I cannot praise this tutor enough. Highly skilled in mathematics, teaches at the student's pace, patient and super efficient. A conscientious person who responds immediately to messages. The lesson was tailor made to suit my own needs. Mayur is excellent at pin pointing math problems quickly - taking things back to basic principals where needed. A most enjoyable tutorial - I would highly recommend this very gifted tutor to students of every ability.

Elizabeth, 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

Parent review - it's a relief to see my son start to understand GCSE maths topics that he has struggled with. His confidence is growing with Chris's help. Chris isn't phased by any questions or topics thrown at him, and is able to clearly explain the methods needed to answer maths questions.

Kelly, Parent from Kent

Joseph has been the most fantastic help with GCSE revision this Easter. He's given up a huge amount of time, is encouraging, flexible, easy to deal with, focussed, targeted - the all round perfect tutor. His greatest achievement is to have taught my son how to enjoy maths. A lifetime's terror of the subject has turned into enjoyment. My son said he'll miss maths and he thinks he has a good chance of getting a high grade. Unthinkable before Joseph started helping him a few months ago.

Alison, Parent from Hertfordshire

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.

x³-x=x(x²-1)=x(x+1)(x-1)

Answered by Jia Qin L.

Studies Statistics with finance at LSE

Let us use Line 1 and Line 2 as an example.

Line 1: y = 2x + 3

Line 2: y = -0.5x + 7

Step 1 - Try and plot the two functions on a x, y coordinate axes.

Remember y = mx + c. Using Line1, we are given the y intercept (c) which is 3. We are also given the gradient (m) which is 2.

We can make one mark at the point (0,3) which corresponds to the y intercept. Another way to think of this: what is y when x=0?

Following on from this: what is x when y = 0 ? To work this out just substitute y = 0 into Line 1. After rearranging (to make it x = all other stuff ) we find that x = -3/2. We now have another point to mark on our axes (-3/2, 0).

Now draw a line between both marks and continue the line in both directions! Repeat for Line 2. It is not vital to plot the lines, but by doing it correctly we can already answer by inspection of the graph " Do these lines intersect?". By doing the graph axis perfectly with sufficient divisions you can even read of the point of intersection (not recommended). In this case the lines do intersect. So what’s next? Let’s prove it using algebra.

Step 2 - What are the coordinates of the point where Line 1 and Line 2 intersect?

Let’s call the point of intersection P with the coordinates (x,y) on our graph.

A good way to think about this is to realise that Line 1 and Line 2 both contain the two same variables, x and y.

At the point P, it is the only position where those two variables x & y must satisfy both Lines!!

The way to prove this is by making the two Line equations equal to each other.

They both start with y = (rest of stuff). At point P, y must be the same for both so we can literally make them equal to each other.

y=y so therefore Line 1 = Line 2

2x + 3 = -0.5x + 7 ( We now have one equation and one variable! We can solve this)

Rearrange (So x = all other stuff) and we find that x = 1.6! We have the first coordinate of P (1.6, y)

Now let’s find y! Pick either of the Lines, it does not matter because at point P, x is going to be the same in both.

Picked Line 1: y = 2x + 3

Substitute in our lovely x value we have just worked out (x = 1.6). And we get:

y = 2*(1.6) + 3

y = 6.2

Now we have the second coordinate of P!

P = point of intersection of Lines 1 & 2 = (1.6, 6.2)

Answered by Ivan J.

Studies General Engineering at Durham

Pythagorus' Theorum enables us to calculate the length of any side of a right angled triangle provided we know the other two lengths. This is done by substituting in the appropriate values into the equation a^{2}=b^{2 }+ c^{2}, where a is the hyponenuse, or the longest side.

Answered by Sam D.

Studies Mechanical Engineering at Birmingham

When you are told to solve a quadratic equation and when factorising is not possible, quadratic formula can be used to find out the solution. First you will need to get the coefficients from the quadratic equation and then plug in these values into the equation appropriately and solve!

Answered by Darem A.

Studies Advanced Chemical Engineering at Manchester

This is similar to other factorisation problems such as x^{2 }+ 5x - 6. In this problem, you would find prime factors of -6 that add up to 5 (In this case -3, -2).

Now, do the exact same with this problem! This is the same as x^{2 }+ 0x - 4, so you need to find factors of -4 that add up to 0. In this case these two primes will be -2 and 2. Using these factors, you can reach the answer: (x - 2)*(x + 2).

To check your work, multiply (x - 2)(x + 2) to get x^{2} + 2x - 2x - 4 = x^{2 }- 4.

The mean of a group of numbers is found by dividing their sum by how many numbers there are.

Mean=(1+7+25+9+34)/5=76/5=15.2

Answered by Isabel R.

Studies Mathematics at Manchester

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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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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.