Parent review - it's a relief to see my son start to understand GCSE maths topics that he has struggled with. His confidence is growing with Chris's help. Chris isn't phased by any questions or topics thrown at him, and is able to clearly explain the methods needed to answer maths questions.

Kelly, Parent from Kent

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Usually we cover both subject knowledge and exam technique, although that can change depending on each individual student. Then we go through diagrams, and they ask questions, and we go from there.

Lots of students say that the classes are too big in school, or that they don't have time to ask teachers after lessons. In my tutorials, we take time to explore things in a little in a bit more detail.

I always look up the board my students are taking so the lessons are really relevant. Then we go through past papers or set texts, whatever the student finds helpful.

I use the shared whiteboard. We make diagrams together and label them, and often the student prints it off because they know it's right and they completely understand it.

After tutoring one girl went and told all her friends the new explanation I gave her. And she was so excited about what she wrote in the exam she emailed me immediately afterwards.

There was one girl who had her exam on Monday. She wanted tuition on Friday, Saturday and Sunday beforehand. It was very intense, but she said the exam went well.

The quadratic equation is x=(-b+-SQRT(b^2-4ac))/2a.
In this instance, a = 3, b = 2, and c = 15. Simply putting the numbers in place of the letter counterpart, gives an answer of 1.93 and -2.59 to two decimal places.

Answered by Callum F.

Studies MEng Civil Engineering at Bath

First of all, you would label the equations, so let's call 3x +4y = 18 equation A, and 5x - 2y = 4 equation B.
Then you would try to find a way to eliminate one of the variables (either x or y). We can do this by multiplying equation B by 2. Therefore the new equation 2B would equal 10x -4y = 8.
We can then eliminate the variable 'y' by adding equation A and 2B together. We would add each individual component, therefore the 'xs' would add together to make (3x+10x) 13x, the 'ys' would add together to make (4x-4x) 0, and the integers would add together to make (18 +8) 26.
You would then have the equation 13x = 26 which can then be solved for x (dividing both sides of the equation by 13) to get x = 2.
This value for x can then be subbed into either equation A or B in order to find the value of y. Therefore subbing into equation A would give 3(2) + 4y = 18, which rearranging and solving would give a value of y = 3.
You can then check your results by subbing both x and y values into the other equation, and seeing if the equation works:
5(2) - 2(3) = 4, which does indeed satisfy the equation.
Therefore x = 2 and y = 3

Answered by Rob S.

Studies Aerospace Engineering at Sheffield

I have labelled the two separate equations A and B so that it is easier to talk about them. There are two ways in which you can do these equations but I am going to explain the method using substitution.
As y and x are in both of the equations we can try to eliminate at least one of these unknowns for the moment. So, if we rearrange [B] so that y=2x-1 we can then substitute this value of y into [A]. This will give us:
2(2x-1) + x =8.
By multiplying this out we get:
4x-2+x=8
By grouping the x values together:
5x-2=8
Then placing all the unknowns to one side of the equation
5x=10 and then dividing both sides by 5 we get:
x=2.
So we have found the value for x! We would then substitute this into [B]:
1+y=2(2),then multiplying this out
1+y=4, placing all the unknowns onto one side:
y=3
So we have a solution for y!
Just to check that our answers are correct we can substitute our two values into [A]:
2(3)+(2)=6+2
=8
and 8 is the correct answer so we know our solutions are correct!

Answered by Ciara D.

Studies BA English Literature and French at Southampton

Although this may look more complicated than normal, this is just a normal qudratic equation. First by rearranging the equation we can get it into a simpler form, and then we can go about solving it.
The first thing to remember is whatever operation you apply to one side of the equation, you must do to the other! So first of all, we can multiply the whole equation by 2 to get simplify the x/2 term.
The equation becomes:
2x^2 +x=10
Now we can subtract 10 from both sides, to bring all of the terms to one side.
The equation becomes:
2x^2 +x -10=0
This is now much nicer to work with!
Now to solve the equation, we can use two methods. We can either use the quadratic formula (whereby we plug all of the coefficients into the equation to find the roots) or we can factorise it.
Luckily this equation factorises nicely into:
(x-2)(2x+5)=0
Since one of the brackets must equal 0, this gives the solutions:
x=-5/2 or x=2

Answered by Hannah J.

Studies Mathematics at Durham

15:
(x+y+z)/3 = 6 so, x+y+z = 18.Range = highest number - lowest number.
For the max range, at least 1 of the numbers must be 1.
1,1,16. 16-1 = 15

Step 1: Start off by labeling the 2 equations.
3 x + y = 7 will be equation 1.
3 x - y = 5 will be equation 2.
Step 2: rearrange equation 1 to make y the subject.
3 x + y = 7 --->
y = 7 - 3 x (here 3 x is now negative as it has switched sides, in other words 3 x has been subtracted from both sides of the equation).
Step 3: rearrange equation 2 to make y the subject.
3 x - y = 5 --->
- y = 5 - 3 x --->
y = -5 + 3 x --->
y = 3 x - 5 --->
*** top tip *** keep your '=' signs in line with each other for nicer presentation and to stop you getting muddled!
so now y = 7 - 3x and y = 3 x - 5
step 4: substitution
If y = 7 - 3 x and 3 x - 5
then:
7 - 3 x = 3 x - 5.
Step 5: put all your like terms on the same side .
7 + 5 = 3 x + 3 x --->
12 = 6x --->
(12/6) = x --->
x = 2
step 6: substitute x=2 into equation 1.
y = 7 - 3x --->
y = 7 - 3(2)--->
y = 7 - 6 --->
so y = 1 .
Answer: x = 2 and y = 1
Step 7 is optional but definitely worth it!!! - check your answer by substituting x = 2 into the equation 2 as well and see if you get the same answer.
y = 3 x - 5 --->
y = 3(2) - 5 --->
y = 6 - 5
again y = 1 so you can be sure you have got the right answer!

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With such a variety of questions, GCSE Maths can be a real headache for students. In individual sessions with a GCSE Maths tutor you can focus on exactly the topics you need help with. In our online classroom, tutors use diagrams, graphs and illustrations to enhance their lessons.

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