Currently unavailable: for regular students
Degree: Mathematics and Physics (BSc MMathPhys) (Masters) - Warwick University
I study Maths and Physics at University of Warwick. I have always had a passion for understanding how and why the world behaves as it does, and so I particularly love phyiscs and the maths behind it all.
I am very patient and understanding. I have tutored at my Secondary School for Year 8/9 maths. More recently, I have externally tutored A Level Maths.
We will build basic knowledge on topics to reach an understanding ready to tackle exam questions. When this point is reached, we will
Most importantly, the tutorials will be directed to meet your needs. I will adapt my teaching methods and ask for your feeback, as we need to find the best approach for your learning.
To consolidate topics, I will ask you to explain them to me and think about examples where they can be applied, as I have found that teaching is the best form of learning.
I aim for the sessions to be enjoyable, so that you look forward to them. Learning shouldn't be a bore, should it?!
If there are any questions you'd like to ask me, send me a WebMail or book a Meet the Tutor Session through the site.
I look forward to hearing from you!
|Maths||A Level||£20 /hr|
|Physics||A Level||£20 /hr|
|Maths||13 Plus||£18 /hr|
|Maths||11 Plus||£18 /hr|
|.STEP.||Uni Admissions Test||£25 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Here, we have two equations involving two variables, x and y.
The question asks us to 'solve' the set of simultaneous equations.
Equivalently, it is asking us to find the x values (and corresponding y values) which satisfy both of the equations.
We can approach this by substituion:
We will rearrange one equation so that one of the variables, for example y, is the subject of the equation (i.e it is on one side of the equation with only x terms on the other).
Looking at the two equations, the first one is already in this form.
So now we can substitute y in terms of x into the second equation, giving:
(x+3)2 - x2 + 3 = -6x
Now we have an equation in terms of one variable, x, which we should be able to solve.
Expand out the terms and group like terms so that we have the equation in a form that we know we can solve from:
x2 + 3x + 3x + 9 - x2 + 3 = -6x
Collecting like terms and simplyfying (x, x2 and constants) on the Left:
6x + 12 = -6x
Bringing all terms to the LHS:
12x + 12 = 0
Factorising out the 12:
12(x + 1) = 0
Now, x + 1 = 0 as 12 doesn't equal zero
x = - 1
Substituing into the simpler equation for y (y = x + 3) gives:
y = - 1 + 3 = 2
We have now finished as we have the x and the corresponding y value that satisfy the simultaneous equations.see more