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**With exams getting closer, we looked at the most-requested subject areas our 10,000+ tutors cover with their students.** This guide (also available to download as slides below) is a breakdown of those GCSE maths topics, together with step-by-step explanations and top tips from our maths tutors. It also makes a handy revision checklist for your students!

**Lowest common multiples and highest common factors****Expand and simplify equations in brackets****Converting decimals to fractions****Finding nth terms****Solving quadratic equations****Rearranging equations to find the subject****Prime factorisation****Intersections of a curve****Fractional indices (high level topic)****Rationalising denominators (high level topic)**

*Thanks to our fantastic maths tutors who contributed to this guide. *

**Top tip:** As there are different methods to approach these questions, check students fully understand the purpose of each step before they begin to tackle the question.

**Example question:**** What is the difference between lowest common multiples (LCM) and highest common factors (HCF)?**

**Answer and explanation:**

LCM stands for Lowest Common Multiple, and HCF stands for Highest Common Factor. The key to telling the difference between these two things is knowing the difference between a multiple and a factor. A multiple of an integer (whole number) is any integer that appears in its times table. For example, the multiples of 3 are 3, 6, 9, 12, and so on. A factor of an integer is any integer that divides the integer with no remainder. For example, the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. We use LCM and HCF to compare two (or more) integers. The LCM of two integers is the smallest whole number that appears in both of their times tables, that is, the smallest integer that is a multiple of both numbers.

For example, the LCM of 4 and 5 is 20. To see this, look at the multiples (times table) of 4:

4, 8, 12, 16, 20, 24, 28, …

and of 5:

5, 10, 15, 20, 25, …

The LCM is 20 because this is the first number that appears in both lists. The HCF of two integers is the largest whole number that divides both numbers without leaving a remainder. For example, the HCF of 16 and 24 is 8. Again, we can look at both sets of factors and compare. The factors of 16 are:

1, 2, 4, 8, 16

and the factors of 24 are:

1, 2, 3, 4, 6, 8, 12, 24.

Although 2 and 4 are common factors (that is, they appear in both lists), we are looking for the highest common factor. **The answer is 8.**

**Top tip: **Don’t forget the rules of ‘signs’ when multiplying the number outside the bracket with each term inside the bracket. E.g – 2X (4-3x)= (-2x*4) + (-2x*-3x).

**Example question: **** Expand and simplify 3(2x + 5) – 2(x – 4)**

**Answer and explanation:**

Firstly, to expand an equation like this, you must multiply the brackets by the number outside of the brackets. Make sure that you multiply every number inside the bracket by the number directly outside, and remember the signs:

3(2x + 5) – 2(x-4) becomes

(3 * 2x) + (3 * 5) + (-2 * x) + (-2 * -4) =

6x + 15 – 2x + 8 (remember that ‘-‘ * ‘-‘ = ‘+’)

Then you need to do something called ‘collecting the like terms’. This means collecting together all the ‘x’ terms and all of the ‘number’ terms, like this:

6x + 15 – 2x + 8 becomes

(6x – 2x) + (15 + 8), working this out means: **The answer is: 4x + 23.**

**Example question: ****How would you convert the decimal 0.125 into a fraction in its simplest form?**

**Answer and explanation:**

0.125 is the same as 125/1000. We have to remove the decimal point before creating a fraction. 0.125 multiplied by 1000 is 125 and at this point the decimal point has been removed, so 0.125 is equal to 125/1000. But 125/1000 is not the simplest form that the fraction could be in. If we divide both the top and the bottom by 5 since we know 125 is a multiple of 5, we get 25/200.

Divide both by 5 again to get 5/40.

Divide by 5 one more time to get 1/8, and since the numerator is 1, the fraction cannot be simplified any further.

**Therefore, the answer in its simplest form is 1/8.**

**Top tip:** It’s easy to forget the different methods for linear sequences and quadratic sequences. As it’s a quadratic sequence it needs extra steps!

**Example question: ****How do you find the nth term formula for a sequence with non-constant difference?**

**Answer and explanation:**

Take the sequence;

9, 12, 19, 30, …

(1) The first step is always to look at difference between the terms;

9, 12, 19, 30, …

+3, +7, +11, …

+4, +4, …

We can see the difference is not constant, (2) so we looked at the change in the difference each term. This gives a constant change in the difference of an extra +4 each term. The fact that we needed to take 2 turns to find the constant difference means we are dealing with a quadratic sequence. (3) Furthermore, because the difference is +4, we are dealing with a 2n2 sequence. If the change in the difference is (a) then the nth term follows a (1/2a)n2 pattern. (4) Now we can rewrite the sequence as follows;

n n2 2n2

9 1 1 2

12 2 4 8

19 3 9 18

30 4 16 32

(5) We need to find the difference between the sequence and 2n2.

2n2 d

9 2 -7

12 8 -4

19 18 -1

30 32 +2

(6) The difference here will either be a constant number, in which case the nth term is (1/2a)n2 +d. Or like this case, will itself follow a linear sequence with constant difference, which we should know how to solve.

1 2 3 4

-7, -4, -1, +2

+3 +3 +3

This gives 3n – 10.

**Therefore, the whole formula for the nth term is;**

**(7) 2n2 + 3n – 10**

**Example question: ****Solve the quadratic equation x^2 + 4x +1 = 0 by completing the square.**

**Answer and explanation:**

Completing the square means to put our equation into a slightly different form which looks like this, where a and b are real numbers:

(x+a)2 + b = 0

From here, we can rearrange the equation and directly solve for x. Let’s have a look at our specific example:

x2 +4x +1 = 0

The first step is to divide the coefficient of x by 2, and add this to x (this is our value of ‘a’ to go inside our bracket). We then square this value of a and subtract it outside the bracket.

In our example it will look like this:

(x+2)2 – 4 + 1 = 0

(x+2)2 – 3 = 0

We have our equation in completed square form. [There is a quick way to check we’ve got this right by expanding out this equation quickly:

(x+2)(x+2) – 3 = 0

x2 + 4x + 4 – 3 = 0

x2 + 4x +1 = 0

We’re back to our original equation, so we know we’ve got it right. Let’s go and solve our equation in completed square form.]

We simply rearrange for x:

(x+2)2 – 3 = 0

Add 3 to both sides.

(x+2)2 = 3

Take the square root of both sides. This splits into two possible cases:

Case 1: Positive square root of 3

x+2 = + sqrt(3)

x = – 2 + sqrt(3)

Case 2: Negative square root of 3

x+2 = – sqrt(3)

x = – 2 – sqrt(3)

So our final answer is…

x = – 2 + sqrt(3)

x = – 2 – sqrt(3)

**Top tip:** Don’t forget to reverse the processes when questions involve fractions and indices!

**Example question: ****How do you change the subject of the formula?**

**Answer and explanation:**

All that changing the subject of the formula means is basically getting a letter on its own on one side of the equation. To begin, let’s take a relatively simple example.

Make x the subject of the formula:

4y + 2 = x – 4

What this question basically means is get x on its own. At the moment on the right-hand side we have x – 4. So, because we want x on its own what we can do is add 4 to this side. However, we must then add 4 to the left-hand side to balance the equation (remember what we do to one side when changing the subject of the formula WE MUST DO TO THE OTHER SIDE). This then gives us the following:

4y + 2 +4 = x

So, our final answer is:

x = 4y + 6

We have successfully made x the subject of the formula. Now let’s try something a little bit more challenging.

**Make y the subject of the formula.**

2y/5 – 3x = 2

First, we must isolate the 2y/5. To do this we can add 3x to both sides to get rid of it on the left side and get 2y/5 on its own. This gives us:

2y/5 = 2 + 3x

Now, the 2y is divided by 5, therefore by doing the inverse function of division which is multiplication, we can get 2y on its own. So, we can multiply both sides by 5.

2y/5 * 5 = 5 * (2 + 3x)

2y = 5(2 + 3x)

It’s probably easier if these brackets are expanded so:

2y = 10 + 15x

**Top tip:** Remember to put brackets around 2 + 3x because both the 2 and the 3x are being multiplied by 5.

Finally, we must get y on its own. At the moment the y is being multiplied by 2. So, we must do the inverse function and divide 2y by 2. Remember to divide by 2 on both sides.

2y / 2 = (10 + 15x) / 2

y = (10 + 15x) / 2

This can be simplified to:

y = 5 + 15x/2

This is because 10/2 = 5 and 15×2 =is just 15x/2. We have now successfully made y the subject of the formula. The key things to remember when changing the subject of the formula are:

All it means is get a letter on one side of the equation on its own

- Use inverses functions (+ and – are inverses of each other and so are * and /) to isolate the letter

*Whatever you do to one side you must do to the other.*

**Top tip:** The tree/branch method is ideal to find all the prime factors of an integer. Once understood and mastered, this tree can be used for LCM and HCF questions also.

**Example question:**** Express 300 as a product of its prime factors**

**Answer and explanation:**

Factors of a number are: numbers that can be multiplied (by a whole number/ integer) to make the original number. For example, 6 is a factor of 42, because 6 can be multiplied by 7 to make 42. Prime numbers are numbers that cannot be divided by anything except themselves and 1 to give a whole number answer; for example 3 and 11. Each time you find a prime factor, circle it in your working, so you do not end up missing one out in your final answer – in this example I underline each prime factor I find.

In order to split 300 into its prime factors, first of all think of a number than goes into 300:

-Let us start with 30. 30 goes into 300 at total of 10 times, as 30×10=300.

-So, we now have two factors of 300: 30 and 10. Now, we must find the factors of these two numbers:

- Factors of 10: 5 x 2 = 10. As the numbers 5 and 2 are both prime numbers, they are counted as prime factors for 300.

-Factors of 30: 3 x 10 = 30. As 3 is a prime number, we have found another prime factor of 300.

However, 10 is not a prime factor, and so we must keep going: 5 x 2 = 10, and therefore we have found another two prime factors: 5 and 2. Looking back over your working, write down every number that is circled (or in this case, underlined): 5, 2, 3, 5, 2. So, we can write 300 as a product (multiplication) of its prime factors by multiplying them all: 2 x 2 x 3 x 5 x 5

**This can be rewritten as 22 x 3 x 52.**

**Top tip:** Remember the difference between linear and quadratic equations!

**Example question:**** How do I find the intersection of a line and a curve?**

**Answer and explanation:**

Say you are given the equations of both a line and a curve, for example y=x2+8x-1 and y=3x-7, and asked to find where these two intersect. This just means where the two lines would cross or touch if drawn on the same graph. To find these points you simply have to equate the equations of the two lines, where they equal each other must be the points of intersection.

For this example this would mean x2+8x-1=3x-7

Collecting like terms leads to x2+5x+6=0

And from then this is a simple case of solving the quadratic. This expression factorises to (x+2)(x+3)=0 which implies either x=-2 or x=-3. To find the corresponding y coordinates for each point simply input these x values one at a time into either one of the original equations.

For x=-2, we get y=3(-2)-7=-13 so the point is (-2,-13)

For x=-3, we get y=3(-3)-7=-16 so the point is (-3,-16)

Both of these are valid intersection points for the line and curve given.

**Top tip:** These are the more difficult questions in the GCSE exams and students often confuse whether one roots the number by the numerator or the denominator- a technique would be to reassure them that the numbers would be simple, an integer and easy to work with, especially in non-calculator papers. So, if they need to root 4 by 3, they are most likely to be using the numbers the wrong way around.

**Example question: How do you calculate a number to the power of a fraction? (8^2/3)**

**Answer and explanation:**

When a number is raised to the power of a fraction, it is the same as rooting that number to the denominator (the number on the bottom). So if the number is raised to ½ it is the same as √ and if it is raised to ⅓ it is the same as ³√. The situation becomes more complicated when the number is raised to another power such as 2/3 but there is a simple solution to working out this problem. If we take the example 8^2/3 we can make it much easier to solve by changing the power to 1/3 by putting the whole thing in brackets with the 2 outside giving us (8^1/3)^2.

We can start off by solving what is inside the brackets which is 8^1/3. We already know that when something is to the power of ⅓ it is the same as cube root so ³√8 = 2. This leaves us with 2^2 to solve which is 4.

**Therefore 8^2/3 = 4**

**Top tip:** Practice makes perfect! These higher-level questions can be tough, but with enough practice of the methodology students will learn exactly what to do.

**Example question: ****Rationalising the denominators (Surds)**

**Answer and explanation:**

When asked to rationalise simple Surd (square roots that cannot be reduced to a whole number) fractions in the form a/√b we are aiming to remove the surd in the denominator (bottom).

e.g 1. Rationalise 3/√2

Answer: We multiply the entire fraction by the denominator √2/√2 (this is equivalent to 1). Let us first consider what happens to the denominator:

√2 x √2 = 2 (any simple surd multiplied by itself equals the number inside)

The numerator (top) becomes 3 x √2 or 3√2

So the fraction rationalises to 3√2 / 2, the surd has now been removed from the denominator. Typically you will be asked to simplify the fraction which is just asking you to rationalise it. The rationalised fraction can be used more easily in further calculations.

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