Currently unavailable: for regular students
Degree: Financial Mathematics (Masters) - Exeter University
I have recently graduated from the Univeristy of Exeter with a degree in BSc. Mathematics. I am passionate about maths, and helping others understand and gain confidence in themselves is a rewarding experience and why I enjoy tutoring.
During the sessions, you will guide what we study. Although it is important to look at the syllabus and follow the text book in order to prepare for an examination, it is also beneficial to go beyond the curriculum and explore other ways of approaching, learning and tackling mathematical problems. These will help with developing problem solving skills which is a vital skill to have in mathematics (and in other areas too!).
The digital white board will allow me to draw diagrams and graphs to explain difficult concepts visually.
I plan on delivering sessions that are enjoyable as well as productive. It is very important that you feel comfortable to ask questions and can learn in a relaxed atmosphere.
If you have any questions or are interested in having me as your tutor, dont hesitate to email me via the ‘webmail’ or book a ‘meet the tutor session’ which you can do through this website.
Let me know what you are struggling with and we will work on it together. I look forward to meeting you!
|Maths||13 Plus||£18 /hr|
|Maths||11 Plus||£18 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
James (Student) July 4 2016
James (Student) June 28 2016
James (Student) July 5 2016
James (Student) July 1 2016
We can start by rearranging the equation such that the modulus (also called the absolute value) is the only component on the right-hand-side of the equation.
This gives us: x-4=-|2x+1| (by subtracting 4 from each side) and we shall call this equation (1)
The modulus of a number may be thought of as its distance from zero e.g. |x|=x for a positive x, |x|=-x for a negative x.
Therefore, equation (1) can be written as two separate equations:
x-4=-(2x+1) (2) & x-4=(2x+1) (3)
We can now rearrange equation (2) and solve for x as follows:
x-4=-2x-1 (we can then plus 2x to each side)
3x-4=-1 (we can then plus 4 to each side)
3x=3 (we can then divide each side by 3)
We will now do the same with equation (3):
x-4=2x+1 (we can then minus x from each side)
-4=x+1 (we can then minus 1 from each side)
We have now solved the equation x=4-|2x+1| and found that the values of x which satisfy this equation are x=1 and x=-5.see more