Algebra Basics
In this video, we are going to be starting with the Algebra Basics.
You will be learning how positive and negative unknowns interact with each other, be it adding them, subtracting them, dividing, or multiplying them.
We’ll also be looking at some basic terminology and how to clean things up when you’ve got a page full of unknowns. This is graded as level 3, or between a D and an E for us older people.
Terminology
First of all, let’s first go over some basic terminology – when we refer to unknowns, all it means is there is an unknown number. The various things we do with these unknowns are essentially all that Algebra is.
These unknown numbers are expressed using a letter or a shape. The most common letters are ‘\(x \)’ or ‘\(y \)’ but it can really be anything. We’ll be using ‘\(x \)’ just to keep things simple.
It’s really important to note that when you’re looking at a problem involving an ‘\(x \)’, the value of that \(x \) will be constant – this means that if \(x \) represents 3, every single \(x \) in the given question would be \(x \). \(x \) cannot be given different values in the same question.
Adding, Subtracting, Multiplying and Dividing
So now we’ve got the introduction out of the way, let’s start with the basics:
As we have already mentioned, unknowns are numbers. This means that we treat them exactly the same as any other number.
For example, if we add 2 of the same positive number together, we would get two times that number. If we add an ‘\(x \)’ to an ‘\(x \)’, we get 2\(x \).
The same happens if you subtract – \(x \) from 2\(x \), you’re just left with \(x \).
Division and Multiplication follow the same process, just as if you were to multiply a positive with a negative, you’d end up with a negative. If you were to multiply \(x \) by -1, you’d get -\(x \).
But what happens if I divide or multiply unknowns against each other I hear you scream – we just need to follow the same logic as adding and subtracting.
As \(x \) represents 1 number, if you were to multiply \(x \) with \(x \) then you would get \(x^2 \) just like if you were to multiply any number by itself, it would be that number… squared.
For division, we know that if you divide any number by itself, you will always have an answer of 1. This is the same for \(x \). \(x \) divided by \(x \) will always be 1.
If we were take things to the next level and were to divide \(x^2 \) by \(x \), we could write this out as \(\frac{x \times x}{x}\) which will just give us \(x \). This is because division is the opposite of multiplication and the last 2 \(x \)’s cancel each other out.
Terms
In Algebra, you will often have expressions that will have a variety of different variations of an unknown – such as \(x^2 + x + 4 \). Each component is called a ‘Term’. So the first part is the ‘\(x^2 \)’ term, the middle is the ‘\(x \)’ term and the final part is the ‘number’ term.
This is some important terminology for what we are going to cover in this final part of the video.
Collecting the Terms
Something called collecting the terms. You do this to keep everything neat and tidy. A good way to think about this is if you were to have a big box full of cutlery. Your forks would be your \(x^2 \)s, your knives would be your \(x \)s, and your numbers would be your spoons.
If you had a drawer with everything all mixed up, it would be really hard to find how many forks you had or how many spoons you had. But if we were to collect all of the ‘like terms’ up, this will make our lives so much easier.
So let’s have a quick look at a worked example.
\(x^2 + 3x -2 +4x^2 + 4-2x^2 \)
Just looking at that expression looks a bit daunting but as we collect the terms up, you’ll see it’s not really that bad.
So first we tackle the ‘\(x^2 \)’ terms – we’ve got a \(x^2 \) which can be seen as \(1x^2 \), a \(4x^2 \) and a \(-2x^2 \). All we simply do is \(1 + 4 – 2 = 3 \). Therefore we can rewrite our expression as \(3x^2 + 3x – 2 – 4 \) – that’s looking a little bit nicer.
We then look at our ‘\(x \)’ terms – there is only 1 term here which is \(3x \) so we don’t need to do anything here.
Finally we move onto our ‘numbers’ terms – we have \(-2 +4 \). This is simply 2. So once we put all of this together, our final expression is \(3x^2 + 3x + 2 \).
If you’re feeling happy with this – go ahead and try out some questions from our question practice section below.