simple Equations Using Addition or Activity No.: 03Subtraction Property of Equality (APE or SPE)Value/s: EqualityReference/s: Oronce, O.A., Mendoza, M.O., (2013) e-Math 7, Manila, Rex Bookstore pp. 235 - 241Designer:Miss Mary Jane A. Palattao RCBN-ES School:OLLCS


1. Solve 1stdegree equations in one variable using APE or SPE.II. LEARNING EXPERIENCES

A.Exploration (Do not copy just study and answer!!!)

Mark has been asked to solve the equation x + 7 = 12. He explains to the class

Fill in both flowcharts for each equation. Use the steps in the inverse flowchart to find the value of the variable. Number1 is done for you.1. x9 = 17 Solution: 26 3. b18 = 30 Solution: ______

2. a + 12 = 53 Solution: _____ 4. c + 25 = 62 Solution: ______

B. Concept NotesIn algebra, there are operations that undo each other, such as addition and subtract ion. For example, if you startwith 5 and add 2, you can get back to 5 by subtracting 2.

1. The equation x + 2 = 6can be represented by algebra tiles as follows.

Note:Do not copy the tiles anymore just the solution

X

- 9

17 17

+ 9

26

x + 22 = 6 - 2

+

+

+

++

+

+ +

+

An addition equationis an equation involving the sum of a number and a variable. An addition equation can be solved by

subtracting the same number from each side of the equation so that the variable is isolated on one side of the equation.

To get the variable x alone on the left side, subtract 2 on both sides. The circled tiles can be removed.

K:_______S:_______U:_______

P:_______

SCORE

+ +

+

+ +

+

+

+

+


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2. The equation x3 = 5can be modelled as follows.

Since there is only one x-tile on the left side of the equation, the number of tiles on the right side is the solution to the equation.

C.Firm-Up Activity (Knowledge)What number would you add or subtract on both sides of the given equation to find the value of the variable(solution)?1. x - 15 = 17 4. 8.5 + y = 13.22. 19 = m3 5. x5 = 103. 19 + a = 19

D.Deepen Activity (Skills): Solve for the value of xand evaluate your answer.a. x + 5 = 9 d. x + 4 = - 5b. x6 = 14 e. x8 = 0c. 7 = x10

E. Synthesis (Understanding)1. One student solved the equation 4x = 12 by subtracting 4 from both sides and got 8 as a result. What is wrong

with the procedure? Explain and show the correct solution.Criteria 0 1 2 3

Explanation(3pts.)

Was not able to identifyand explain the error inthe solution.

Incorrectly identifies andexplain the error in thesolution.

Correctly identifies andexplains the error in thesolution but did not show thecorrect solution.

Correctly identifies andexplains the error in thesolution and shows thecorrect solution.

2. Create two equationseachthat can be solved by using: [8 points]a. Addition property of equality b. Subtraction property of equality

Solve your own equationCriteria 0 1 2

Solution

Solution does not correctly applyaddition or subtraction property ofequality

Solution contains minor error in applyingaddition or subtraction property of equality

Solution correctly applies addition or subtractionproperty of equality

Example Not able to set-up an equation Set-up 1 addition/subtraction equation Set-up 2addition/subtraction equations

x = 4+

+

+ +

+

A subtraction equationis an equation involving the difference between a variable and a number. You can solve a subtraction

equation by adding the same number on each side of the equation. The goal is the same: to isolate the variable on one side of

the equation.

x = 8

-

-

+-

+ +

+

+ +

To get the variable x alone on the left side, add 3 on both sides. The circled tiles can be removed.x

3 + 3 = 5 + 3

+

+ +

+

+ +

+

+

+

A process of subtracting the same number/term on both sides of

the equation is subtraction property of equality(SPE)

x + a = b x + aa= b - a

-

-

-+

+

+

+ +

+

+ +

+

+

+

The process of adding the same number on both side of an equation

to produce a new equation having the same solution as the original is

addition property of equality(APE).

xa = b xa + a = b + a

Solving Addition and Subtraction Equations

1. Undo each operation performedon the variable by using the inverse operations.2. Continue until the variable is isolated on one side of the equal sign.3. Check your solution by substituting it in the original equation.


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Telephone 430-7819



Activity Title: Solving Equations Using Addition or Subtraction Property of Equality (APE or SPE) Activity No.: 04Value/s: EqualityReference/s: Oronce, O.A., Mendoza, M.O., (2013) e-Math 7, Manila, Rex Bookstore pp. 235 - 241Designer:Miss Mary Jane A. Palattao RCBN-ES School:OLLCS


1. Solve 1stdegree equations in one variable using APE or SPE.II. LEARNING EXPERIENCES

A. Concept NotesWe can also solved 1st degree equations containing variable on both sides by applying APE or SPE.

Examples: Solve each equation.1. 3(m4) = 2m + 3Solution:

3m12 = 2m + 3 Simplify the equation by applying distributive property3m2m12 = 2m2m+ 3 Eliminate first the variable on one side with the lesser numerical coefficient(SPE)

m12 = 3 Combine similar termsm12 + 12= 3 + 12 Isolate the variable on one side of the equation by applying APE

m = 152. -5x = 186xSolution:

-5x + 6x= 186x + 6x Add 6x on both sides (APE)

x = 18 Combine similar termsB. Firm-Up Activity (Skills)

Solve for the value of xto make the equation true.a. 9x10 = 8x d. 6x7 = 5x + 2b. 2x + 4 = x + 7 e. 2 + 7x = 3 + 6xc. 3x = 2x + 8

C. Deepen Activity (Skills)1. Solve for the value of x to make the equation equal.

a. 7x = 6x + 5 b. 5 + 8x = 7x + 11 c. 10x4 = 9x + 72. Write two equationscontaining variable on both sides that can be solved by using addition or subtraction

property of equality in which the solution is: a. 3 b.6

D. Synthesis (Understanding)Haley and Andy each solved the equation -3x - 6 = - 4x + 8 on the board. Whose process and answer is/are correct?Explain the reasoning for your choice and why the others is/are incorrect?

Criteria 0 1 2 3

Explanation(3pts.)

Was not able to explainthe given answers.

Correctly identifies andexplains only the correct orincorrect answer.

Correctly identifies the correctand incorrect answer butexplains incompletely.

Correctly identifies andexplains the correct andincorrect answer.

K:_______S:_______U:_______

P:_______

SCORE

Andy

-3x - 6 = - 4x + 8-3x + 4x - 6 = -4x + 4x + 8

x - 6 = 8x6 + 6 = 8 + 6

x = 14

Haley

-3x - 6 = - 4x + 8-3x + 3x - 6= -4x + 3x + 8

-6 = x + 8-6 - 8 = x + 8 - 8

2 = x or x = 2


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Telephone 430-7819



Activity Title: Solving Equations Using Addition or Subtraction Property of Equality (APE or SPE) Activity No.: 05Value/s: EqualityReference/s: Oronce, O.A., Mendoza, M.O., (2013) e-Math 7, Manila, Rex Bookstore pp. 235 - 241Designer:Miss Mary Jane A. Palattao RCBN-ES School:OLLCS


1. Solve problems involving 1stdegree equations in one variable using APE or SPE.II. LEARNING EXPERIENCES

A. Concept Notes

Solution:1. Representation:Let x be the weight of the marbles

Equation: x + 1.5 = 8.6Solution: x + 1.5 = 8.6

x + 1.51.5= 8.61.5 SPEx = 7.1 pounds

2.

Representation:Let n be the total number of students in the roomEquation: n5 = 21Solution: n5 = 21

n5 + 5= 21 + 5 APEn = 26 students

B. Firm-Up Activity (Skills)

Write an algebraic equation for each sentence. Then solve the equation. Use any variable to represents the unknown.1. The sum of a number and 30 is1.2. The difference between a number and 24 is24.3. A book was on sale for 450. The sale price was 125 lower than the regular price. Find the regularprice.4. May has read 210 pages of novel. That is 25 more pages than Bob has read. How many pages has Bob read?

C. Deepen Activity (Skills)

Create one real life problemthat can be solved using the equation x + 25 = 75and solve your own problem. Show all yourworkings.Criteria 0 1 2

SolutionSolution does not correctly apply addition orsubtraction property of equality

Solution contains minor error in applyingaddition or subtraction property ofequality

Solution correctly applies addition orsubtraction property of equality

ProblemCreates a real-life problem using inappropriatequantities which do not match the givenequation

Creates a real-life problem but somequantities do not match those (values isoff) in the given equation

Creates a completely correct real-lifeproblem using appropriate quantitiesthat can be solved using the givenequation

D. Synthesis (Understanding)Create one problemeach that forms an equation that can be solved using addition and/or subtraction properties of equality andsolve your own problems

Criteria 0 1 2

SolutionSolution does not correctly apply additionor subtraction property of equality

Solution contains minor error in applying

addition or subtraction property of equality

Solution correctly applies addition or

subtraction property of equality

Equation Was not able to provide an equationCreates an incorrect equation that do notmatch the given problem

Creates a correct equation that matchthe given problem

Problem Was not able to create a real-life problemCreates a correct real-life problem but someunclear quantities.

Creates a completely correct real-lifeproblem using appropriate quantities.

K:_______S:_______U:_______

P:_______

SCORE

Write an algebraic equation for each problem and then solve the equation.1. An empty box weighs 1.5 pounds. When the box is filled with marbles, the total weight is 8.6 pounds. Find the weight of the

marbles.2. Five students left a room and 21 students remained. How many students were in the room before any student left?3. Elena has 5 500 in her savings account. This is 2 100 more than Dennis. Find the amount Dennis has in his savings

3. Representation:Let a be the amount Dennis has inhis savings accountEquation: a + 2 100 = 5 500Solution: a + 2 100 = 5 500

a + 2 1002 100= 5 5002 100 SPE

a = 3 400


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Activity Title: Quiz # 1 Activity No.: 06Reference/s: Oronce, O.A., Mendoza, M.O., (2013) e-Math 7, Manila, Rex Bookstore pp. 224 - 241Designer:Miss Mary Jane A. Palattao RCBN-ES School:OLLCS

I. LEARNING TARGETSAt the end of the lesson the students should be able to:

1. Give an example of conditional, inconsistent and identity equations2. Solve problems involving 1stdegree equations in one variable using APE or SPE.

II. LEARNING EXPERIENCESA. Give two examples each of conditional, inconsistent and identity equations. Explain your answer. [ 9 pts.]

Criteria 1 2 3

ExplanationCorrectly explains only 1 type ofequation

Correctly explains the 2types ofequations

Correctly explains the 3types ofequations

Example(per type)

Correctly gives only 1 example ofequationper type

Correctly gives 2examples ofequationsper type

B. Solve for the value of the variable. [2 pts. Each]

1. c + 5 = - 2 3. 5x + 3 = 4x - 72. k1.6 = -0.4 4. 710x = - 9x - 11

C. that can be solved using the equation 2x = 3x

24and solve your own problem. Show all your workings.

Criteria 0 1 2

Solution

Solution does not correctly applyaddition or subtraction property ofequality

Solution contains minor error inapplying addition or subtractionproperty of equality

Solution correctly applies additionor subtraction property of equality

ProblemCreates a real-life problem usinginappropriate quantities which do notmatch the given equation

Creates a real-life problem butsome quantities do not matchthose (values is off) in the givenequation

Creates a completely correct real-life problem using appropriatequantities that can be solvedusing the given equation

K:_______S:_______U:_______

P:_______

SCORE


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++

++

++

++

++

+++

+ +



Telephone 430-7819



Activity Title: Solving Simple Equations Using Multiplication or Division Activity No.: 07Property of Equality (MPE or DPE)

Value/s: EqualityReference/s: Oronce, O.A., Mendoza, M.O., (2013) e-Math 7, Manila, Rex Bookstore pp. 242 - 247Designer:Miss Mary Jane A. Palattao RCBN-ES School:OLLCS

I. LEARNING TARGETAt the end of the lesson the students should be able to:1. Solve 1stdegree equations that can be solved using MPE or DPE

II. LEARNING EXPERIENCESA. Exploration Activity

To solve multiplication and division equations, a flowchart and its inverse flowchart can be used.

Equation:7x = 84 Equation: 95

n

Complete the flow chart and the inverse for each given equation.

1. 12b = 48 2. 11c = 99 3. 98

x 4. 12

10

a

B. Concept Notes

Using algebra tiles, represent the equation 3x = 12

X

Multiplied by 7

8484

Divided by

x

X = 12

84 7 = 12

n

divided by 5

9 9

Multiplied by 5

n

9 x 5 = 45

b = ______ c = ______x = ______

a = ______

A multiplication equationis an equation involving a product of a variable and a number. You can solve a multiplication equation

by dividing the same nonzero number on each side of the equation. The goal is the same: to isolate the variable on one side of

the equation.

+ +

++

++

++

++

+++

++

K:_______S:_______U:_______

P:_______

SCORE

n = 45


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Divide each side of the equation into three equal parts.3

12

3

3

x Each part has a single x-tile on the left.

Just as we have un-added or un-subtracted to undo what had been done to a variable in an equation, we can also un-multiplyby dividing both sides of an equation by a number other than zero. We can also un-divide by multiplying both sides of the equationby any nonzero number.Example:

1. 4x = 124

Solution:

4x = 124

4

124

4

4

x Divide both sides by 4

x = 31 Simplify

Examples:

1. 403

x

Solution:

403

x

)(x 4033

3 Multiply both sides by 3

x = 120 SimplifyTo Check:

403

120

40 = 40

C. Firm-Up Activity (Knowledge)

Write the number by which you could multiply or divide both sides to solve the equation. Do not solve.

1. -9y = 81 4. 86b

2. 4.3a =12.9 5.10

17

5

h

3. 15d = 60 6.7

5

3

x2

D. Deepen Activity (Skills)Solve and check the following equations.

1. 3x = 63 3. 15x4

3

2. 8x = 17 4. 74x

E. Synthesis (Skills)Write a multiplication and division equation that has the given number as its solution. Solve the equation.

a. 3 b. 11

++++ + x = 4

To check:

4(31) = 124

124 = 123

A division equationis an equation involving a quotient of a variable and a nonzero number. You can solve a division equation by

multiplying the same nonzero number on each side of the equation. The goal is the same: to isolate the variable on one side of

the equation.

The process of dividing the same nonzero number on

both sides of an equation to produce a new equation

having the same solution as the original is division

property of equality(DPE).

ax = b 0awherea

b

a

ax

2. 127

3x

Solution:

1207

3x

)(x 123

7

7

3

3

7

Multiply both sides by 3

7

x = 28 Simplify

The process of multiplying the same nonzero number on both sides of an

equation to produce a new equation having the same solution as the

original is multiplication property of equality(MPE).

0awhereabaa

xb

a

x


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Telephone 430-7819



Activity Title: Solving Simple Equations Using Multiplication or Division Activity No.: 08Property of Equality (MPE or DPE)

Value/s: EqualityReference/s: Oronce, O.A., Mendoza, M.O., (2013) e-Math 7, Manila, Rex Bookstore pp. 242 - 247Designer:Miss Mary Jane A. Palattao RCBN-ES School:OLLCS

I. LEARNING TARGETAt the end of the lesson the students should be able to:1. Solve problems involving 1stdegree equations that can be solved using MPE or DPE

II. LEARNING EXPERIENCESA. Concept Notes

Examples:1. The height of Building A is 10 times greater than the height of Building B. Find the height of Building B if Building

A is 170 ft.

2. The family business had an average monthly profit of 23 000 for the first quarter of the year. Find the totalprofit for the quarter.

Solution:1. Representation:Let x be the height of Building B

Equation: 10x = 170Solution: 10x = 170

10

170

10

x10 DPE

x= 17 ft.

B. Firm-Up Activity (Skills)

Write an algebraic equation for each sentence. Then solve the equation. Use any variable to represents the unknown.1. The product of a number and -6 is30.2. Fifteen equals the quotient of a number and -15.3. A rectangular garden with an area of 120 m2has a length of 15 m. Solve its width.4. Mrs Centeno bought paper and pencils for her pupils. She bought papers and pencils of the same quantity. The pencils

cost 8 each and the papers cost 11 each. She spent 722 in all. How many of each did she buy?

C. Deepen Activity (Understanding)You are given an equation like this 5x

2 = 33. Make a problem that will lead to the information of the given equation and then

solve your own problem. Let your creativity flow!Note:Do NOT use x as part of your problem.

Criteria 0 1 2

SolutionSolution does not correctly apply addition orsubtraction property of equality

Solution contains minor error in applyingaddition or subtraction property ofequality

Solution correctly applies addition orsubtraction property of equality

ProblemCreates a real-life problem using inappropriatequantities which do not match the givenequation

Creates a real-life problem but somequantities do not match those (values isoff) in the given equation

Creates a completely correct real-lifeproblem using appropriate quantitiesthat can be solved using the givenequation

K:_______S:_______U:_______

P:_______

SCORE

2. Representation:Let xbe the total profit per quarter

Equation: Average profit per monthquarterainmonthsofNumber

quarterperprofitTotal

3

x00023

Solution:

3

x)00023(3

3

x00023

69 000 = x


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Telephone 430-7819



Activity Title: Solving Multi-step Equations Activity No.: 09Value/s: EqualityReference/s: Oronce, O.A., Mendoza, M.O., (2013) e-Math 7, Manila, Rex Bookstore pp. 248 - 256Designer:Miss Mary Jane A. Palattao RCBN-ES School:OLLCS


1. Solve 1stDegree Equations in one variable using two or more properties of equality.II. LEARNING EXPERIENCES

A. Concept Notes

Examples:Solve each equation

1. -3

x7 - 2 = 3

-3

x7- 2 + 2 = 3 + 2 APE

-3

x7= 5

-3

x7 -

7

3= 3 -

7

3 Multiply both sides by -

7

3

x = -7

9 3.

4

m54

3

m35

124

m54

3

m3512

Multiply both sides by LCM (12)

4(53m) = 3(45m) Simplify

2012m = 1215m Use the Distributive property

2012m + 15m= 1215m + 15m Add 15m on both sides

20 + 3m = 12 Simplify

20 - 20+ 3m = 12 - 20 Isolate the variable on one side by subtracting 20 on both sides

3m = - 8

3

8

3

m3 Divide both sides by 3

3

8m

Solving Equations with Parentheses1.Combine similar terms that are either free (not within the parentheses) or within the same parentheses.2.Multiply to clear parentheses looking for similar terms after each clearing.3.After the elimination of all parentheses and the combination of similar terms on both sides of the equation, collect all terms

containing the variable on one side of the equation and all numerical terms on the other side.4.Un-add and/or un-subtract (or use transposition)5.Un-divide (by multiplication)6.Un-multiply ( by division)

K:_______S:_______U:_______

P:_______

SCORE

2. 5 (2f1) + 6 = 7f - 8

10f5 + 6 = 7f8 Use the Distributive property10f + 1 = 7f8 Simplify the left side expression

10f7f+ 1 = 7f7f8 Subtract 7f from both sides (SPE)3f + 1 = - 8 Simplify

3f + 1 - 1= - 81 Subtract 1 from both sides (SPE)3f = - 9

3

9

3

f3 Divide both sides by 3 (DPE)

f = - 3


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B. Firm-Up Activity (Skills) Solve each of the following equations.

1. 5x + 2(4x) = 38 3.7

5x2

3

1x

2. 11x5

122x

5

2 4.

7

4

5x3

1x2

C. Deepen Activity (Skills)Make a 1stdegree equations with:a. involving two operationsb. with fractions involving two or more operationsSolve the equations.

Criteria 0 1 2

SolutionSolution does not correctly applyproperties of equality

Solution contains minor error in applyingproperties of equality

Solution correctly appliesproperties of equality

Equation

Not able to set-up two 1stdegreeequations:a. involving two operationsb. with fractions involving two or more

operations

Set-up (one) 1stdegree equationsinvolvinga. two or more operations (or)b. with fractions involving two or moreoperations

Set-up two 1stdegree equationsa. involving two operationsb. with fractions involving two ormore operations


14/14




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Activity Title: Solving Multi-step Equations Activity No.: 10Value/s: EqualityReference/s: Oronce, O.A., Mendoza, M.O., (2013) e-Math 7, Manila, Rex Bookstore pp. 248 - 256Designer:Miss Mary Jane A. Palattao RCBN-ES School:OLLCS


1. Solve 1stDegree Equations in one variable using two or more properties of equality.II. LEARNING EXPERIENCES

Note: Provide a review.A. Firm-Up Activity (Skills)

Solve for the value of x.

1. 3725

x 4.

3

5x1

4

3x2

2. 40 + (4x + 5) = 9x 5. x5

3295x

5

12

3. 3x2 ( x + 3 ) = 92xB. Deepen Activity (Skills)

1. Write two-step equation involving addition and division that has -10 as its solution.2. Write two-step equation involving subtraction and multiplication that has 4 as its solution.

C. Synthesis (Understanding)1. Your friend did the following solutions on the board:

a. 54x

3x

54x

3x)4x(

x3 = 5x = 2

a. Comment on your friends solution.

b. How would you solve the equations? Explain.Criteria 0 1 2

SolutionSolution does not correctlyapply properties of equality

Solution contains minor error in applyingproperties of equality

Solution correctly applies properties ofequality

CommentNot able to identifyinconsistencies in the solutionand no explanation

Identifies inconsistencies in the solution butwas not able to explain what and why (e.g.incorrect properties of equality used forcomputation)

Correctly explain and justifyinconsistencies in the solution (e.g.identifies and explain incorrect propertiesof equality used for computation)

K:_______S:_______U:_______

P:_______

SCORE

b. 2[ 3(3 + 2x)(3x)] = 5(2x + 3)(12x)2[ 3(3 + 2x)3x] = 5(2x + 3)12x

2[ 3x] = 5(2)6x = 10x = 10 - 6x = 4

3rd quarter math g7 las (week 1 - 2).pdf

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