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Introduction

Hello and Good Day Everybody !
I am a new teacher in the school named Sekolah Menengah Kebangsaan Datuk Bendahara. This school is situated in Jasin, Melaka.
I am accepted to be one of the Guru Sandaran Terlatih (GST) here. There are 10 of us, my batch joined the schools in 27th June 2011.
In this blog, I share with you my experience and great things that are happening to me in the school.
There are also notes and things for good English class usages, so feel free to browse them whenever there are gaps between classes.
My mentor here is Hajjah Zaorah and she has been guiding me to be a better English teacher.
Thank you !
.
In this blog, I have uploaded
Literature lesson plans for form 4 and 5
working papers, reports after programmes, yearly uniform body reports, weekly reports, sponsorship letters, sample of memos and thank you notes, short stories and others.



Happy teaching, teachers !
Here is a good link to English Materials:

http://www.ppdjasin.edu.my/unitakademik/index.php?option=com_content&view=article&id=47&Itemid=54

Pengikutku, Sila jadi pengikut untuk blogs saya !

Wednesday, June 6, 2012

HBMT 3403 TEACHING MATHEMATICS IN FORM TWO(2)







FACULTY OF EDUCATION AND LANGUAGES


HBMT 3403 TEACHING MATHEMATICS IN FORM TWO
OCTOBER 2011















Tutor
MR.
 




Table of Contents


1.0        Introduction

2.0        How to Solve the Linear Equation
2.1       Equation
2.3       Equation Involving Addition
2.4       Equation Involving Subtraction
2.5       Equation Involving Multiplication
2.6       Equation Involving Divide
2.7       Let’s Solve Equations With Two Unknowns
(a)                  We Try to Solve Linear Equation By Using the Elimination Method
(b)                  We Try to Solve Linear Equation By Using the Replacement Method

3.0  Reflection
4.0  Conclusion
5.0  References


1.0            Introduction

Mathematics is knowledge that come from observation on nature. Mathematics is a logic system that shows the formula, that creates the language of Mathematics such as symbols, rules and operations.
Mathematics is a way of thinking that is used to expand on reasoning and reach to a conclusion from the existence of the universe.Mathematics is also interpreted as an art form, because it contains its language and patterns in an interesting shapes.
A character in Mathematics is Mathematics contains language, symbols and operations. The basic of the language system is the grammar. In the Mahematics language, the grammar consists of rules, theorems and formula that connects the symbols.

According to Piaget, the mathematics language is abstract and only secondary school students can understand it. Students in primary schools cannot understand the mathematics symbols that are abstract without relating it to their concrete experience.  Thus, teaching the Mathematics language must be related to their experience , relate it with teaching aids and concrete items. For example, when we want to teach on Mathematics, we relate to the spupil’s daily experience, for example, I am purchasing three items at the store, at these prices, RM 19.95, RM9.98 and RM 29.97. About how much money am I spending ? The fastest way to solve this problem is to round off and approximate. The first item costs about $20, the second about $40, and the third about $30; therefore, I am spending about $90 on your shopping spree. Rounding is often an excellent heuristic for arriving quickly at approximate answers to mathematical problems.
At school, students typically get far more practice solving specific problems rather than general ones, and they are taught many more algorithms than heuristics. For example, they are likely to spend more time learning problem-solving strategies that they cannot use, such as determining the length of planks needed for a village house than strategies applicable to the problem of deforestation. Many real-world problems cannot be solved with this algorithms. Algorithms exist for solving problems outside the domains of mathematics and science.
Problem-solving strategies, algorithms and heuristics alike, are often specific to particular content domains. But here are several general problem-solving heuristics that students may find helpful in a variety of contexts: Identify subgoals. Break a large, complex task into two or more specific subtasks that can be more easily addressed. Use paper and pencil. Draw a diagram, list a problem’s components, or jot down potential solutions or approaches. Draw an analogy. Identify a situation analogous to the problem situation, and derive potential solutions from the analogy. Brainstorm. Generate a wide variety of possible approaches or solutions—including some that might initially seem outlandish or absurd—without initially evaluating any of them. Once a lengthy list has been created, evaluate each item for its potential relevance and usefulness.  “Incubate” the situation. Let a problem remain unresolved for a few hours or days, allowing time for a broad search of long-term memory for potentially productive approaches.
Problem solving refers to the process of moving from the given state to the goal state of a problem. It can be summarised that problem solving is a series of mental operatiomns that are directed towards some goals. (Mayer, 1983).
Dewey(1910) mentioned that problem solving involves some form of information received perceptually, physiologically and via sensory utilisation to reach a solution. Later, Polya (1945) dexcribed propblem solving process as a “grain of discovery in which curiosity and inventive facilities one will be able to solve the problem and may experience the tension and triumph of discovery”.

John Dewey. "What is thought?" Chapter 1 in How we think. Lexington, Mass: D.C. Heath, (1910): 1-13

Some problems can be successfully solved by following specific, step-by-step instructions—that is, by using an algorithm. We can correctly assemble the pieces of a new bookcase by following the directions for assembly that come with the package. We can calculate the length of a slanted roof by using the Pythagorean theorem. When we follow an algorithm faithfully, we invariably arrive at a correct solution.

However, the world presents many problems for which no algorithms exist. There are no rules we can follow to identify a substitute metal ship, no list of instructions to help us address the destruction of rain forests. In the absence of an algorithm, learners must instead use a heuristic, a general problem-solving strategy that may or may not yield a successful outcome. For example, one heuristic that we might use in solving the deforestation problem is this: Identify a new behavior that adequately replaces the problem behavior (i.e., identify another way that peasant farmers can meet their survival needs).


HEURISTICS FOR PROBLEM SOLVING
This is a set of heuristics that has been proven to be successful with students and teachers at all levels of instructions

Read the problem

  1. Note key words
  2. Describe the problem
  3. Visualise the action
  4. Restate the problem in your own words
  5. What is being asked for ?
  6. What information is given ?

Explore

  1. Organise the information
  2. Is there enough information ?
  3. Is there too much information ?
  4. Draw a diagram or construct a model
  5. Make a chart or table

Select a strategy

  1. Pattern recognition
  2. Working backwards
  3. Guess and test
  4. Simulation or experimentation
  5. Simplify the problem
  6. Organised listing or make a table
  7. Logical deduction
  8. Using formula

Solve

  1. Carry out the strategy
  2. Use computational skill
  3. Use geometric skills
  4. Use algebraic skills
  5. Use elementary logic

Look back

  1. Check your answer
  2. Find another way
  3. What if... ?
  4. Extend
  5. Generalize

Krulik, S.Redneck,JA (1989) “Problem Solving A Handbook for Senior high School Teacher”.

(J. R. Anderson, 1990; J. E. Davidson & Sternberg, 1998, 2003; H. C. Ellis & Hunt, 1983; Halpern, 1997a)
.

Dewey J., 1910, How We Think, Dover, New York.

Another strand of thinking that appears to support the ‘conjecture-first’ approach to thinking may be drawn from the words of that classic problem solving text by Polya, 1945)  ‘How To Solve It’. This booklet is aimed at mathematicians. It provides a suggested methodology to tackle the creative process of solving mathematical and geometric problems.

His ‘steps’ are:
“First. You have to understand the problem”

Second. Find the connection between the data and the unknown.
(You may be obliged to consider auxiliary problems if an immediate connection cannot be found. You should obtain eventually a plan of the solution)

Third. Carry out your plan

Fourth. Examine the solution obtained” [1945, p. xvi]

Polya then provides further elaboration of what he meant under the second step in terms of needing to seek analogous problems that have been solved. He suggests using existing solutions to old problems. This is further understood by his elaboration of the third step which he states as testing to see if the old analogous solution or at least similar concepts work for the new problem. Polya, who is a well respected author on problem solving, is therefore not suggesting that solutions come after reasoning about alternatives. Rather, he suggests that you search for a conjecture solution (from an analogous problem) and then think about its usefulness for the current problem. This again seems very much like Dewey’s advice and the advice of argumentative inquiry: conjecture something that might work, and then think about the feasibility of this solution carefully.

Polya G., 1945, How To Solve It, Princeton University Press, New Jersey.

(Unknown=anu)
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2.0            How to Solve The Linear Equation

            For this task, I have taken Chapter 4, Linear Equations topic equality (page 70) , linear equations in one unknown (72) , solutions of linear equations in one unknown(75). The textbook that I am using is Mathematics Textbook Form 2 Volume 1 2003, Arus Intelek Sdn Bhd. Cheang Chooi Yoong, Khau Phoay Eng, Yong Kien Cheng ISBN 983-150-241-8.

2.1       Equation
            Linear equation is used everyday in students’ lives, but the pupils cannot utilise the knowledge of linear equation because the approaches in schools are different from their lives. Equations enable us to visualize complex problems easily. They are formed from numbers, equations and the symbol of equation =.  For example, Salleh and Jamil both have the same weight. We can write the equation like this: 40 kg=4000 grams. It is an equation. 2+8=10 is an equation.
            If the value of x is 2, then x can replace 2 and we can write this equation as x+8=10.
This is an equation that has x. The value for x is 2(x=2)
Let’s have a look at a scale, the scale becomes balance if we put 2 kg of flour on each side. Example: Consider this simple equation:
x=5

Look at the equation  as a balanced scale with x and 5 is measured in kg.
5 kg
=
x kg
2
 







Salleh puts 5 kg of chicken meat on the right hand of the scale, and the scale becomes balanced when he puts 5 kg of fish on the left hand, that is x+3= 5


5 kg
=

(x+3)kg

 






When we subtract the same weight, let’s say 3 kg, the scale becomes balanced.x-3+2.



2 kg
=

(x-3)kg

 








2.2       Finding Nemo

Teacher needs to guide the students to find the value of an equation.

                          I.          Equal number can be subtracted from both equations.
                        II.          Similar number can be added to both equations.
                      III.          Both equations dan be divided with similar numbers.
                      IV.          Both equations dan be multiplied with similar numbers

Operations such as addition, subtraction, multiplication and divide are used to build an equation. To solve an equation , we use operations so the unknown is the only thing that is left on the left handside.
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2.3       Equation Involving Addition

The opposite operation for addition is minus. So, to solve an equation that involves addition, we resolve addition with subtraction, the same number for both equations.
Example:        Solve the equation  x + 6 = 14.
Method:
                        x + 6 = 14
                        x + 6 - 6 = 14 - 6        (Subtract 6 from both sides)
                        x + 0 = 8
                        x = 8


2.4       Equation Involving Subtraction
            The opposite operation for subtraction is addition. So, to solve an equation that involves subtraction, we resolve subtraction with addition, the same number for both equations.

Example:        Solve the equation  x - 9 = 17.
Method:
                        x - 9 = 17
                        x - 9 + 9 = 17 + 9        (Adding 9 to both sides )
                        x + 0 = 26
                        x = 26




2.5       Equation Involving Multiplication
The opposite operation for multiplication is divide. So, to solve an equation that involves multiplication, we subtraction both equations with the same number.

Example:        Solve the equation  9x = 72.
Method :
                        9x = 72
                        So,   =    (Divide both sides with 9)
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                        x = 8
2.6       Equation Involving Divide
The opposite operation for divide is multiplication. So, to solve an equation that involves subtraction, we multiply both equations with the same number.

Example:        Let’s solve this equation   = 8.
Method:
             = 8
So we know that 5 x    = 8 x 5   (We multiply this equation with 5)
                        x = 40

First question

The Maths teacher shows the students of 2A class all possible means in solving the linear equations.The Maths teacher will show the class how to perform these methods. The first method is the elimination method, and the second method is called the replacement method.Let us look at one example.then, the class is given a task, that is to work in groups of two and solve a problem.

Solve the linear equations

3p – q = 11
4p + 5q = -17

(c)            We Try to Solve Linear Equation By Using the Formula of Elimination Method
Look at the linear equation below, we need to eliminate q. That is to multiply 5 to the equation 3q – q = 11

3p – q = 11   ................................. (1)  x  5
4p + 5q = -17  ................................(2)
15p – 5q = 55 .................................(3)
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Then, we add equation two with equation one
            4p + 5q = -17  ................................(2)
15p – 5q = 55 .................................(3)


                        19p        = 38
                            p       =38/19
                             p       =2
    
So, we get the value of p is 2

Now, we already know the value of p, so we replace p with 2 into all equations above

We replace p with 2 in equation two

4(2) + 5q = - 17
8 + 5q = - 17 .................. we subtract 8 in equation (8 – 8 + 5q = - 17 – 8)
5q = - 25
 q = - 5

so from the equations that we have solved, we get the values, it shows that p is 2 and q is 5.


(b)       We Try to Solve Linear Equation By Using the Formula of Replacement method
             

3p – q = 11 .................................. (1)
4p + 5q = -17 ................................ (2)

 Let’s find q value first.
From equation number one,
            3p – q = 11
-q = 11 – 3p .............. x – we multiply this equation with negative(-)
q = 3p – 11 ............................. we form the equation number three
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we replace the q value with 3p-11 in equation number three to equation number 2
            4p + 5q = - 17
            4p + 5(3p – 11) = -17
            4p + 15p – 55 = - 17

Let’s solve this:
,           19p = - 17 + 55
            19p = 38
                p    = 2

Now, we can replace the p value that we have obtained into all equations.
We insert 2 in to equation number 1
            3(2) – q = 11

Let’s solve this:
           
6 – q = 11
                        -q = 11 – 6
                        -q = 5  ........................ we multiply this equation with negative (-)
                        q = - 5
So, from what we have done, we find that p is 2 and q is -5.
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Second question

Salleh  and Karim both have RM 510 in their savings accounts now. They opened their savings account on the same day in which Salleh  started with twice as much money as Angela. From then on Salleh  added RM 10 to her account each week and Angela put in RM 20 each week. How much money did Salleh  bank in when she open her saving account ?

We Try to Solve Linear Equation By Using the Guessing Method

Guess
(Guess how many months have they banked in)
e.g. 15 months
16 months
17 months
Salleh  (Add RM 10 every month)
RM 150
RM 160
RM 170
Balance, (RM 510 - ?)
RM 510 – RM 150=
RM 510 - RM 160=
RM 510 - RM 170= RM 340


Guess
(Guess how many months have they banked in)
e.g. 15 months
16 months
17 months
Karim (Add RM20 every month)
300
RM 320
RM340
Balance, (RM 510 - ?)
RM 510 - RM 300=
RM 510 - RM 320=
RM 510 - RM 340= RM 170

Salleh  started with RM 340, and Karim started with RM 170, when they opened their savings account 17 months ago.

Method two:Formula
2x + (n-1)(10)
=510
2x + 10n-10
=510
2x + 10n
=500
X + 5n
=250
This is the first equation

X+(n-1)(20)
=510
X+20n-20
=510
X=20 n
=490


15n
=240
n
=240/15
n
=16


x
=250-5(16)

=170
2x
=170 / 2
x
=340


This is the second equation







Page 57 of oum

Making a table
Haris and Faris
Formula

Second qns

We Try to Solve Linear Equation By Using the Formula Method

Miss Saleha posted altogether 32 hari raya cards and letters. She has to pay RM 8.35 for the postal fees at the post office. The postcards cost 20 cents each and the letters cost 33 cents each. How many postcards and letters had she sent  ?

  1. Understand the problem
Total cards and letters=32
32= RM 8.35
One postcard costs 20 cents, one letter costs 33  cents.
We need to determine how many postcards and letters are sent.

2 Devise a plan
We need to put this into equations:

0.2 x + 0.33 y

=8.35


y
=32-x 


3 Carry out the plan



0.2 x + 0.33 y

=8.35
0.2x
=8.35-0.33 y
         0.2

x
=8.35-0.33 y                0. 2
X+y
= 32
Y
=32-x 

this is the equation number 2

Substitute equation number 2 into equation number 1
0.2x+0.33 y
=8.35

x

= 8.35-0.33y                  0.2

32-y
=8.35-0.33y
         0.2

6.4-0.2y
=8.35-0.33y

0.13y
=1.95

y
=15

Put in equation number 1

y

=32-x 
15
=32-x
x
=32-15
x
=17


So, x
=17
y
=15

Look back
0.2x+0.33 y
=8.35

x

= 8.35-0.33y                  0.2

32-y
=8.35-0.33y
         0.2

6.4-0.2y
=8.35-0.33y

0.13y
=1.95

y
=15


The number of cards sent is 17 and 15 letters are sent.
The second method is by guess and checking
  1. Understand the problem
Total cards and letters=32
32= RM 8.35
One postcard costs 20 cents, one letter costs 33  cents.
We need to determine how many postcards and letters are sent.

2.Devise a plan
Suppose we guess that the number of postcards is 10. It means the number of letters is 32-10=22. If this is true, then the total bill will be (10 X 20 cents) +(22 X 33 cents)=RM 9.26.

This tells us that the 10 cards and 22 letters would cost more, so we need to guess one more time.
3       Carry out the plan. Let us guess
Suppose we guess that the number of postcards is 16. It means the number of letters is 32-16=16. If this is true, then the total bill will be (16 X 20 cents) +(16 X 33 cents)=RM 8.48
Suppose we guess that the number of postcards is 17. It means the number of letters is 32-17=15. If this is true, then the total bill will be (17 X 20 cents) +(15 X 33 cents)=RM 8.35
  1. Check the answer

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4.0       Reflection

I have shown these examples to two classes, that is 2A and 2F. These two methods have been shown to them and i have explained it differentlly for both classes. The first class, ihave explained in detail, and the second class, I explain the basics only, I don’t want them to get confused which method to use.
These two classes have different capabilities from each other. 2A can understand the two methods, whereas 2F needs more explanations from the teacher. It looked 2F needed a double period for linear equation topic, and I wish to include this into their end year final examination, and I want both classes to be able to answer the linear equation problems and get A One for Maths
The students are encouraged to apply both methods so they can tell which method is better in a given problem.Usually they like to use the replacement method because the method is simpler and not confusing. If we compare with the elimination method, there are many students who are confused with the subtraction method that involves addition and subtraction. They will be able to use the method for simpler questions such as x + y = 8,  x – y = 2. It shows that they can still answer question using the elimination method because the question does not require multiplication.

Conclusion
It is shown that by only using three methods the studnets are able to solve 2 problems easily and quickly. Not only that, The Polya’s and Dewey’s methods which is proven will guide the students in making better judgments in facing their daily problems, and they can calculate using one of the three methods, how much money they have spent for that week, even though they do not have the information organised neatly. Gthey also can create new statistics on the aspect of time management or how to manage time better using the heuristics. They can solve their problem of not having much time on studying at home. Time can be calculated better using the maths formula they have learnt in class.


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References

Mohd Nasir Mahmud.(2010). HMBT3403 Teaching Mathematics in Form Two Kuala Lumpur:OUM






References

Bull, Ian,1995.Mathematics Revision and Practice, Book 1. Australia:Phoenix Education.
Coffey,Lynch Picking Anders, 1981.Maths 7.Australia:Longman Cheshire Pte Limited.



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