This was my son's final Math's Tutorial with Jonny - GCSE today! Jonny has been a huge help over the past months, and with his patient and supportive teaching style has, I'm certain, contributed to the potential for a much better grade pass for my son. Totally and utterly recommended. Thank you Jonny.

Emma, Parent from Cornwall

A very patient tutor. Mayur will pre-plan a lesson in advance if required, this way the lesson begins exactly where you need. Response to messages is immediate and efficiency at the very highest. A most enjoyable and informative lesson. I would highly recommend this tutor to students of all levels; this was my third successful lesson.

Elizabeth, Student

Callum has become so much more confident with his maths since taking you on as his maths tutor. He not only enjoys your tutorials he says he has gained a lot from them since starting your tutalage. Thank you so much for your time, effort and flexibility over the last several months. Regards Stephen Brown (Callums dad)

Callum, Student

Jack has been tutoring my 12 year old son for about 2 months now and is really beginning to improve his confidence and mathematical ability. The lessons are fun and very thorough, often reinforcing a topic covered in classwork or tricky homework, leaving him confident and ready to move on to the next one. Unlike in school, during sessions with Jack my son is never afraid to say when he doesn't fully understand something, ensuring 100% confidence in a topic. Jack has been very flexible with the timings of our sessions and has endless patience. This is a great way to sign up to a tutor and if you are lucky enough to have Jack you will not be disappointed.

Julia, Parent from Hampshire

Why limit yourself to someone who lives nearby, when you can choose from tutors across the UK?

By removing time spent travelling, you make tuition more convenient, flexible and affordable

We've combined live video with a shared whiteboard, so you can work through problems together

All your tutorials are recorded. Make the most out of your live session, then play it back after

Usually we cover both subject knowledge and exam technique, although that can change depending on each individual student. Then we go through diagrams, and they ask questions, and we go from there.

Lots of students say that the classes are too big in school, or that they don't have time to ask teachers after lessons. In my tutorials, we take time to explore things in a little in a bit more detail.

I always look up the board my students are taking so the lessons are really relevant. Then we go through past papers or set texts, whatever the student finds helpful.

I use the shared whiteboard. We make diagrams together and label them, and often the student prints it off because they know it's right and they completely understand it.

After tutoring one girl went and told all her friends the new explanation I gave her. And she was so excited about what she wrote in the exam she emailed me immediately afterwards.

There was one girl who had her exam on Monday. She wanted tuition on Friday, Saturday and Sunday beforehand. It was very intense, but she said the exam went well.

Firstly number the two equations 1 and 2 for simplicity. To solve this we want to eliminate one of the variables, x or y, so we start by looking at the equations and seeing if we can add or subtract them from each other to get rid of one of the variables. From these equations it doesnt look like we can do this immediately. What we do now is multiply equation two (x-y=5) by 5, to get our new equation three : 5x-5y =25. REMEMBER to multiply the right hand side as well... Now we see we can add equation three to eqaution one and this removes y...we are left with 9x =63. Divide by 9 to get x=7. Now plug x = 7 into one of the original equations, for simplicity use equation two. And this gives us y=2

Answered by Ali A.

Studies Mathematics at Bristol

Rule: An equation must be balanced. Therefore, in order to keep the equation balances, what you do to one side you must do to the other side.
Change of subject questions require you to isolate the required letter on one side of the equation.
In change of subject question such as the one above 'v = u + at', first you need to realise that the equation is the same as +v= +u + ((+a)x(+t)).
To begin with to remove the 'u' from the right hand side of the equation we should subtract the 'u' and thus given our rule of balanced equations we should do the other side as well. (v - u = u - u + at ) and this is equal to (v-u=at).
Then to remove the 't' from the right hand side of the equation we should divide the 't'. We divide because to get rid of a 'x t' we need to divide 't' and thus given our rule of balanced equations we should do that to the other side as well.
(v - u)/t = (at)/t >>>>>>>>>> t/t =1
((v - u)/t = 1a
(v - u)/t = a

Answered by Sherin W.

Studies Biomedical science at Sussex

4x^2+10x+6 We can see that all the coefficients and constants here can be divided by 2, which simplifies the quadratic for us: 2(2x^2+5x+3) Next, we need to factor the quadratic by making it into two pairs of brackets. The only way we can get 2x^2 is from multiplying 2x by x, so we know that these must go in the brackets. Similarly, the only way to get 3 is by multiplying 3 by 1, so these also must go in. We then need to work out the arrangement for these values, so that we can get the 5x. 5x = 2x+3x so we should let the 3 be multiplied by the x and the 2x be multiplied by the 1. Hence we get: 2(2x+3)(x+1) = 0 We can divide by 2 to get: (2x+3)(x+1)= 0 There are always two solutions to a quadratic equation, which can be found here by equating each of the brackets with 0, as anything multiplied by 0 is 0. If 2x+3 = 0, then 2x = -3 and therefore x = -3/2. If x+1 = 0, then x = -1. Our solutions are x = -3/2 and x = -1

1. x^{2} + 15 x + 50 = 0 (rearrange by adding 50 to both sides)
2. (x + 10)(x + 5) = 0 (factorise the quadratic equation)
3. x + 10 = 0 or x + 5 = 0 (for the above to equal zero, at least one of the bracketed terms must equal zero)
4. x = -10 or x = -5 (solving the two equations that result from step 3)
5. The solution is x = -10, x = -5

Answered by Julia S.

Studies Economics and Management at Oxford, St Edmund Hall

fg(x) = 2(cx+5) + c
= 2cx+10+c [1 mark]
fg(x) = 6x+d. fg(x) = 2cx+10+c
6x = 2cx and d = 10+c [2 marks]
3 = c and therefore d = 13 [3 marks]

Answered by Heba K.

Studies Medicine at University College London

x^2= 4(x^2 - 6x + 9)
x^2= 4x^2- 24x + 36
3x^2 - 24x + 36= 0
(x-2)(x-6)=0
x= 2 or x=6

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With such a variety of questions, GCSE Maths can be a real headache for students. In individual sessions with a GCSE Maths tutor you can focus on exactly the topics you need help with. In our online classroom, tutors use diagrams, graphs and illustrations to enhance their lessons.

From algebra to indices,

they will explain concepts at your own pace and in new ways so that you fully understand them.So when your GCSE exams come round, you'll be able to tackle even the most difficult questions on the paper.