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‘A’ is for Algebra

I asked my good friend, Dr David Perkins, to contribute a post called ‘A’ is for Algebra. This is an introduction to to algebra. His Bio Details are at the end of the post.

“If we are to believe television sitcoms, algebra is a branch of mathematics to be feared; to understand it is to unlock the intricacies and mysteries of a topic comprehended only by an elite few. The subject, for some reason, holds a special place in the lexicon of script writers as the go-to topic when something in their plot is difficult to understand – they liken it to learning algebra. However I would like to spend the following few paragraphs debunking the myth that it is a difficult subject.

First some history. Algebra comes from the Arabic word al-jabr meaning the reunion of broken parts. The word appears in the title of the first known book on the topic written in Baghdad about 825 C.E. by the Arab mathematician Mohammed ibn-Musa al-Khowarizmi titled Hidab al-jabr wal-muqubala which means “The Book of Restoration and Balancing.” In this book, al-Khowarizimi describes a method for piecing together the value of an unknown quantity using some available information.

The concepts are deceptively simple, and easy to understand if some basic principles are understood first.
To understand algebra there are two basic concepts that need to be grasped. The first is very simple:
Let’s start with a simple concept: the equals sign.

2 = 2
It is difficult to disagree with the above mathematical statement, because it is self-evident. Two equals two. But once that truth has been established there are certain things I am not allowed to do. For example I cannot add three to one side, because it renders the equals sign as no longer true.
2 + 3 = 2.
This statement is not true, (partly because it isn’t true – but mostly because we had already established the equality of both sides, and then did something to one side which we didn’t do to the other. If I add three to one side I must add it to the other. This is the first and most fundamental rule of algebra. What I do to one side of an equation I MUST ALSO DO TO THE OTHER.
2+3 = 2+3
This is now true again, because when I added 3 to the left hand side I also added 3 to the right hand side.
OK. We will come back to this point later, but I’ll sum it up here so we can use it again later.

To understand our second rule I am going to describe a scenario. (Although this scenario involves a casino, in no way do I wish to condone gambling, I simply find it a very concrete example of what I wish to illustrate.

The scenario:
A man walks into a casino and makes a bet which triples all his money. With his next bet he loses $15. With his third bet he doubles his money. With his 4th bet he wins $10. With his 5th bet he halves his money. If he ends up with $50, how much money did he start with?
Most people will realise that to solve this problem, we have to work our way backwards from the end, doing each step in reverse. So we might ask, if he ended with $50 by halving his money, we need to double that $50 to find out what he had before he halved his money. (The opposite of halving is doubling – so we needed to double his money going backwards)
This means he had $100 before he made his last bet.
But he won $10 to have $100 and we need to take that $10 away to find out how much he had before he won the $10. (The opposite of adding 10 is subtracting 10.)
This means he had $90 before he won the $10.

Now he had doubled his money to get to the point of having $90. So we need to halve that to figure out how much he had before he doubled his money. (The opposite of doubling is dividing by two – or halving). As you can see we are simply working backwards through the story and doing the opposite of what actually occurred.
This means he had $45 before he doubled his money.
But he lost $15 to have the $45. So we need to add $15 to the $45 (the opposite of subtracting $15 is adding $15.)
This means he had $60 before he lost the $15.

Finally (or rather, initially) he tripled his money to get the $60, so we need to divide 60 by 3 to find how much he started with. (The opposite of multiplying by 3 is dividing by 3).
So he started with $20.
Algebra works in exactly the same way, and in a moment we will solve the exact same scenario using algebra, but the equations will look a little complicated at first so we’ll work up to it. But first we have our second rule of algebra.

Rule 1.
And now
In an algebraic situation the unknown number (in the above example, it’s the amount he started with) is represented by some symbol (usually the letter x) and we simply work backwards with opposite operations to find that unknown from an equation.

A couple of house-keeping issues. In algebra when a letter is multiplied by a number – say for example, x is multiplied by 3 – it is written as 3x. This means 3 lots of x. It saves some time, stops confusing x with the multiplication symbol, x, and is how we would say it aloud. We don’t say “Can I have three times apples please?” We simply ask for 3 apples.

OK, now we are ready to solve some algebraic equations. A number is multiplied by 4 and the answer is 12. We should realise fairly readily that to find the answer we need to divide 12 by 4. We find the number we multiplied was 3.
Let’s use the two algebra rules to do the same problem. We’ll use x for our unknown number.
4x = 12
(This is the algebraic equivalent of: A number is multiplied by 4 and the answer is 12.)
Algebraically we have to work backwards, doing the opposite of what was done, well x was multiplied by 4, and we need to do the opposite. So let’s divide by 4.
4x ÷ 4 = 12

Now 4 lots of x divided by 4 is just x or 1x but we don’t write the 1, as it is just x but we’ve broken our first rule and done something to one side that we didn’t do to the other. The sides are no longer equal. So I must also divide the other side by 4.
4x ÷ 4 = 12 ÷ 4

Since we already know the left hand side is just x and the right hand side is 3 we get
x = 3
Let’s do one more then tackle our gambling problem 
3x + 2 = 17
So now we have a number that is multiplied by 3 then has 2 added and the result is 17. So first we need to take the 2 away as that added last, then we need to divide by 3 as that was done first. But we need to do these things to both sides.
3x + 2 – 2 = 17 – 2
This gives
3x = 15
Now we divide BOTH SIDES by three:
3x ÷ 3 = 15 ÷ 5
And we get:
x = 5

OK now to tackle the gambler’s problem. Remember He had some money to start with x And it was tripled. Which is represented as 3x. He then lost $15, so our equation becomes 3x – 15. The entire amount was doubled. Which means our equation becomes 2(3x – 15). He then won $10 which leads to 2(3x – 15) + 10. All of this was halved. Which becomes (2(3x – 15) + 10) ÷ 2 and the final amount was $50
(2(3x – 15) + 10) ÷ 2 = 50
Now we solve it
Multiply both sides by 2
(2(3x – 15) + 10) ÷ 2 x 2 = 50 x 2
2(3x – 15) + 10 = 100
Subtract 10 from both sides
2(3x – 15) + 10 – 10 = 100 – 10
2(3x – 15) = 90
Divide both sides by 2
2(3x – 15) ÷ 2 = 90 ÷ 2
3x – 15 = 45
Add 15 to both sides
3x – 15 + 15 = 45 + 15
3x = 60
Divide both sides by 3
3x ÷ 3 = 60 ÷ 3
x = 20

If you look at the right hand sides all the way down the answer undergoes the same evolution as it did in our worded version a few paragraphs earlier.
That is essentially the basics of algebra in a nutshell. Remember those two basic rules: I must always do to one side what I did to the other or my equals sign will no longer be true, and I must always work in reverse order using the opposite operation to get the unknown on its own.” And thank you David. I am sure parents and students will find it useful.

Bio: Dr David Perkins is a teacher of mathematics, physics and chemistry at Kelvin Grove State College in Brisbane. He has previously worked with the National Institute of Biomolecular Sciences in conjunction with the Victor Chang Heart Research Institute designing molecules to mitigate heart pump function for reducing blood pressure. Prior to that he gained his PhD in supramolecular architecture from the University of Sydney designing molecular cage compounds that could potentially act as molecular switches and logic gates in molecular computer components in roughly 200 years’ time. But he didn’t really want to wait 200 years, so now he teaches others so the process can be accelerated.