Reece K. A Level Maths tutor, GCSE Maths tutor, A Level Physics tutor...

Reece K.

Currently unavailable: for new students

Degree: Aerospace Engineering (Masters) - Bristol University

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About me

Who am I?

My name is Reece, I'm 21 and I study Aerospace Engineering at the University of Bristol. I've just finished my 3rd year and am about to embark on a year in industry with Airbus, working in Flight Physics, before returning to Bristol to finish my masters. 

Why am I tutoring? 

I believe that with hard work and the right attitude you can do anything, and taking the step to do extra tutoring outside of school hours shows the right attitude for success. I'm more than willing to go the extra mile for those with the right attitudes and help them as much as possible in their drive for bettering themselves. 

What interests me? 

As you may have guessed from my degree and year in industry I'm quite interested in planes. I really enjoy going to airshows and watching some awesome machines do their thing! 

Alongside my degree I'm quite interested in 3D printing, having built 2 3D printers of my own. I enjoy taking things apart, fixing things up, and just generally learning about stuff. To balance the time I spend sitting down doing academic fun stuff, I also enjoy playing rugby, running, and going to the gym! 

Tutoring methods

I believe that the best way to tackle exams is to make sure you've got a sound understanding of the basics and key ideas behind the questions. By the end of one of my sessions I'd like you, the tutee, to be able to explain the concepts we've discussed back to me and feel confident in your abilities to be able to tackle any tricky exam questions that may come your way! 

I'm a really patient, easy going guy who enjoys breaking things down and giving explanations which show why theory works and how we can directly see the results of the topics we will discuss in the real world. 

So book a session with me and continue your push to success! I look forward to working with you! :)

Subjects offered

SubjectLevelMy prices
Computing A Level £20 /hr
Maths A Level £20 /hr
Physics A Level £20 /hr
Computing GCSE £18 /hr
Electronics GCSE £18 /hr
ICT GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr
Science GCSE £18 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
ICTA-LevelA*
PhysicsA-LevelA
ChemistryA-LevelB
Disclosure and Barring Service

CRB/DBS Standard

11/06/2014

CRB/DBS Enhanced

No

Currently unavailable: for new students

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Ratings and reviews

5from 4 customer reviews

Joanne (Parent) August 20 2016

Really systematic and giving my son confidence in this subject thank you!

Joanne (Parent) August 16 2016

Really managed to engage my son who was not at all keen at the start!

Joanne (Parent) September 2 2016

Joanne (Parent) August 23 2016

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Questions Reece has answered

How do I solve a pair of simultaneous equations?

An example of a pair of simultaneous equations is as follows: 4x + 3y = 10        (1)   and 2x - 3y = 14         (2) Now to solve these equations we can simply add them and solve for x, then sub that value of x in to one of the original equations and solve for y, but how and why do we do tha...

An example of a pair of simultaneous equations is as follows:

4x + 3y = 10        (1)   and

2x - 3y = 14         (2)

Now to solve these equations we can simply add them and solve for x, then sub that value of x in to one of the original equations and solve for y, but how and why do we do that?

Firstly we can try to eliminate one of the unknowns from equation (1) by using equation (2), in this case we can simply add them to eliminate y. So upon adding (1) and (2), we are left with:

6x = 24        (3)

As you can see, upon adding 3y to -3y, the unknown y has been taken out of the equation all together, which now allows us to find a value for x. Now dividing both sides of (3) by 6, we are left with:

x = 4

Hooray, we've manipulated our two equations and managed to find a value for x, now how can we find y?

To find the value of y, we can simply put our newly found value of x back in to (1). So lets try that and see what we get:

4(4) + 3y = 10, which can also be written as:

16 + 3y = 10       (4)

So now we can take the 16 from both sides of (4) which gives us:

3y = -6

Now dividing through by 3 we are left with:

y = -2

Hooray! So by using the two equations we were originally given and manipulating them in a certain way, we have solved them and found the values of the unknown variables x and y.

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1 year ago

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