# The Ultimate Guide To Algebra

Algebra, the area of Maths that uses:
NO NUMBERS!

But how does maths work with no numbers? Well, in place of numbers, it uses letters. x and y instead of 1 and 2. So how does one go about this? Well, first you have to know various rules...

## BIDMAS

BIDMAS is the order that operations are done: Brackets, Indices, Division, Multiplication, Addition, Subtraction. Operations inside brackets also happen in the order of BIDMAS.
Writing Algebraic Terms

It is important to remember that x is often used so writing x in order to mean "multiplied by" would be very confusing so axb is written as ab.

## The Laws Of Indices

You've probably seen $\inline x^{2}$ written down before but what does it mean? Well it means $\inline xx$. $\inline ^{2}$ is what's called an index (plural indices). $\inline ^{3}$($\inline xxx$) is also an index but they aren't the only indices. Any number or letter can be an index. A number like 2 always has index of 1. However there are certain rules that control the use of indices:
$\inline x^{a}x^{b}=x^{a+b}$
$\inline \frac{x^{a}}{x^{b}}=x^{a-b}$
$\inline \left ( x^{a} \right )^{b}=x^{ab}$
$\inline x^{-a}=\frac{1}{x^{a}}$
[[image:http://latex.codecogs.com/gif.latex?%5Cinline%20x%5E%7B%5Cfrac%7B1%7D%7Ba%7D%7D=%5Csqrt[a]%7Bx%7D]]
$\inline x^0=1$
So, what do you think of the laws of indices? Well, they're going to be pretty useful later on.

## Basic Equations

Before working out any equations, you have to know that WHATEVER YOU DO TO ONE SIDE, YOU HAVE TO DO TO THE OTHER SIDE AS WELL!
How would you work out what $\inline x$ is if you're given the statement $\inline 2x=10$? Well, you'd divide by 2 to get $\inline x=5$.
How about $\inline 2x-3=7$. Add 3 and you're back to $\inline 2x=10$. Divide by 2 and you once again have $\inline x=5$.
How about $\inline \frac{2x-3}{7}=1$? Multiply by 7 and you're back at $\inline 2x-3=7$. Etc. etc.

Simplifying Equations

$\inline x+x+2+4+y+3$. Quite complicated, isn't it? But it doesn't have to be. You could add all the like terms together ($\inline x$s, $\inline y$s and numbers). This would give you
$\inline 2x+y+9$. Remember though, you can't simplify something like $\inline x^{2}+x+2y+3$ because none of them are like terms.

## Intermediate Equations

What if you had $\inline x$ in both sides? $\inline 3x+2=x+6$ for example. What you have to do is get all the letters on one side and the numbers on the other. So what you'd do is subtract $\inline x$ to get $\inline 2x+2=6$, then subtract 2 to get $\inline 2x=4$. From here, divide by 2 to get $\inline x=2$.
But hold on. What if its something like $\inline -x+12=2-3x$? Well then you subtract 2 ($\inline -x+10=-3x$), add $\inline x$ ($\inline 10=-2x$) and divide by -2 to get $\inline -5=x$. However, it is better to have $\inline x$ on the left, so you switch the sides to get $\inline x=-5$. Problem solved.

Next, what if you have a fraction with $\inline x$ in both sides? What if you had something like $\inline \frac{x^{2}}{2x}=3x+4$? Well first you have to get rid of the fraction, so multiply by $\inline 2x$ and you'll have
$\inline x^{2}=6x^{2}+8x$. Now, you would divide by $\inline x$ and have $\inline x=6x+8$. Now you can subtract $\inline 6x$ and be left with $\inline -5x=8$ divide by -5 and you'll have $\inline x=-\frac{8}{5}$.

## Expansion

What is $\inline 2(x+3)$ without the brackets? $\inline 2x+3$? No, it's $\inline 2x+6$ because you have to multiply everything inside the brackets by the thing you're multiplying the brackets by. So how about $\inline x(x+y+3)$? You'd get $\inline x^{2}+xy+3x$. How about $\inline 4x(x^{2}+3y+\sqrt{z})$? You'd get $\inline 4x^{3}+12xy+4x\sqrt{z}$.

## Factorisation

Factorising is the opposite of expanding. When you factorise, you put the terms into brackets. So, imagine you had to factorise $\inline 2x^{2}+8xy+2x$. How would you do it? Well, first, you have to find the highest common factor of all of those terms, so for this it is $\inline 2x$. Now you divide all of the terms by that factor: $\inline x+4y+1$. Now you put brackets around that and write next to it the factor: $\inline 2x(x+4y+1)$.

## Basic Simultaneous Equations

$\inline 2x+3y=10$
$\inline x+y=4$
How would you work out what $\inline x$ and $\inline y$ are? First, you have to work out what one letter is in terms of the other one. So, in this case we can rearrange the second equation to get $\inline x=4-y$. Now, we can substitute that into the first equation to get $\inline 2(4-y)+3y=10$. Expand the brackets and you'll have $\inline 8-2y+3y=10$. This simplifies down to $\inline y+8=10$. From here you subtract 8 and you'll have $\inline y=2$. Now, you can use that to work out $\inline x$ by replacing $\inline y$ in the first equation with 2: $\inline 2x+6=10$. Surely, you can work that out: $\inline x=2$. So now, you have:
$\inline x=2$
$\inline y=2$

## Basic Graphs

$\inline y=2$. Fairly simple: it's an equation. But it can also be used to plot a line on a graph. This line, for example is like so:

Anything that is $\inline y=n$ is a horizontal line, just like $\inline x=n$ is a vertical line. But what about a diagonal line? Well that is $\inline y=mx+c$. Here, $\inline m$ is the gradient of the line and $\inline c$
is the point at which the line intersects the $\inline y$ axis. So $\inline y=2x+1$ looks like this:

## Intermediate Expansion

What if you had to do something like $\inline (x+2)(y+3)$? For this, you have to use FOIL (First, Outside, Inside, Last) . This means that you multiply the first term in each bracket, the two outside terms, the two inside terms and then the last term in each bracket. This would give you $\inline xy+3x+2y+6$. Now, obviously, you can't simplify that, but often, you can simplify the result, such as $(x+12)(x+3)=x^{2}+3x+12x+36=x^{2}+15x+36$.

If you have three sets of brackets, e.g.$(x^{2}+12)(3y+16)(\sqrt{z}+20)$, you expand the first two sets of brackets first. This gives you $3x^{2}y+16x^{2}+36y+192$. Now you put this in brackets and multiply it by the third set of brackets, so $(3x^{2}y+16x^{2}+36y+192)(\sqrt{z}+20)$. But wait, you can't use FOIL here. Just multiply each term in the first set of brackets by both terms in the second set of brackets:$3x^{2}y\sqrt{z}+60x^{2}y+16x^{2}\sqrt{z}+320x^{2}+36y\sqrt{z}+720y+192\sqrt{z}+3840$.

## Intermediate Factorisation

$4x^{2}+8x+4y$. Obviously, this can be factorised or else it wouldn't be in this section. This factorises into $4(x^{2}+2x+y)$. But do you see the problem here? It can be factorised again. Look at what's inside the brackets. You can't factorise all of it, but you can get$4(x(x+2)+y)$. Always watch out for things like that.

$\frac{1}{x}=\frac{x^{2}}{5}$. Oh no! Fractions in both sides: what do I do?! Cross-multiply. What's that? It's when you multiply the denominators together and the multiply both sides by the result, so in this case you'd multiply by $5x$ and get $x^{3}=5$.Now cube root and get [[image:http://latex.codecogs.com/gif.latex?x=%5Csqrt[3]%7B5%7D]].

## Basic Rearranging Equations

What is you had $v=u+at$ and were told to rearrange it? What is rearranging an equation? Rearranging an equation is when you change the subject of an equation. So, in this case $v$is the subject but in the example question you need to make$u$the subject. You need to change it to $u=at-v$ so that the equation is equal.
Now, how about rearranging $s=ut+\frac{1}{2}at^{2}$to make $a$the subject. You would need to change it to $a=\frac{s-ut}{\frac{1}{2}t^{2}}$. Remember, the key to rearranging equations is to keep the equation equal.

$(x+y)^{2}$. Expand. The answer is:
$x^{2}+y^{2}$. No. You need to think of it as
$(x+y)(x+y)$. This means that using FOIL you can get
$x^{2}+xy+xy+y^{2}$ which simplifies to become:
$x^{2}+2xy+y^{2}$.