I'm a final year Masters student in Chemistry at UCL. I have been tutoring for MyTutor for three years now and I'm extremely delighted to say that all my students so far have thoroughly enjoyed themselves, some of which come back year after year! I have experience tutoring Maths for GCSE and FSMQ examinations, in Chemistry for all GCSE and A level syllabi as well as Pre-U, and French GCSE. I love Physical Chemistry and I just wrote my dissertation on plasmonic enhancement of photocatalysis - if that sounds like jibberish then it's one way of improving the efficiency of electrochemical reactions, for example to produce hydrogen. I can't deny my passion for the arts - in fact this year I produced a sell-out West End dance production for my Dance Society and I coach the GB tap dancing team! More than anything, this brings a lot of creativity to my tutorials, where I am always happy to improvise and think of new ways to present a topic. What's more I volunteer for CoachBright, an amazing organisation that promotes academic coaching. I try to use some of these techniques in my tutorials to improve students' independence and initiative.

I'm a final year Masters student in Chemistry at UCL. I have been tutoring for MyTutor for three years now and I'm extremely delighted to say that all my students so far have thoroughly enjoyed themselves, some of which come back year after year! I have experience tutoring Maths for GCSE and FSMQ examinations, in Chemistry for all GCSE and A level syllabi as well as Pre-U, and French GCSE. I love Physical Chemistry and I just wrote my dissertation on plasmonic enhancement of photocatalysis - if that sounds like jibberish then it's one way of improving the efficiency of electrochemical reactions, for example to produce hydrogen. I can't deny my passion for the arts - in fact this year I produced a sell-out West End dance production for my Dance Society and I coach the GB tap dancing team! More than anything, this brings a lot of creativity to my tutorials, where I am always happy to improvise and think of new ways to present a topic. What's more I volunteer for CoachBright, an amazing organisation that promotes academic coaching. I try to use some of these techniques in my tutorials to improve students' independence and initiative.

I like the students to direct the topic for each of our tutorials. I encourage students to choose between: - revising a topic studied in class - going through a particularly tricky homework question - discussing their answers to a past paper - introducing a new topic from their syllabus/specification From there, I normally ask them what they know about the topic in question, which quickly turns into some problem-solving. I like to use real-life applications for the students to be able to connect the subjects they are studying to their everyday life and career prospects, for example for organic chemistry we have discussed how not to make TNT and in Maths, we calculated the volume and surface area of a Terry's Chocolate Orange from only the information on the box! I always encourage students to fully understand something, even if that goes beyond what's required of the specification. Learning should not be restricted to the exam requirements, but more discovering why things happen and how they work. It is important, however, to try past papers and learn how to apply their knowledge in the exam, so I make an emphasis on the importance of exam technique.

I like the students to direct the topic for each of our tutorials. I encourage students to choose between: - revising a topic studied in class - going through a particularly tricky homework question - discussing their answers to a past paper - introducing a new topic from their syllabus/specification From there, I normally ask them what they know about the topic in question, which quickly turns into some problem-solving. I like to use real-life applications for the students to be able to connect the subjects they are studying to their everyday life and career prospects, for example for organic chemistry we have discussed how not to make TNT and in Maths, we calculated the volume and surface area of a Terry's Chocolate Orange from only the information on the box! I always encourage students to fully understand something, even if that goes beyond what's required of the specification. Learning should not be restricted to the exam requirements, but more discovering why things happen and how they work. It is important, however, to try past papers and learn how to apply their knowledge in the exam, so I make an emphasis on the importance of exam technique.

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29/01/20154.9from 62 customer reviews

Alex (Student)

Imogen provided my son with some excellent strategies to overcome exam hurdles and find quick routes to problems... all around a good experience and achieved the required grade for university D3!

Ealasade (Student)

May 26 2017

Alway a great lesson

Vickie (Parent)

May 9 2017

yay! I understand converting between units AND she helped me go through a very hard multi-step question! 100% recommend!!

Jan (Parent)

April 27 2017

Very helpful.

**Question:**

**Line A has a gradient of 4 and passes through point (5,6).**

**Line B passes through points C (0,5) and D (2,0).**

**Find the coordinates of the point where the two lines intersection.**

**Solution:**

*First of all find the equation of line A:*

Using y= mx + c,

Applying the gradient, line A has equation, y = 4x + c

To find c, substitute in the coordinates of point P,

6 = (4x5) + c

6 = 20 + c

c = 6 - 20 = -14

Therefore the equation of line A is y = 4x - 14

*Now find the equation of line B:*

Using ( y_{2} - y_{1} ) / ( x_{2} - x_{1} ) = gradient of a line

Substitute in coordinates of points C and D,

( y_{C} - y_{B} ) / ( x_{C} - x_{B }) = ( 5 - 0 ) / ( 0 - 2 ) = 5/-2 or -5/2

Using y = mx + c

Applying the gradient found, line B has the equation, y = -5/2 x + c

To find c, substitute in the coordinates of point C,

5 = ( -5/2 x 0 ) + c

c = 5

Therefore the equation of line B is y = -5/2 x + 5

This can be rearranged,

(multiply everything by 2) --> 2y = -5x + 10

(rearrange) ---> 5x + 2y = 10

You can check your answer by using the coordinates of point D,

( 5 x 2 ) + ( 2 x 0 ) = 10 ---> Yes

*Finally find the coordinates where the lines intersect:*

A y = 4x - 14

B 5x + 2y = 10

A x2 2y = 8x - 28

Rearrange 8x - 2y = 28

Using simultaneous equations, add A x2 and B, to eliminate y,

5x + 8x + 2y - 2y = 10 + 28

13x = 38

x = 38/13

Substitute in x to A to find y,

y = ( 4 x 38/13 ) - 14

y = 152/13 - 182/13

y = -30/13

Put these coordinates into the equation for line B to check it works,

( 5 x 38/13 ) + (2 x -30/13 ) = 10

190/13 - 60/13 = 130/13 = 10 ----> Yes

**Answer:**

The lines cross at coordinate ( 38/13, -30/13 )

**Question:**

**Line A has a gradient of 4 and passes through point (5,6).**

**Line B passes through points C (0,5) and D (2,0).**

**Find the coordinates of the point where the two lines intersection.**

**Solution:**

*First of all find the equation of line A:*

Using y= mx + c,

Applying the gradient, line A has equation, y = 4x + c

To find c, substitute in the coordinates of point P,

6 = (4x5) + c

6 = 20 + c

c = 6 - 20 = -14

Therefore the equation of line A is y = 4x - 14

*Now find the equation of line B:*

Using ( y_{2} - y_{1} ) / ( x_{2} - x_{1} ) = gradient of a line

Substitute in coordinates of points C and D,

( y_{C} - y_{B} ) / ( x_{C} - x_{B }) = ( 5 - 0 ) / ( 0 - 2 ) = 5/-2 or -5/2

Using y = mx + c

Applying the gradient found, line B has the equation, y = -5/2 x + c

To find c, substitute in the coordinates of point C,

5 = ( -5/2 x 0 ) + c

c = 5

Therefore the equation of line B is y = -5/2 x + 5

This can be rearranged,

(multiply everything by 2) --> 2y = -5x + 10

(rearrange) ---> 5x + 2y = 10

You can check your answer by using the coordinates of point D,

( 5 x 2 ) + ( 2 x 0 ) = 10 ---> Yes

*Finally find the coordinates where the lines intersect:*

A y = 4x - 14

B 5x + 2y = 10

A x2 2y = 8x - 28

Rearrange 8x - 2y = 28

Using simultaneous equations, add A x2 and B, to eliminate y,

5x + 8x + 2y - 2y = 10 + 28

13x = 38

x = 38/13

Substitute in x to A to find y,

y = ( 4 x 38/13 ) - 14

y = 152/13 - 182/13

y = -30/13

Put these coordinates into the equation for line B to check it works,

( 5 x 38/13 ) + (2 x -30/13 ) = 10

190/13 - 60/13 = 130/13 = 10 ----> Yes

**Answer:**

The lines cross at coordinate ( 38/13, -30/13 )