Form 2 Assigment

8/3/2019 Form 2 Assigment

1/22

Azyyati bte A.Rahim 761022017628

1.0 INTRODUCTION

A "system" of equations is a set or collection of equations that you deal with all together

at once. Linear equations (ones that graph as straight lines) are simpler than non-linear

equations, and the simplest linear system is one with two equations and two variables.

Think back to linear equations. For instance, consider the linear equationy = 3x 5. A

"solution" to this equation was anyx,y-point that "worked" in the equation. So (2, 1) was

a solution because, plugging in 2 forx:

3x 5 = 3(2) 5 = 6 5 = 1 =y

On the other hand, (1, 2) was not a solution, because, plugging in 1 forx:

3x 5 = 3(1) 5 = 3 5 = 2

which did not equaly (which was 2, for this point). Of course, in practical terms, you did

not find solutions to an equation by picking random points, plugging them in, and

checking to see if they "work" in the equation. Instead, you picked x-values and then

calculated the corresponding y-values. And you used this same procedure to graph the

equation. This points out an important fact: Every point on the graph was a solution to the

equation, and any solution to the equation was a point on the graph.

A system of linear equations can be solve by using three methods. They are Elimination,

Substitution and Graphing. When we solve by Elimination method we have to make the

coefficient of one the variable into equal and then add both the equations. Now one of the

variables will get cancel and then one of the variables is left out. After that, solve for the

left variable. Finally substitute the result of one of the variable in any one of the equation

we will get the result for another variable. When we solve by substitution method

substitute one of the equation in another and then solve. If we are solving by graphing

method draw a graph for both the equation in a single graph. The intersection point of the

lines is a solution for the given system of linear equation.

HBMT 3403 TEACHING MATHEMATICS IN FORM TWO 1
http://www.purplemath.com/modules/graphlin.htmhttp://www.purplemath.com/modules/graphlin.htmhttp://www.purplemath.com/modules/graphlin.htmhttp://www.purplemath.com/modules/graphlin.htm


2/22


2.0 Elimination Method

In the elimination method for solving simultaneous equations, two equations are

simplified by adding them or subtracting them. This eliminates one of the variables so

that the other variable can be found.

To add two equations, add the left hand expressions and right hand expressions

separately. Similarly, to subtract two equations, subtract the left hand expressions from

each other, and subtract the right hand expressions from each other. The following

examples will make this clear.

Example 1: Consider these equations:

2x 5y = 1

3x + 5y = 14

The first equation contains a-5yterm, while the second equation contains a+5yterm.

These two terms will cancel if added together, so we will add the equations to eliminate

y.

To add the equations, add the left side expressions and the right side expressionsseparately.

2x 5y = 1+ 3x + 5y = + 14

(2x 5y) + (3x + 5y) = 1 + 14

Simplifying, -5y and +5y cancel out, so we have:

5x = 15

Therefore x is 3.

By substituting 3 for x into either of the two original equations we can find y.

2.1 Procedure involved in solving the problem using the elimination method



3/22


Step 1 : Enter the linear equations to solve

Label the equations as follows:

equation 1 - 3p q = 11

equation 2 - 4p + 5q = -17

Step 2 : If necessary, prepare either or both equations for eliminating a

variable.

Decide which variable you are going to eliminate from the equations and whether you are

going to add the equations or subtract them.

For the example equations, we will multiply the first equation through by

5 and eliminate variable q by adding the both equation.


equation 2 - 4p + 5q = -17

To modify an equation, use a copy of the original equation.

For the example, copy the equation:

3p q = 11

To multiply the bottom equation by 5:

Multiplying by '5' and expanding both sides

5(3p q) = 11 x 5

Simplifying, both sides

15p 5 q = 55

Step 3 : Eliminate the first variable



4/22


Write the two equations that you are going to add or subtract to the bottom of the

working page, using the copying technique described above.

For the example, the last few lines of the display will now look like this:

Simplifying, both sides ...

15p 5q = 55

15p 5q = 55

4p + 5q = -17

Add the equations.

15p 5q = 55

+ 4p + 5q = -17

19p = 38

Simplify all. You will now have an equation in only one variable.

19p = 38 19

p = 2

Step 4 : Substitute to find the other variable



5/22


4p + 5q = -17

Select the equation for the known variable.

For the example, select p= 2

You will now have an equation in the other variable. Solve this equation to find the value

of this other variable.

For the example, the equation is:

4p + 5q = -17

4(2) + 5q = -17

8 + 5q = -17

In this case, we add 4 to both sides and simplify:

5q = -17- 8

q = -255

q = -5

Step 5 : Validate the solutions by substitution

You can test your answers to make sure they are correct. Type both solutions with a

comma between them:

For the example, type: p = 2, q = -5

State one of the original equations to make it the target data set:

equation 1 - 4p + 5q = -17

4(2) + 5 (-5) = -17

8 -25 = -17

-17 = -17




6/22


3(2) - (-5) = 11

6 + 5 = 11

11 = 11

If you get equations that are true in both cases, then your solutions are correct!

For the example, we get these equations:

-17 = -17

and

11 = 11

So the solutions:p = 2, q = -5 are correct.

3.0 Substitution Method

The method of solving "by substitution" works by solving one of the equations (you

choose which one) for one of the variables (you choose which one), and then plugging

this back into the other equation, "substituting" for the chosen variable and solving forthe other. Then you back-solve for the first variable. Here is how it works.

2x 3y = 2

4x + y = 24



7/22


The idea here is to solve one of the equations for one of the variables, and plug this into

the other equation. It does not matter which equation or which variable you pick. There is

no right or wrong choice, the answer will be the same, regardless. But some choices may

be better than others.

4x +y = 24

y = 4x + 24

Now I'll plug this in ("substitute it") for "y" in the first equation, and solve forx:

2x 3(4x + 24) = 2

2x + 12x 72 = 2

14x = 70x = 5

Now I can plug thisx-value back into either equation, and solve fory. But since I already

have an expression for "y =", it will be simplest to just plug into this:

y = 4(5) + 24 = 20 + 24 = 4

Then the solution is (x,y) = (5, 4).



8/22


3.1 Procedure involved in solving the problem using the substitution method

Step 1 : Enter the linear equations to solve

Label the equations as follows:


equation 2 - 4p + 5q = -17

Step 2 : Solve one of the equations for one of the variable with a numerical

coefficient of 1, if possible, to try to avoid working with fraction.


-q = 11 - 3p

So, q = - 11 + 3p

Step 3 : Substitute the expression found in step 2 into the other equation,

yielding an equation, in terms of a single variable.

equation 2 - 4p + 5q = -17

q = - 11 + 3p 4p + 5(-11 +3p) = -17

4p 55 + 15p = -17

19p 55 = -17

19p = -17 + 55

19p = 38

p = 38 19

p = 2



9/22


Step 4 : Substitute the value found in step 3 into the equation rewritten in

step 2 and solve the other variable.


p = 2 3 (2) q = 11

6 q = 11

-q = 11-6

-q = 5

q = -5

Step 5 : Validate the solutions by substitution

You can test your answers to make sure they are correct. Type both solutions with a

comma between them:

For the example, type: p = 2, q = -5

State one of the original equations to make it the target data set:

equation 1 - 4p + 5q = -17

4(2) + 5 (-5) = -17

8 -25 = -17

-17 = -17


3(2) - (-5) = 11

6 + 5 = 11

11 = 11



10/22


4.0 Volume of a Prism

A rectangular prism is one type of polyhedron. The name of the prism is depends upon

the shape of its base. For example, if the prism has rectangular base then it is called

rectangular prism. If the prism has triangular base then it is called triangular prism. Thespace occupied by the prism is called volume of the prism. To solve the volume of prism

easily, formulas are used. The list of various kind of prism is shown in below.

1. Triangular Prism

2. Rectangular Prism

3. Cube

4. Pentagonal Prism

5. Hexagonal Prism

Types of prisms:



11/22


Formulas to Volume of Prism:

By using the formula we can solve volume of a prism.

1. Rectangle prism:

The volume of rectangular prism can be calculated by multiplying the length, width and

height of rectangular solid.

Volume of rectangular solid (V) = length x width x height Cubic units

= l x w x h cubic unit.

2. Triangular prism:

The volume of triangular prism triangular prism is half the product of length, width andheight if the prism.

Volume of triangular prism (V) = (length x width x height) cubic units

= (l x w h) cubic units.

3. Cube:

The volume of cube can be calculated by multiplying the length of three sides. (All sides

are equal in length)

Volume of cube (V) = a x a x a cubic units

= a3 cubic units

4. Pentagonal prism:

The volume of pentagonal prism is calculated by multiplying the area of the base and

height.

Volume of pentagonal prism (V) = A x h cubic units

A Area of the base.



12/22


5. Hexagonal prism:

The volume of hexagonal prism is calculated by multiplying the area of base of hexagon

and height of the hexagon.

Volume of hexagonal prism (V) = area of the base x height of prism cubic units

= A x H cubic units.

4.1 Steps involved in calculating the volume of prism

The volume of any type of mass can be measured by how much space it takes up.

Volume is most often measured by using some kind of container and taking stock of how

much it can hold. The volume of a prism is measured as a three dimensional object, with

mass, as explained in the following steps.

The formula for such an equation is: v=L x W x H.


Height (H)

Length (L)

Width

(W)


13/22


i) First method to calculating the volume of prism.

Step 1 - Find the length of the prism.

The length = 18 cm

Step 2 - Find the width of the prism.

The width = 12 cm


L

W


14/22


Step 3 - Find the height of the prism.

The height = 9 cm

Step 4 - Multiply the 3 dimensions together. It does not matter the order in

which you multiply the numbers.

The volume of triangular prism triangular prism is half the product of length, width and

height if the prism.

Volume of triangular prism (V) = (length x width x height) cubic units

= (l x w h) cubic units.

The volume of a prism = x L x W x H

= x 9cm x 12cm x 18 = 432cm


H

H

W

L


15/22


ii) Second method to calculating the volume of prism

Step 1 - Find the length of the prism.

The length = 18 cm

Step 2 - Find the width of the prism.

The width = 12 cm

Step 3 - Find the height of the prism.

The height = 9 cm


L

W

H


16/22


Step 4 - Find the area of the base face. Obviously, different shapes will need different

formulas.

Formula for the area = x H x W

x 9cm x 12cm = 54cm

Step 5

Multiply the area of the base with length.

Volume of prism = Base area x length

54cm x 18 cm = 432cm


H

W

L


17/22


5.0 Summary of students view of the methods explained

By using the two methods is the elimination and substitution, the teacher gives students

the freedom to choose which way is more easily understood. Teachers will act as

instructors and mentors to students to choose according to their rules. Therefore, to obtain

feedback methods which are preferred by students, a group of students was given an

evaluation form to be filled and the questions are linear and volume of prisms to be

completed by the students. Students are also required to give a little review or comment

on the form. (Form and the student answers, please refer to appendix)

Based on the assessment carried out to a group of students consisted of 10 students, most

of them prefer the method of substitution against elimination. This finding is reflected in

Table 1.1 below. Comments given by students to state substitution method is easier and

clearer to understand. This is because this method does not require a lot of operations and

the way it works very simple. Substitution method using the existing equations to solve a

given problem without the need to multiply or divide first linear equation.

Finding a formula for the volume of the prism is very important in mathematics. Students

are also exposed to this formula while in primary school. So this is easy for students to

master these skills the better. Based on the reviews of my students are very interested in

the two methods introduced by me to find the volume of prisms. Some of them are not

exposed to a variety of methods to find the volume of prisms. however, they are more

comfortable using their control of the primary school. It can be seen in Table 1.2 below

where all the students in my research using the first method for finding the volume of

prisms.

Two methods are given a real big help because it shows that students in solving

mathematical problems, especially these equations there are several ways and is not tied

to one way. Through research done on 10 students showed that both methods actually

help the students because they have their own advantages. According to their measures in



18/22


solving linear equations and find the volume of the prism is given by the teacher was very

effective and help them in this topic.

To find the volume of the prism it is much easier with the steps given by the teacher. By

following the steps given they have no problem to find the volume of prisms. Students

were shown how to determine the height, width and length are given in the form of a

prism, it was indirectly introduced to students the concept of the volume or prism.

Apparently the students more easily understand mathematical concepts to be included in

the calculation steps and systematic progression.

BIL NAMEMETHOD

ELIMINATION SUBSTITUTION1. Muhammad Amri bin Saleh /

2. Iqram bin Jazman /

3. Nur Aimy Mahirah bte Mohd Rais /

4. Shafiq Aiman bin Razali /

5. Ahmad Irfan bin Rahim /

6. Roshini a/p Kumaresan /

7. Nurul Izzati bte Khamis /

8. Syed Jaafar bin Syed Ali /

9. Rashika bte Rashid /

10. Hadif Danish bin Hanafi /

Table 1.1 Elimination and Substitution Method

BIL NAMEVOLUME OF PRISM

FIRST METHOD SECOND METHOD

1. Muhammad Amri bin Saleh /

2. Iqram bin Jazman /

3. Nur Aimy Mahirah bte Mohd Rais /

4. Shafiq Aiman bin Razali /

5. Ahmad Irfan bin Rahim /6. Roshini a/p Kumaresan /

7. Nurul Izzati bte Khamis /

8. Syed Jaafar bin Syed Ali /

9. Rashika bte Rashid /

10. Hadif Danish bin Hanafi /

Table 1.2 Volume of prism



19/22


6.0 Reflection on the method and steps

The first step to choosing a teaching method is to assess the students. This assessment can

be formal or informal. Formal assessments include standardized tests, tests from the

textbook orcurriculum being used, or teacher-created tests. These assessments can give

you an idea of the previous instruction that the students have received as well as their

academic level. The students in your class may have undergone various teaching methods

and quality of instruction in previous years.

As any good teacher knows, all students do not learn in the same way. In addition, it is

common for a class of students to be at a variety of levels in any particular subject.

Teachers need to use different teaching methods in order to reach all students effectively.A variety of teaching strategies, a knowledge of student levels, and an implementation of

which strategies are best for particular students can help teachers to know which teaching

methods will be most effective for their class.

They are usually not grouped by ability, but put in a group with children at a variety of

levels. The students are then given tasks to accomplish together. Teachers may need to

monitor these groups carefully, to make sure they are staying on task and that all students

are participating. This form of instruction also lends itself well to differentiation, because

the teacher can assign specific tasks to children at different ability levels.

Good teachers know their students. If you have been teaching a particular group of

students for some time, you probably already know quite a bit about their interests, ability

levels, and learning styles. If the group of students is new to you, you can make a point of

asking them, individually or in a group, about their interests and academic strengths.

Depending on the age of the children, they may also be able to write about this, or answer

some form of questionnaire about their hobbies, interests, previous instruction, strengths,

and weaknesses. Students generally enjoy talking about themselves and having their

teacher get to know them well, as it makes them feel special, as well as directing you in

choosing your teaching methods.

http://www.wisegeek.com/what-are-standardized-tests.htmhttp://www.wisegeek.com/what-is-curriculum.htmhttp://www.wisegeek.com/what-are-learning-styles.htmhttp://www.wisegeek.com/what-are-standardized-tests.htmhttp://www.wisegeek.com/what-is-curriculum.htmhttp://www.wisegeek.com/what-are-learning-styles.htm


20/22


As a teacher I was able to introduce some method to my students to help them develop

mathematical concepts are further substantiated. Students are not necessarily tied to just

one method to solve mathematical problems. This is because in mathematics, he has

several methods that can be used to solve a given problem. Students can choose which

way they are easier to understand and convenient to take as a way of calculating them.

Teachers can only act as mentors and demonstration only.

The process of calculating a more regular basis to comprise the simple steps and

accompanied with examples to facilitate students' understanding of how such

calculations. These measures must include instructions that are given clear and easily

understood by students. It must be shaped progression from easy to difficult to attract

students to find the answer to the problem. If a student is to understand the process ofteaching and learning well, they will be more motivated to learn mathematics. They will

be more excited to learn mathematics more broadly.

Increased professionalism in teaching comes from reflective practice. Teachers who think

meta cognitively about what works for them, and what needs improvement, become more

thoughtful in their day-to-day implementation of instructional strategies and classroom

management techniques. Approaching one's own teaching from an action research stance

allows teachers to objectively pinpoint areas that interest them and that need work. The

following methods are suggested for incorporating more systematic reflection into the

teaching day. Both new and veteran teachers become more effective when they regularly

reflect on their own practices and set personal goals for professional development.

http://www.suite101.com/content/action-research-strategies-for-k12-classrooms-a90193http://www.suite101.com/content/action-research-strategies-for-k12-classrooms-a90193


21/22


REFERENCE

http://www.suite101.com/content/reflective-teaching-strategies-for-more effective-k-8-

instruction

http://www.onlinemathlearning.com/volume-formula.html

http://www.sparknotes.com/testprep/books/sat2/math2c/chapter7section1.rhtml

http://www.tutorvista.com/math/linear-equations-elimination-method

http://www.suite101.com/content/reflective-teaching-strategies-for-more%20effective-k-8-instructionhttp://www.suite101.com/content/reflective-teaching-strategies-for-more%20effective-k-8-instructionhttp://www.onlinemathlearning.com/volume-formula.htmlhttp://www.sparknotes.com/testprep/books/sat2/math2c/chapter7section1.rhtmlhttp://www.suite101.com/content/reflective-teaching-strategies-for-more%20effective-k-8-instructionhttp://www.suite101.com/content/reflective-teaching-strategies-for-more%20effective-k-8-instructionhttp://www.onlinemathlearning.com/volume-formula.htmlhttp://www.sparknotes.com/testprep/books/sat2/math2c/chapter7section1.rhtml


22/22



Documents

Form 2 Assigment