Hi! I'm Catherine and I'm a Mathematics student at the University of Warwick. I love studying maths - it's a subject that is so often described as "boring" and "difficult", but I truly believe that with the right teachers and tutors, everyone can enjoy maths (or at the very least, understand it!)
I'm friendly, encouraging and professional. I spend time with my students, so that they really understand the material. I am also a Level 2 trampoline coach at my university club and at a local school. This has given me the chance to gain valuable teaching experience to lots of different people, with many different learning styles and needs. A large part of my training as a coach involved understanding how people learn, so I really know what I'm talking about.
The sessions are really all about you, and what you are struggling with. When it comes to maths, nothing is better than PRACTICE! We will work through examples together to really develop your understanding of concepts. We can work together to find what style of teaching and learning is best for you (whether that be spoken explanations, diagrams, activities, to name a few), so you can learn effectively and efficiently.
Feel free to get in contact with me and ask me any questions! I'm available weekday evenings and weekend mornings, for meet-the-tutor sessions and tutorials.
|Maths||A Level||£22 /hr|
|Maths||13 Plus||£20 /hr|
|Maths||11 Plus||£20 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Mahnaz (Parent) September 6 2016
Habiba (Student) September 6 2016
Liv (Student) June 16 2016
Lola (Student) June 5 2016
Method 1 (algebraically):
Since we have "y=" at the start of both equations, we can substitue one into the other to get
Then we want all the x terms on one side of the equation, and all the number terms on the other side. So we can subtract 2x from both sides to get
then add 10 to both sides to get
Now we can substitute this x value back into our first equation to find a value for y!
Then we check that our x and y values (x=15 and y=35) satisfy our second equation.
They do, so we have our answer!
Method 2 (graphically):
If you have graph paper, and can draw accurately enough, you can draw the two equations as graphs (on the same piece of paper) and find the co-ordinates of the point where they cross.
Ideally, we want you to able to solve equations like this using algebra, however being able to draw the graph allows you to visualise what is happening.see more