Dreaded algebra. It doesn’t have to be all bad! Algebra is simply the use of letters rather than numbers to help describe a value that can vary (a variable) or for a value that is simply not known. I hope this makes algebra seem less scary!

In this section we’re going to take a look at algebraic expressions. An expression is a set of terms that can be added, divided, multiplied or subtracted together. And an equation states that two expressions are equal in value.

When working out equations we are able to collect together like terms in order to simplify an expression.

For example, you may be asked to simplify an expression similar to the following:

7a + 4b + 2a – 3b.

7a +2a can be grouped as can 4b – 3b.

Simplified the expression is 9a – b.

Simplifying made simple!

Next up we’re going to look at expanding double brackets. If you see two brackets written next to each other it means you’re supposed to be multiplying them together.

To do this, follow the four steps of first, outer, inner and last! Not got a clue what’s happening, here’s an example to help you out:

Expand the brackets: (y + 4)(y + 2)

First: Multiply the first items in the brackets: y x y = y^{2}

Outer: Multiply the outer terms of the brackets: y x 2 = 2y

Inner: Multiply the inner terms of the brackets: 4 x y = 4y

Last: Multiply the last terms in the brackets: 4 x 2 = 8

And add the similar terms (4y and 2y) and you reach the answer: y^{2} + 6y +8.

That wasn’t so bad, was it?

Lastly we’re going to take a look at solving equations. This is often where a question will ask you to find the value of x or some other variable.

This can take on a few different forms so let’s work through a couple of examples to illustrate the different questions you may come across.

Find x in the following equation: ^{x}/_{2} = 3

Multiply both sides of the equation by 2. This gives x = 6.

Find x in the following equation: ^{2x}/_{3} = 4

Multiply both sides of the equation by 3 to get 2x = 12.

Divide both sides by 2 for the answer. x = 6

Find x in the equation: x + ^{x}/_{3} (x – 2)= 4 x = 3

Multiply both sides by 3 to get 3x + x = 12.

4x = 12 so x = 3.

Solve for x: ^{x}/_{2} = ^{2}/_{x}

Multiply both sides by 2x to get x^{2} = 4.

So, x = √4.

Remember to use the positive and negative roots. x = ±2

And just for good measure let’s finish off with a slightly tricker one!

Solve: x – 2y = 1, 4y^{2} – 3x^{2} = 1 (-1, -1) or (0, -1⁄2 )

From the first equation x = 2y + 1.

Substitute this into second: 4y^{2} – 3(2y + 1)^{2} = 1.

Expand, simplify and rearrange to get 4(2y^{2} + 3y + 1) = 0.

Factorise: 4(2y + 1)(y + 1) = 0 therefore y = –^{1}/_{2} or y = -1.

If y = –^{1}/_{2} x – (-1) = 1 so x = 0.

If y = -1 then according to first equation: x – (-2) = 1 so x = -1.

And if you can understand all of that you’re practically a maths genius!

So far that’s probably all been pretty similar to the stuff you studied at GCSE, however there are of course a few extra bits to be aware of!

The first is the use of proofs. A mathematical proof is a sequence of statements that follow on logically from each other, showing that something is always true.

Using letters to stand for numbers means that we can make statements about all numbers in general, rather than specific numbers in particular.

At A level we also need to take a look at factorising quadratics when the coefficient of x squared ≠ 1.

It sounds worse than it is, I promise!

We know that quadratic expressions can be written as: ax^{2} + bx + c.

Here’s an example to help you out when it comes to factorising quadratics!

You’re given the quadratic expression: 6x^{2} + 13x + 6. a = 6, b = 13 and c = 6. Firstly multiply the coefficient of x^{2} by the constant term, c.

6 × 6 = 36. Find two numbers which have a product of 36 and a sum of 13. These are 4 and 9 as 4 × 9 = 36 and 4 + 9 = 13.

Rewrite 13x as 4x + 9x as these are the two numbers found using the product and sum rule.

Factorise the first two terms and the last two terms separately. The bracket created should always be the same. The two brackets have now been found.

he first bracket is the common factor of (3x + 2). The second bracket is the factorised terms outside of each bracket (2x + 3).