David M. A Level Maths tutor, GCSE Maths tutor, A Level Further Mathe...

David M.

Unavailable

Mathematics (Bachelors) - Warwick University

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About me

Me:

I am a second year maths student at Warwick University. I have always loved maths and very much enjoy finding solutions for problems and also alternative ways to approach them. 

I am a calm individual who has helped a number of students both peers and those younger than me with maths at varying levels. The joy from helping people understand and learn is something I have found I enjoy immensely. 

What I will do:

My aim is to help you understand the subject and therefore improve your confidence and skill with it, ultimately this will help with exams and also whatever comes after. My hope is that after each session you will come out with a greater enjoyment and understanding and if you have any suggestions on things that work better for you I will glady listen.

What should you do:

If you have any questions just send me a message via 'Webmail' or book a 'Meet the Tutor session' and I will do my best to answer them fully. 

Hopefully hear from you soon.

Me:

I am a second year maths student at Warwick University. I have always loved maths and very much enjoy finding solutions for problems and also alternative ways to approach them. 

I am a calm individual who has helped a number of students both peers and those younger than me with maths at varying levels. The joy from helping people understand and learn is something I have found I enjoy immensely. 

What I will do:

My aim is to help you understand the subject and therefore improve your confidence and skill with it, ultimately this will help with exams and also whatever comes after. My hope is that after each session you will come out with a greater enjoyment and understanding and if you have any suggestions on things that work better for you I will glady listen.

What should you do:

If you have any questions just send me a message via 'Webmail' or book a 'Meet the Tutor session' and I will do my best to answer them fully. 

Hopefully hear from you soon.

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Personally interviewed by MyTutor

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
Further MathematicsA-level (A2)A*
PhysicsA-level (A2)A
ChemistryA-level (A2)A
STEP 1Uni admission test1
STEP 2Uni admission test1
STEP 3 Uni admission test3

General Availability

Pre 12pm12-5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
Further MathematicsA Level£20 /hr
MathsA Level£20 /hr
MathsGCSE£18 /hr

Questions David has answered

Find the set of values for which: 3/(x+3) >(x-4)/x

First we must consider for which values of x the equation: 3/(x+3) = (x-4)/x is undefined. In this case, x=-3, and x=0. 

Now we must consider each of the cases, x<-3, -3

Case 1 (x<-3): if we assume 3/(x+3) > (x-4)/x then it follows, as x<-3, 3x > x2-x-12. Which implies (x-6)(x+2) < 0. For this to hold exactly one of (x-6) and (x+2) must be less than 0 and the other greater than 0. which implies -2

Case 2 (-32-4x-12 > 0, which implies (x-6)(x+2) >0. Therefore for x2-4x-12 > 0, either both (x-6) and (x+2) must be less than 0 or greater than 0. therefore x<-2 or x>6. Therefore as we know -3

Case 3 (x>0): Therefore x2-4x-12 < 0, which implies (x-6)(x+2) < 0. For this to hold exactly one of (x-6) and (x+2) must be less than 0 and the other greater than 0. which implies -2

Finally as we have considered all cases the final answer is -3

First we must consider for which values of x the equation: 3/(x+3) = (x-4)/x is undefined. In this case, x=-3, and x=0. 

Now we must consider each of the cases, x<-3, -3

Case 1 (x<-3): if we assume 3/(x+3) > (x-4)/x then it follows, as x<-3, 3x > x2-x-12. Which implies (x-6)(x+2) < 0. For this to hold exactly one of (x-6) and (x+2) must be less than 0 and the other greater than 0. which implies -2

Case 2 (-32-4x-12 > 0, which implies (x-6)(x+2) >0. Therefore for x2-4x-12 > 0, either both (x-6) and (x+2) must be less than 0 or greater than 0. therefore x<-2 or x>6. Therefore as we know -3

Case 3 (x>0): Therefore x2-4x-12 < 0, which implies (x-6)(x+2) < 0. For this to hold exactly one of (x-6) and (x+2) must be less than 0 and the other greater than 0. which implies -2

Finally as we have considered all cases the final answer is -3

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2 years ago

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