O-Level Additional Math

[ADoc]

Herbie:

hi adot, thanks for ur explamation. can give example on the factorisation by group and factors? Can? Yq

Here are two trivial examples. You will be able to find plenty of worked examples from the lower sec textbooks and the school notes (if your child is from one of the IP schools).

By Group:

a + ac + b + bc = a(1+c) + b(1+c) [by factoring \"a\" & \"b\", we have created new \"grouped\" factor of (1+c)]

= (1+c)(a+b)
Factorisation by group usually involves at 3 unknowns.


\"Normal\" Factorisation
x + xy = x(1+y)

[Herbie]

Hi Adot,


Thanks for yr reply.
Can help to show the step in solving the qn below? tq

(7a+5b)(3c-5d)-(5d-3c)(2a+3b)

I have a qn on mode.

The no.of bags owned by a group of 9 students are 5,5,47,9,2,6 5 and 2.

State the no. of bags owned by a new member of the group such that there are now 2 modes for the group.

can explain what the qn meant by 2 modes??

[CoffeeCat]

Herbie:

I have a qn on mode.

The no.of bags owned by a group of 9 students are 5,5,47,9,2,6 5 and 2.

State the no. of bags owned by a new member of the group such that there are now 2 modes for the group.

can explain what the qn meant by 2 modes??

Currently there is only 1 mode, \"5\" which occured 3 times. The next candidate is \"2\" which occured 2 times. 2 modes just means there are 2 objects that occured the highest number of times (means they are sorta tied together for the winner). So the new member of the the group must owned \"2\" bags, so that now the frequency of \"5\" and \"2\" are 3.

[ADoc]

Herbie:

Hi Adot,


Thanks for yr reply.
Can help to show the step in solving the qn below? tq

(7a+5b)(3c-5d)-(5d-3c)(2a+3b)

Hi. This question is similar to the previous that you have posted. The technique is to \"switch\" the sign of (5d-3c) to become (3c-5d).

I will run through the first couple of steps again. I'll leave the rest for you to solve.

(7a+5b)(3c-5d)-(5d-3c)(2a+3b)
= (7a+5b)(3c-5d)-(-1)(-5d+3c)(2a+3b) [take note of the plus & minus signs]
= (7a+5b)(3c-5d)+(3c-5d)(2a+3b) [we can now proceed to extract the factor (3c-5d)]
= ... ...

cheers!

[ADoc]

Hi, I am no guru. Just wanna to share a little something with our aspiring Sec Ones. It is assumed that you are already comfortable with product of plus & minus, i.e. [plus] x [minus] = [minus], etc.


(a). Invisible Plus
Every number or unknown (for algebra) is always \"paired\" with either a positive or negative sign, such as (+)3, -10, etc. Notice that I wrote the + in parenthesis, because we don't, and it's not necessary, write positive as such. It's common knowledge that 3 means positive 3.

(b). What is the negative or minus sign?
Consider 10 - 5. It can mean two things: one, positive ten minus positive five, (+)10 - (+5); two, positive ten plus, negative five, (+)10 + (-5).

The \" - \" in case one is the mathematical operator for subtraction, whereas in case two, it is to denote that the number 5 is a negative instead of a positive number.

You must be wondering why I am going through these seemingly trivial concepts. The understanding of (a) & (b) will help you in your factorisation, as well as rationalising when and how you can manipulate the plus & minus sign when expanding and simplifying those hideous brackets. Foundation at this early stage is critical so you won't find yourself having difficulty in your subsequent years of algebra. Our teachers nowadays may not always appreciate the importance of making this distinction to our Sec1. There's only a short little para in the textbooks.

(c) The Invisible Negative 1
Having understood (b), it shouldn't be difficult to appreciate that, say -5 is made up of two very important factors, which are (-1) & (+5), i.e. (-1)(5). More importantly, any positive number can be re-written as a \"negative\".

5 = (-1)(-5)

Example 1
Trying reordering x - y such that y comes first on the left.
Again, it may sound trivial but please bear with me.

From above, we know every number or unknown is always paired with a sign. Hence whenever we manipulate an equation, we must always carry the signs together with the unknowns.

x - y = (+)x + (-)y = -y + x

Example 2
x - y = (-1)(-x) + (-1)(y) [factorising the invisible negative 1]
= (-1)(-x + y)
=-(-x + y) or -(y - x)

This technique is particular useful for Factorisation by Group

Example 3
Factorise 2a(b - c) + d(c - b)

2a(b - c) + d(c - b) = 2a(b - c) + d[(-1)(-c) + (-1)(b)]
= 2a(b - c) + d(-1)(-c + b) = 2a(b - c) - d(b - c) [+ve & -ve = -ve]
= (b - c)(2a - d)

Note that these steps aren't required in your workings. They are just for explanation.

Do leave a note if you think it's useful. And if it isn't, do post a reply saying so as well. No hard feelings at all... :xedfingers: Else I will continue to post a few more of such explanations on other operations. And that was Part [1]...boys & girls...

My students found these trivial explanations useful in guiding their algebra. Hope you'll find them useful too. Cheers! Basics are super important in order to breeze through the rest of your mathematics career for the next few years.

[Herbie]

HI ADoc,


Thanks for yr explanation. I find them very useful. Many thanks!

Can show some light on how to have a better understanding on question such as intersection, union and B’ etc?
Cos many a times we dun know which one to shade?

[ADoc]

Herbie:

HI ADoc,


Thanks for yr explanation. I find them very useful. Many thanks!

Can show some light on how to have a better understanding on question such as intersection, union and B' etc?
Cos many a times we dun know which one to shade?

Hi! I can only explain in words here but a Venn diagram (involving shading) would be most useful. Let me try to help you understand better. Other than shading, your child must be able to list the items in the sets as well. This isn't a S1 topic.

Firstly, let's straighten out a few definitions under the topic \"Set Language & Notation\" in a layman manner which, IMO, is easier to appreciate:

(1) Intersection
This means overlap or common area.

Example:
Think of sets as bags containing a number of items. They can be objects or numbers, etc. Let's use objects instead.

Set A or Bag A contains table, chair, student.
Set B or Bag B contains chair, blackboard.

A intersect B (AnB) = { chair } because chair is the only item that is common to BOTH set A & B. Hence we shade the overlap area ONLY.

(2) Union
This means everything that each and every set contains even if they don't overlap. Treat this as the \"mother of all sets\", not exactly so but let use this crude definition for now.

Example:
Using the above example, A union B (AUB) = everything in A plus everything in B = {table, chair, student, blackboard}.
There's no need to write chair twice. Hence we shade the entire Set A and B, even if they don't overlap.

(3) Complement Set
The complement of A is denoted by A' (reads A prime).
This means everything else other than A. So in terms of shading, we shade the area that is outside of A.

Let's do slightly more advanced examples:

(AnC)'
Tell your son to read this out in words if he gets confused, so that he can hear for himself what the question is asking.
We read as A intersect C, prime. This means everything other than the common area of A & C.

For such operations, a quick guide is to operate from inside out, just like the 4 orders of operations.
- brackets first (inside out)
- left to right

In terms of shading, to avoid confusion, [1] start by shading what is in the bracket. In this case, we shade the common area of A & C first, say diagonally left to right.
The next operation is \"prime\", [2] so we shade everything else, say using diagonally right to left. Erase the strokes in [1] and you have the required answer for (AnC)'.

Example:
(AnBnC)'
-brackets first: start by shading the common area of A, B & C.
-inside out: this is the next operation which is Prime. So we shade everything else that does not overlap the common area of A, B & C.


(AnBnC)' n (DnE)
-brackets first: start by shading the common area of A, B & C.
-inside out: this is the next operation which is Prime. So we shade everything else that does not overlap the common area of A, B & C.

-brackets first: Next shade the common area of D & E.
-left to right: now that we have two shaded areas, highlight or shade the area that is common to these two areas. And that's your answer.

Hope this isn't confusing. There are a number of interactive websites that can aid visual understanding. Here's one:

http://www.saskschools.ca/curr_content/mathb30/prob/les2/notes.html

cheers!

[Herbie]

HI Adoc,


Many thanks for yr explanation on union and intersection.

It is possible explain on the topic "Direct and Inverse Proportion’? Can?


Many thanks!

[ADoc]

Here’s a quick guide to understanding proportionality & solving formulating / solving proportion statements/equations (E Math Paper I)


By definition, a proportion is a mathematical statement or equation stating that two ratios are equal, i.e. a/b = c/d

But this is definition isn’t really useful at first glance to understand and solve proportionality.

Let’s understand these first:

(1) Direct Proportion
when we say A is proportional to B, we mean as A increases, B increases as well, vice versa.

Example: The more money I have, the more I-phone apps I can buy.

(2) Inverse Proportion
when we say A is inversely proportional to B, we mean as A increase, B decreases.
Or, as A decreases, B increases.

Example: The more time I spend playing on my I-phone, the lesser time I have for revision.

Now that we have sorted out definitions (1) & (2), next is about the proportionality constant (usually denoted as k).

Example:
y is directly proportional to x,
this means y ∝ x
to formulate a proportion statement, we rewrite as y = kx
Why is there a need for the proportionality constant k?
We only know that as y increases, x increases, however we do not know the exact magnitude of increment (or decrease). The constant k is the unknown that will allow us to equate y to x. So now we can happily remove the ∝ symbol and replace it by =

Example:
if y is inversely proportional to x³
we rewrite as y ∝ (1/x³). Convince yourself that as x increases, we are dividing by a larger number, hence y decreases. Thus an inverse proportional relationship.

similarly, we can arbitrarily insert the constant k to formulate an equation such that, y = k(1/x³)

Solving for K usually requires the question to provide a pair of values of x & y for example.

Example:
given that y = 2 when x = 4, and y is ∝ x
y = kx –> 2 = k(4) –> k = 1/2

given that y = 2 when x = 4, and y is ∝ 1/x
y = k(1/x) –> 2 = k(1/2) –> 2 = k/2 –> k = 4 (cross-multiplication is a common technique for solving inverses)

Hope these basics are useful. Cheers! There are many more variations and techniques, and styles of questions. I find it difficult to explain thru typing. Besides, the post will get incredibly long and boring! ha! Tks for the understanding.

[ADoc]

Just an add-on. The understanding of proportional is exceptionally useful for Physics (and other subjects of cos).


For example, the all-time favourite of D-S-T.
we know that D = S x T
with a constant S, the longer the D, the more T is required.
or, S = D / T, with a constant D, the longer T one takes to travel, it translate to a lower average S.

Density = Mass / Volume

For a constant M, the bigger the V, the lower the D, for example.

Pressure = Force / Area
The larger the A with a constant F, the lower the P.

Essentially, students must learn to appreciate the physics formula in this manner to be effective and efficient in their learning.