I study General Engineering at the University of Durham. I have always enjoyed the maths and science subjects, and hope to help students who also share my love for them too! I have experience in working with young pupils and tutoring students in their GCSEs and A Levels.

I study General Engineering at the University of Durham. I have always enjoyed the maths and science subjects, and hope to help students who also share my love for them too! I have experience in working with young pupils and tutoring students in their GCSEs and A Levels.

Before a session, you can let me know specific topics/areas you need help with, and I will prepare a session based on what **you** would like to cover. This can be starting from scratch, or usual revision and practice, or even exam questions and techniques! We will start by focussing on understanding the concepts, until you are confident enough to move on to answer exam questions, and we can learn to revise using visual aids, diagrams, mnemonics, brainstorms, and more! **If you have any questions, send me a 'WebMail' or book a 'Meet the Tutor Session'. Please remember to tell me what exam board you're on and what you are struggling with.**

Before a session, you can let me know specific topics/areas you need help with, and I will prepare a session based on what **you** would like to cover. This can be starting from scratch, or usual revision and practice, or even exam questions and techniques! We will start by focussing on understanding the concepts, until you are confident enough to move on to answer exam questions, and we can learn to revise using visual aids, diagrams, mnemonics, brainstorms, and more! **If you have any questions, send me a 'WebMail' or book a 'Meet the Tutor Session'. Please remember to tell me what exam board you're on and what you are struggling with.**

Enhanced DBS Check

22/12/20165from 131 customer reviews

Sufi (Parent from London)

February 27 2018

Really helpful. Thanks a lot!

Sufi (Parent from London)

February 20 2018

Your really encouraging I feel more enthusiastic and I feel like It is easy if you try especially after this lesson. Thank you Sidra Sufi

Sufi (Parent from London)

February 13 2018

Great session! Really helped me with circumference,Radius and Diameter. Thanks a lot!

Sufi (Parent from London)

February 6 2018

Great session! Really helped me with prime number percentages Thanks a lot! Sidra

This is an algebraic fraction. There is more than one way of solving this expression, but the simplest is to factorise both the top and the bottom quadratic expressions.

Firstly we shall simplify the top expression, 3x^2-x-2. This means putting it in the form (ax + b)(cx + d) where a, b, c, and d are constants.

Now, the only way to get 3x^2, you can see that a multipled by c must equal to 3. Therefore, (3x + b)(x + d). Now you can also see that b multiplied by d must be -2, but also, 3 multipled by d plus b multiplied by x must also be -1. Therefore it can be seen that b is 2, and d is -1. So now the top is (3x + 2)(x -1).

Next, we factorise the bottom of the fraction. At first glance, it looks like it is the simplest it can be. However, x^2 - 1 is actually the difference of two squares. This is (x + 1)(x - 1).

Now, the full factorised expression becomes (3x + 2)(x - 1)/(x + 1)(x - 1). As (x - 1) is on the top and the bottom, this is equal to 1, and can be cancelled out.

So now we have the answer! (3x + 2)/(x + 1)

This is an algebraic fraction. There is more than one way of solving this expression, but the simplest is to factorise both the top and the bottom quadratic expressions.

Firstly we shall simplify the top expression, 3x^2-x-2. This means putting it in the form (ax + b)(cx + d) where a, b, c, and d are constants.

Now, the only way to get 3x^2, you can see that a multipled by c must equal to 3. Therefore, (3x + b)(x + d). Now you can also see that b multiplied by d must be -2, but also, 3 multipled by d plus b multiplied by x must also be -1. Therefore it can be seen that b is 2, and d is -1. So now the top is (3x + 2)(x -1).

Next, we factorise the bottom of the fraction. At first glance, it looks like it is the simplest it can be. However, x^2 - 1 is actually the difference of two squares. This is (x + 1)(x - 1).

Now, the full factorised expression becomes (3x + 2)(x - 1)/(x + 1)(x - 1). As (x - 1) is on the top and the bottom, this is equal to 1, and can be cancelled out.

So now we have the answer! (3x + 2)/(x + 1)

This question is about rearranging the equation so that we have all the unknown values on one side, and all the known values on the other side.

To start, we want to put all the x values on one side, and all the known constants on the other side.

First, to get all the x values on one side, we must subtract x from both sides of the equation (so that the equation is still correct, you must remember to do everything to both sides!). Therefore we get 4x + 4 = 14.

Now we want the known constants on the right hand side of the equation, so we subtract 4 from both sides. Now we get 4x = 10.

Now to find x on its own, we divide both sides of the equation by 4. So we get the answer to be x = 2.5!

This question is about rearranging the equation so that we have all the unknown values on one side, and all the known values on the other side.

To start, we want to put all the x values on one side, and all the known constants on the other side.

First, to get all the x values on one side, we must subtract x from both sides of the equation (so that the equation is still correct, you must remember to do everything to both sides!). Therefore we get 4x + 4 = 14.

Now we want the known constants on the right hand side of the equation, so we subtract 4 from both sides. Now we get 4x = 10.

Now to find x on its own, we divide both sides of the equation by 4. So we get the answer to be x = 2.5!

This is an example of differentiation. This can be useful in many concepts, one being finding the gradient of a line or curve at a certain point. To differentiate these types of equations, the rule is to multiply the front by the power and to take one from the power!

y = x^2 - 5x

We will take each part separately. Starting with x^2. We multiply the front (which is 1) by the power (which is 2), therefore the constant at the front is now 2. We take one from the power, so 2 - 1 = 1. Therefore the derivative of x^2 is 2x.

Next we take 5x. Multiply the front (5) by the power (1), and take 1 from the power (1 - 1 = 0). Therefore the derivative of 5x is 5.

Now, we put it all together! dy/dy = 2x - 5!

This is an example of differentiation. This can be useful in many concepts, one being finding the gradient of a line or curve at a certain point. To differentiate these types of equations, the rule is to multiply the front by the power and to take one from the power!

y = x^2 - 5x

We will take each part separately. Starting with x^2. We multiply the front (which is 1) by the power (which is 2), therefore the constant at the front is now 2. We take one from the power, so 2 - 1 = 1. Therefore the derivative of x^2 is 2x.

Next we take 5x. Multiply the front (5) by the power (1), and take 1 from the power (1 - 1 = 0). Therefore the derivative of 5x is 5.

Now, we put it all together! dy/dy = 2x - 5!