Currently unavailable: for regular students
Degree: Mathematics and Philosophy (Bachelors) - Warwick University
Who am I?
Hi I’m Roxane! I’m a Maths and Philosophy student and a native French speaker. I really love both Maths and French and I’m very keen to tutor both!
Why do I want to tutor?
I’ve had experience being a face to face tutor both at school and university, I have tutored a girl regularly for two years in French. She used to struggle but after a lot of practice, she achieved a First in her exams. The feeling of pride and accomplishment I felt when she told me her results are the reason that I am so keen to tutor today.
What about the subjects?
As a Maths student, I am surrounded by Mathematics every day and its strongest feature for me is connectedness. No topic can be viewed completely independently from the rest. So if you are struggling with one topic in particular, I will try to explain to you how it works in relation to other topics and to put it in a new light, to help you better understand, because in Maths, more than in any other subject, understanding will get you a top grade, much more than learning and applying techniques.
I am French, so I have the level required for both GCSE and A-level. I am also learning 3 foreign languages at different levels from beginner to fluent. Hence, whatever your level or difficulties, I can empathise and help you overcome them. My main approach to a language is always to surround myself with it. Therefore I will speak to you in French and try to encourage you to listen to or read as much French as possible.
What would a session look like?
The session will always be centred around your needs. I will aim to create a friendly and relaxed environment, as I find learning is more effective when it’s fun. I am very patient and involved, and I’d be happy to answer questions anytime if you drop me an email.
Don’t hesitate to sign up for a free meet the tutor session! I look forward to meeting you!
|French||A Level||£24 /hr|
Jack (Student) August 8 2016
Brian (Parent) August 8 2016
Jack (Student) August 5 2016
Brian (Parent) July 28 2016
Always follow the same three steps : equating – substracting – substituting but you can have different cases :
Case 1: we only need to multiply one equation
Note : Always check first that the equation is solvable (= there are as many or more equations than there are unknowns.)
We notice that 6x is a multiple of 2x. So we only need to multiply the second equation to equate the coefficients.
Substracting the 2nd equation from 3 times the first one gives 19y=19 so y=1
Now we substitute into the first equation to get
The solution is x=2, y=1
Case 2 : we need to multiply both equations
3x+5y = 14
We proceed the same way as in the first case but to equate coefficients we need to multiply both equations, because the coefficients are not multiples of each other.
Multiplying the first equation by 3 and substracting 5 times the second gives 11x=33 so x=3.
Now we substitute back in to get
The solution is x=3, y=1see more