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My son says he's very lucky to have Joseph's help. He's an excellent tutor, 100% reliable, very generous with his time and patient and willing to direct help exactly where it's needed. My son's made progress after only a few lessons and is motivated to work much harder on a subject he finds hard. Couldn't recommend him more highly. Thank you Joseph.

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.

Firstly, you should aim to eliminate one of the unknown values. As b is positive and negative in each equation, this would be a good value to eliminate. Both equations would have to be multiplied to cancel out one of the values. For example, if equation 1 is multiplied by 2 and eqution 2 is multiplied by 5 you get:
Equation 1: 4a-10b= 22 Equation 2: 15a+ 10b= 35
Then add the two new equations together to cancel out b and simplify, which leaves you with:
19a= 57 therefore a= 3
Then substitute a with 3 in equation 1 or 2 to find out the value of b. For example, if substituted into equation 1 you get:
(2 x 3) -5b=11 therefore b=-1

Answered by Ikraan H.

Studies Biomedical Science BSc at Birmingham

There are a number of steps before sketching a graph. We will look over bits of theory and then do some examples (I'll show the examples in the online lesson space)
First we have to find the x-intercept and y-interecept. That simply means finding the coordinates of the function when either x or y (which is the same as f(x)) are 0.
The we have to find the asymptotes. There are three types of asymptotes:
horizontal - when y remains the same as x is very large
vertical - when y becomes very large near a certain value of x
oblique - similar to the horizontal one (show in the lesson space)
The third step is finding any inflection points. We do this by using derivatives.

Recall that the derivative of exp(x) is exp(x), but notice this question is slightly more complex due to the x^2 term. This is example of differentiationg composite functions, and so the chain rule is required. To begin, we'll set u = x^2, and then compute du/dx = 2x. Furthermore, we observe that y = exp(u) and dy/du = exp(u). Then, by the chain rule, we have dy/dx = dy/du * du/dx = exp(u) * 2x = exp(x^2) * 2x.

Answered by Stuart B.

Studies Statistics and Computer Science at St Andrews

The perimeter is the distance around the rectangle. A rectangle has 4 sides: 2 of them will be the length (8cm) and 2 of them the width (3cm). To get the perimeter, you add up 2 of the length and 2 of the width: 2 x 8 + 2 x 3. The answer therefore is 22cm (don't forget the units!)
The area of the rectangle is worked out by mulitplying the length by the width: 8 x 3. The answer is therefore 24cm^2.

Answered by Emma L.

Studies Psychology at Leeds

Firstly, get all of the x's onto one side of the equation and all of the numbers onto the other. To make it easier at the end, subtract the smaller value of x so that it doesn't become negative: 3x-10=12x-7 (-3x) -10=9x+7 (+7) -3=9x To get what a single value of x equals, divide -3 by 9. Then simplify the fraction (3/9 = 1/3) -3/9 = x = -1/3 (don't forget the negative sign!)

Answered by Emma L.

Studies Psychology at Leeds

To combine these two fractions into one, you have to multiply the equation by the denominators in order to make a common denominator: 5(x-5)/(x+1)(x-5) + 6(x+1)/(x+1)(x-5) Now that the denominators are the same, you can add the two numerators to eachother, and then expand the brackets: [5(x-5)+6(x+1)]/(x+1)(x-5) [5x-25+6x+6]/(x+1)(x-5) Now simplify the numerator: [11x-19]/(x+1)(x-5) (This is a GCSE question though I'd only want to tutor 11+ 13+ for now)

Answered by Emma L.

Studies Psychology at Leeds

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With personalised help from a maths tutor, you can learn to tackle even the trickiest of problems.

Enjoy individual support in the topics you struggle with most, or across the whole syllabus.Our Maths tutors have helped with 13+, GCSEs, IB, A Level and have even given a hand to the odd university student.

Your tutor will be a bright young student at a leading UK university, selected for their subject knowledge and ability to explain tricky concepts. You'll meet in our online classroom, where your tutor will talk through problems with you and use diagrams, graphs and illustrations to bring the subject to life.