Michael F. A Level Maths tutor, 11 Plus Maths tutor, 13 plus  Maths t...
£18 - £20 /hr

Michael F.

Degree: Mathematics (Bachelors) - Oxford, Pembroke College University

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About me

Personal Description:

I have started my first year of studying Maths at Oxford University. I enjoy teaching Maths and have been extremely passionate about it since the start of my academic career. I am here to help students of any age prepare for their exams (GCSE, AS/A Level etc), as well as casual learners who are just looking to improve their skills. It is great to see people succeed, and I would love to be part of your success story!

Tutoring Experience:

I have just finished Sixth Form, and often helped my classmates when they were struggling with classwork and exam questions. During my time at secondary school I was once asked to cover lessons for my own year group!

Tutoring Approach:

As a tutee you will learn the tips, tricks and shortcuts I created and used during my GCSEs and A Levels - I will make sure you understand the necessary concepts so that you can get through your qualification with confidence.

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Further Mathematics GCSE £18 /hr
Maths GCSE £18 /hr
Maths 13 Plus £18 /hr
Maths 11 Plus £18 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
Further MathematicsA-LevelA*
FrenchA-LevelB
STEP Mathematics IUni Admissions Test2 (74 marks)
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

General Availability

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Ratings and reviews

5from 14 customer reviews

Seda (Parent) November 2 2016

My daughter really enjoyed her tutorial with Michael, who is helping her a lot. Michael is a kind and encouraging tutor.

Seda (Parent) November 25 2016

Seda (Parent) December 4 2016

Seda (Parent) November 30 2016

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Questions Michael has answered

Core 3 - Modulus: Solve the equation |x-2|=|x+6|.

Modulus, also known as absolute value, takes whatever's between the |straight brackets| and makes it positive. For example, |3|=3, and |-3|=3. Interestingly, if you have any real number x, then |x|=sqrt(x2). Try putting some numbers in and see! We can't solve a modulus question until we get r...

Modulus, also known as absolute value, takes whatever's between the |straight brackets| and makes it positive. For example, |3|=3, and |-3|=3. Interestingly, if you have any real number x, then |x|=sqrt(x2). Try putting some numbers in and see!

We can't solve a modulus question until we get rid of the straight brackets, but this little trick will do the job every time. If we square both sides of the modulus equation in the last paragraph, we get |x|2=x2. So modulus brackets disappear when we square both sides of our equation. Let's try it...

|x-2|2=|x+6|2

(x-2)2=(x+6)2

x2-4x+4=x2+12x+36

-16x=32

x=-2

This trick is great, because the xterms cancel out and there's no quadratic equation to mess about with. Beware though, this will not be the case in all questions - if you get a quadratic equation to solve, you may end up with more than one solution. Try these bonus questions and see for yourself:

(1) Solve the equation |x+4|=|x-5|.​

(2) Solve the equation |x-3|=|2x|.​

(3) Solve the equation |3x-1|=|3-x|.​

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3 months ago

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