Hello, I’m Harry and I’m currently in my final year reading Mathematics at Durham University and have achieved a First overall so far. I received 3 A*s at A-Level in Maths, Further Maths and Physics.

__My approach__

I have loved helping students with their maths ever since I volunteered as a mentor when I was in the sixth form at Highgate School, regularly helping pupils with their maths problems. I enjoy the challenge of explaining concepts, finding numerous different ways to get them across in a friendly and engaging manner. Above all, I am patient, allowing students to achieve their goals at the best pace for them.

__Sessions__

I am flexible with regard to the session content, allowing students to present particular problems, revise topics or go through past papers.

__What next?__

Please contact me with any questions using the ‘Webmail’ on this site. We can arrange a free 15 minute ‘Meet-the-Tutor’ session to establish what problems you are having and how I can help you.

Hello, I’m Harry and I’m currently in my final year reading Mathematics at Durham University and have achieved a First overall so far. I received 3 A*s at A-Level in Maths, Further Maths and Physics.

__My approach__

I have loved helping students with their maths ever since I volunteered as a mentor when I was in the sixth form at Highgate School, regularly helping pupils with their maths problems. I enjoy the challenge of explaining concepts, finding numerous different ways to get them across in a friendly and engaging manner. Above all, I am patient, allowing students to achieve their goals at the best pace for them.

__Sessions__

I am flexible with regard to the session content, allowing students to present particular problems, revise topics or go through past papers.

__What next?__

Please contact me with any questions using the ‘Webmail’ on this site. We can arrange a free 15 minute ‘Meet-the-Tutor’ session to establish what problems you are having and how I can help you.

No DBS Check

5from 2 customer reviews

Natalia (Parent from Alicante)

June 16 2016

I learnt a lot in his classes. I think he is a great tutor and he helped me for my physics and maths IGCSE.

Natalia (Parent from Alicante)

April 11 2016

I like the lesson, it was useful and he was nice and patience

As the number of equations is the same as the number of unknowns, there is exactly one solution!

We start by labelling the two equations:

2x + 3y = 28 (1)

x + y = 11 (2)

There is more than one way to approach this. We only need to use one approach, but let's consider two different methods here.

**Method 1**

We can use substitution. We start by making the coefficient of one of our unknown values the same in both equations; in the first equation, we have the term "2x" and in the second equation we have the term "x". We can multiply both sides of the second equation by 2 in order to have a term in "2x".

So let's start by multiplying the second equation by 2:

2*(2): 2x + 2y = 22

We can now subtract this from the first equation:

(1)-2*(2): 2x + 3y - 2x - 2y = 28 - 22

so y = 6

We've found y! To find x, we can substitute our value for y into either of the two equations. Let's substitute it into equation (2). We see:

x + 6 = 11

so x = 5

We have now found both unknowns. We know from the previous step that equation (2) is satisfied. In order to check our answer, it is a good idea to substitute both unknowns into equation (1):

LHS (the left hand side of the equation) = 2x + 3y

= 2*5 + 3*6

= 10 + 18

= 28

= RHS (the right hand side of the equation)

Both equations are satisfied, so we know that we have found the correct answer.

**Method 2**

We can solve simultaneous equations by elimination. We start by making one of our unknowns the subject of one of our equations. Let's make y the subject of equation (2). We simply subtract x from both sides, so:

y = 11 - x

We can now substitute this into equation (1); we write (11 - x) instead of y as they are the same. So (1) becomes:

2x + 3(11 - x) = 28

2x + 33 - 3x = 28

33 - x = 28

33 = 28 + x

x = 5

We've found x, and now we can find y and check our answer in the same way as in Method 1.

As the number of equations is the same as the number of unknowns, there is exactly one solution!

We start by labelling the two equations:

2x + 3y = 28 (1)

x + y = 11 (2)

There is more than one way to approach this. We only need to use one approach, but let's consider two different methods here.

**Method 1**

We can use substitution. We start by making the coefficient of one of our unknown values the same in both equations; in the first equation, we have the term "2x" and in the second equation we have the term "x". We can multiply both sides of the second equation by 2 in order to have a term in "2x".

So let's start by multiplying the second equation by 2:

2*(2): 2x + 2y = 22

We can now subtract this from the first equation:

(1)-2*(2): 2x + 3y - 2x - 2y = 28 - 22

so y = 6

We've found y! To find x, we can substitute our value for y into either of the two equations. Let's substitute it into equation (2). We see:

x + 6 = 11

so x = 5

We have now found both unknowns. We know from the previous step that equation (2) is satisfied. In order to check our answer, it is a good idea to substitute both unknowns into equation (1):

LHS (the left hand side of the equation) = 2x + 3y

= 2*5 + 3*6

= 10 + 18

= 28

= RHS (the right hand side of the equation)

Both equations are satisfied, so we know that we have found the correct answer.

**Method 2**

We can solve simultaneous equations by elimination. We start by making one of our unknowns the subject of one of our equations. Let's make y the subject of equation (2). We simply subtract x from both sides, so:

y = 11 - x

We can now substitute this into equation (1); we write (11 - x) instead of y as they are the same. So (1) becomes:

2x + 3(11 - x) = 28

2x + 33 - 3x = 28

33 - x = 28

33 = 28 + x

x = 5

We've found x, and now we can find y and check our answer in the same way as in Method 1.

First we calculate the difference between the original price and the new price. This is £1200 - £970 = £230. We now divide this difference by the original price, and multiply it by 100 to find the percentage reduction. If we type 230/1200 *100 into a calculator, we get 19.16666667 (or something similar, depending on how many decimal places our calculator gives us!). We only want our answer to one decimal place. We look at the digit in the second decimal place, which is 6. Since this is greater than or equal to 5, we round up the first decimal place, so the answer is 19.2%.

First we calculate the difference between the original price and the new price. This is £1200 - £970 = £230. We now divide this difference by the original price, and multiply it by 100 to find the percentage reduction. If we type 230/1200 *100 into a calculator, we get 19.16666667 (or something similar, depending on how many decimal places our calculator gives us!). We only want our answer to one decimal place. We look at the digit in the second decimal place, which is 6. Since this is greater than or equal to 5, we round up the first decimal place, so the answer is 19.2%.