Excellent lesson. Elisha provided work sheets before the session so that my daughter could practice beforehand. She was very thorough, provided feedback so that we knew what stage our daughter was at and then gave her homework to prepare for the next session. Thoroughly recommend Elisha as a tutor. Excellent.

Andrea, Parent from Gibraltar

Thank you so much Sohini for being able to tutor Charley for the past year with lots of patience and for giving her the confidence and knowledge she needed for her GCSE exam. Charley no longer finds maths so daunting and even quite likes the subject now thanks to you. I would highly recommend Sohini as a well organised and very knowledgeable Maths tutor.

Sylvie, Parent from Buckinghamshire

Lloyd is a really good tutor, I would highly recommend him. He has been reliable, available and booked times in to suit us at short notice. He worked hard with my son and helped him achieve a grade higher in his GCSE than in his mocks and made sure he understood a topic before moving on to the next one.

Simone, Parent

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

Why waste time looking locally when itâ€™s easier to find the right tutor online?

Tutoring is easier and more flexible when you remove the need to plan around travel

With live online one-to-one sessions you’re always engaged

Your live sessions are recorded, so you can play tutorials back if you want to revise

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.

Begin by adding the 2 ratio parts (i.e. the numbers which make up the ratio) so 8+5 = 13.

Then divide the total sum by the sum of the ratio parts. Thus £650/13 = £50.

Multiply this by each of the ratio parts.

£50 * 8 = £400

£50 * 5 = £250

Finish by returning the answers to the correct form. Thus the answer is £400:£250

Answered by Harry V.

Studies Politics and Economics at Exeter

So x^{2 }+ 9x + 20 = 0

My preffered way of solving this equation is to factorise the equation. (Though I understand that different students may find other ways easier)

Factorisation is where the above equation is (x+a)(x+b) = 0

So if we times out (x+a)(x+b) we get

x^{2} + ax + bx + ab = 0

therefore

x^{2} + (a+b)x + ab = 0

Therefore we can equate this to the original question, so

x^{2 }+ 9x + 20 = x^{2} + (a+b)x + ab

so now we can see that

9 = a + b and

20 = ab

I would reccomend using trial and error (although I understand that different students may prefer other techniques).

So by trying for multiple values of a and b, we can see that they must equal 5 and 4.

Therefore

x^{2} + 9x + 20 = (x+5)(x+4) = 0

We know that the only way of producing a 0 through multiplication is through multiplying one number by another. Therefore we know that

x+5= 0 or x+4=0

Through rearranging these equations we can conclude that x must equal -4 or -5.

Answered by Tilly P.

Studies Computer Science and Mathematics at Exeter

Let's draw a line. It crosses the point (0,1), intercepting the y-axis (remember, y to the sky!) at 1 and the x-axis at the point (-2,0).

An equation tells us what a line looks like, it's a way of describing a line in a mathematical way. If we do it right, we can draw a line just by looking at the equation. Equations for straight lines are ususally given in the form y=mx+c - this tells us what y values do for different values of x. But what are m and c?

So, in order to find the equation for any straight line, we need two things: the gradient and the y-intercept. We call the y-intercept c and the gradient m. As soon as we know these two things, the rest is easy!

First step: finding the y-intercept - this is the simple bit, all you need to do is look at the y-axis and see where the line crosses it. In this case it is 1, so c=1.

Second step: finding the gradient - this requires a little bit more thought. All gradient means is the steepness of the line, so for every place the line moves to the right, how many places up does it go? In this case the line is not very steep at all and, in fact, you can see that for every place you move to the right it only goes up half a place up. Therefore m=0.5.

The tricky bit is over! Now comes the simple bit. All you have to do now is sub the values you've found for m and c back into the template equation, y=mx+c. So, if c=1 and m=0.5 then y=0.5x+1 - problem solved!

Answered by Dominique G.

Studies Classics at Cambridge

x=0.045 (45 recurring)

10x = 0.45 (45 recurring)

100x = 4.54 (54 recurring)

1000x = 45.45 (45 recurring)

To get rid of the decimals:

1000x-10x = 45.45 - 0.45

990x = 45

x = 45/990

x = 9/198 (simplify by dividing by 5)

x = 1/22 (simplify by dividing 9)

Answered by John T.

Studies Biochemistry at Exeter

We have two algebraic equations and we are trying to find what a and b can equal to fit for both situations.

1. 6a + b = 16

2. 5a - 2b = 19

The easiest method is substitution because we can sub in an equation for b by rearranging it.

therefore: b= 16 - 6a

From there on we can sub in b= 16 - 6a into equation 2. to give us:

5a - 2(16-6a) = 19

we expand the bracket to give:

5a - 32 +12a =19 **BE CAREFUL OF SIGNS**

17a -32 =19

17a = 51

therfore a = 3

We can use a=3 to sub back into equation 1

6(3) + b = 16

18 + b = 16

b= - 2

To check the two values for a and b are correct sub them back into equation 2. Follow the rule: SUB IN 1, CHECK IN 2:

Therefore when a= 3 and b= -2

5a - 2b = 19

5(3) -2(-2) = 19 **BE CAREFUL OF SIGNS**

15 + 4 = 19 which is correct.

Answered by Sophie W.

Studies Political Economy at Kings, London

This is simplified by solving common terms (the x's and the y's). Where X and Y terms can't be solved together.

Splitting it up:

You can see 2x-x= x

and for the Y terms

(+)6y+2y=8y

Therefore the final answer is

x+8y

Answered by Sophie W.

Studies Political Economy at Kings, London

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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.

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