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

Jonny is a 5 star tutor and has gone out of his way to help us at times when things haven't gone quite to plan. His knowledge, patience and delivery of problem solving is excellent. I totally recommend Jonny from our experience of him a my son's math's tutor. Thanks Jonny!

Emma, Parent from Cornwall

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.

example: solve the simultaneous equations 3x+2y=4 and 2x+y=3
__1:__ 3x+2y=4

__2:__ 2x+y=3
want to either get same number of x's or same number of y's in equation __1__ and equation __2__

easiest way to do this is by making both equations contain**2y** by multiplying equation __2__ by 2
__2__x2: 2(2x+y=3)

4x+2y=6 call this equation__3__
now you have the same number of y's in both equations you can subtract one from the other so that there are no y's in the overall equation

choose the equation with the highest number of x's to subtract from so that you will not have a negative number of x's left over**3**-**1**: (4x+2y)-(3x+2y)=6-4

4x+2y-3x-2y=6-4

4x-3x + 2y-2y = 6-4

x+0=2

therefore x=2 now you have found what x equals you can substitute this value back into one of your original equations (equation__1__ or equation __2__) to find the value of y. I will sub into **2** as there is only one y in the equation so less working out is needed
sub x into** ****2**: 2x+y=3

2(2)+y=3

4+y=3

y=3-4

y=-1

easiest way to do this is by making both equations contain

4x+2y=6 call this equation

choose the equation with the highest number of x's to subtract from so that you will not have a negative number of x's left over

4x+2y-3x-2y=6-4

4x-3x + 2y-2y = 6-4

x+0=2

therefore x=2 now you have found what x equals you can substitute this value back into one of your original equations (equation

2(2)+y=3

4+y=3

y=3-4

y=-1

We all make silly mistakes when in comes to Maths, so here are some common mistakes to help you out.
**1. Partial Fractions: ** Many students think x+2/(x+1)^2 becomes A/(x+1) + B/(x+1)
This is however incorrect, as if we were to re-combine the fraction we would not get (x+1)^2 inn the denominator
Therefore, the correct method would be x+2/(x+1)^2 = A/(x+1) + B/(x+1)^2
**2. Fractions** Many students infer a/b+c to become a/b + a/c
This is not the case however, as we can only split a fraction numerators apart
For example: a+b/c = a/c + b/c
**3. Dividing by zero** Take the following, where we assume a=b:
1) a = b
2) ab = a^2
3) ab - a^2 = a^2 - b^2
4) b(a-b) = (a+b)(a-b)
5) b = a + b
6) b = 2b
7) 1 = 2
Do you see the mistake? The problem lies in the step 5. Since we assumed a = b, then a-b=0, and hence when we divided by a-b we were really dividing by zero!

Answered by Calum J.

Studies Computer Science at Durham

Step 1: Rewrite
(x+3)(x+3)
Step 2: Multiply out
x^2 +3x +3x +9
Step 3: Clean up
x^2 +6x +9

Answered by Virad K.

Studies Medicine at University College London

Step 1: Rearrange
x^2 - 4x = 0
Step 2: Factorise
x (x-4) = 0
Step 3: Solve for x
Recall: a(0) = 0
Either x = 0, or x-4 = 0
Therefore x = 0 or x = 4

Answered by Virad K.

Studies Medicine at University College London

(2x+3) (2x+1) = 4x^2 + 8x + 3
(2x-8) (x-4) = 2x^2 - 16x - 32
(4x^2+8x+3) - (2x^2-16x-32) = 1
2x^2 - 8x - 29 = 1
2x^2 - 8x - 30 = 0
x^2 - 4x - 15 = 0
Using the quadratic formula;
x = 6.36 or x= -2.36 to 2dp

Firstly, use the factor theorem to determine one factor. Substitute factors of 24 into the equation, beginning at plus or minus 1 and then increasing. The first factor found will be -2, therefore (x+2) is a factor.
Using polynomial division, we find that (x^{3} + x^{2} -14x-24)/(x+2) = x^{2} - x -12. This can be easily factorised into (x-4)(x+3), so the final answer is (x-4)(x+3)(x+2).
This can be checked by expanding the brackets.

Answered by Scarlet W.

Studies Maths and Statistics at Lancaster

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