Peter H. GCSE Maths tutor, A Level Maths tutor, GCSE Physics tutor, A...

Peter H.

Currently unavailable: for regular students

Degree: Engineering Science (Masters) - Oxford, Balliol College University

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About me

I am a second year Engineering Science student at Balliol College, Oxford University. One of the most appealing features of studying engineering is being able to apply Mathematics and Physics theory to real world problems. I feel that having had previous experience assisting younger students in Maths classes alongside being a swimming teacher for nearly five years, I would be able to offer a friendly and beneficial service.

The Sessions

During the sessions the main focus will be to improve the basic understanding of the topic. This will be done flexibly depending on the student and the ways they learn best. Once the understanding is in place the student will be guided through a large variety of exam questions and how to obtain the maximum number of marks.

Oxbridge Applications

Having been through the application process myself I can offer extensive assistance in writing personal statements and preparing for the PAT entrance exam.

Next Step

If you have any questions send me a Webmail or book a ‘Meet the Tutor’ session. Let me know what topics and questions you require assistance with and we can move on from there!

Subjects offered

SubjectLevelMy prices
Electronics A Level £20 /hr
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Physics A Level £20 /hr
Biology GCSE £18 /hr
Electronics GCSE £18 /hr
Further Mathematics GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr
-Personal Statements- Mentoring £20 /hr
.PAT. Uni Admissions Test £25 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
Further MathematicsA-LevelA*
PhysicsA-LevelA*
BiologyA-LevelA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for regular students

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Ratings and reviews

5from 3 customer reviews

Joshua (Student) July 10 2016

Very helpful and informative.

Joshua (Student) July 6 2016

Joshua (Student) June 29 2016

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Questions Peter has answered

Solve the simultaneous equations 5x+2y=11 and x-y=-2.

Simultaneous Equations are equations involving 2 or more unknowns that are to have the same value in each equation. The aim is to eliminate all except one unknown which will then allow us to find the value of that unknown. From this any other unknowns can be found. There are 2 main methods we ...

Simultaneous Equations are equations involving 2 or more unknowns that are to have the same value in each equation. The aim is to eliminate all except one unknown which will then allow us to find the value of that unknown. From this any other unknowns can be found. There are 2 main methods we can use to solve these problems:

1.Solving by Elimination

This method works due to 2 properties of equations:

1.Multiplying or dividing the expression on each side by the same number does not alter the equation.

2.Addition of 2 equations produces another valid equation.

We must manipulate the 2 equations so that when added or subtracted an unknown is removed. First we must label our equations:

(1)5x+2y=11

(2)x-y=-2

Now we must manipulate one or both of the equations so we can remove an unknown by elimination.

For example here the easiest method would be to multiply equation (2) by 2 to obtain equation (3)2x-2y=-4. Then by adding equation (1) to equation (3) we would eliminate y from our equations.

This would give us 7x=7. Therefore we can see that x=1.

Now we substitute this value of x back into either of our original equations (1) or (2). If we substitute back into (2) we obtain the equation 1-y=-2. From this we can see that y=3.

Therefore we have solved the simultaneous equations and obtained the answer of x=1 and y=3.

2.Solving by Substitution

This involves making one of the variables the subject of the equation then substituting this equation into the other simultaneous equation hence removing one of the unknowns.

For this equation if we make x the subject in equation (2) we obtain x=y-2. Then by substituting this x expression into equation (1) we obtain 5(y-2)+2y=11.

By expanding the bracket and collecting like terms we obtain the equation 7y=21. We can see from this that y=3.

Then similarly to before we substitute this y value back into either equation (1) or (2).

If we substitute back into equation (2) we obtain the equation x-3=-2.

We can see from this that x=1.

Therefore we have obtained the answer of x=1, y=3 which is the same as for solving by elimination.

For simple simultaneous equation problems such as this one, elimination is often the more efficient method. However as problems become more complex substitution is often required as integers may not always be used.

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6 months ago

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