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
problemsolving 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 realworld problems cannot be solved with this
algorithms. Algorithms exist for solving problems outside the domains of
mathematics and science.
Problemsolving strategies, algorithms and heuristics alike, are often
specific to particular content domains. But here are several general
problemsolving 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 longterm 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): 113
Some problems can be successfully solved by following specific,
stepbystep 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
problemsolving 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
 Note key words
 Describe the problem
 Visualise the action
 Restate the problem in your own words
 What is being asked for ?
 What information is given ?
Explore
 Organise the information
 Is there enough information ?
 Is there too much information ?
 Draw a diagram or construct a model
 Make a chart or table
Select a
strategy
 Pattern recognition
 Working backwards
 Guess and test
 Simulation or experimentation
 Simplify the problem
 Organised listing or make a table
 Logical deduction
 Using formula
Solve
 Carry out the strategy
 Use computational skill
 Use geometric skills
 Use algebraic skills
 Use elementary logic
Look back
 Check your answer
 Find another way
 What if... ?
 Extend
 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 ‘conjecturefirst’ 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)
1

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 9831502418.
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.x3+2.
2 kg

=


(x3)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.
3

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)
4

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)
5

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
6

we replace the q
value with 3p11 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.
12

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 + (n1)(10)

=510


2x + 10n10

=510


2x + 10n

=500


X + 5n

=250


This is the first equation



X+(n1)(20)

=510


X+20n20

=510


X=20 n

=490





This is the second equation


Page 57 of oum
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
?
 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

=32x

3 Carry out the plan
0.2
x + 0.33 y

=8.35

0.2x

=8.350.33 y
0.2

x

=8.350.33 y 0. 2

X+y

= 32

Y

=32x

this is the equation number 2
Substitute equation number 2 into equation
number 1
0.2x+0.33
y

=8.35

x

= 8.350.33y 0.2

32y

=8.350.33y
0.2

6.40.2y

=8.350.33y

0.13y

=1.95

y

=15

Put
in equation number 1


y

=32x

15

=32x

x

=3215

x

=17



So,
x

=17

y

=15

Look
back
0.2x+0.33
y

=8.35

x

= 8.350.33y 0.2

32y

=8.350.33y
0.2

6.40.2y

=8.350.33y

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
 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 3210=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 3216=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 3217=15. If this
is true, then the total bill will be (17 X 20 cents) +(15 X 33 cents)=RM 8.35
 Check the answer
1

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.
14

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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