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Degree: Economics BSc (Hons) (Bachelors) - Newcastle University
I am currently in my second year of an Economics Degree at Newcastle University. Maths was always my favourite subject at school and I am very passionate about it. I was in the top set for maths at a selective grammar school and took part in numerous maths challenges throughout my time there.
I am fun and approachable and my aim is to try and make maths as enjoyable as possible, as in my opinion that’s the best way to learn. I previously volunteered at my school’s homework club for students struggling with academic life, allowing me to tutor these students in a range of subjects, particularly in maths and science. I have also individually tutored a student for two years in preparation for GCSE Maths, helping him engage with the subject and improve his grades.
I am also avid sports fan and I enjoy reading.
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x2 + y2 = 36 (1)
x = 2y + 6 (2)
You can solve simulateous equations in three ways: Substitution, addition or subtraction. If one equation contains both x2 and y2 then you should use substitution as with this example.
1.) Substitute equation (2) into equation (1):
x = 2y + 6 therefore x2 = (2y + 6)2 so
(2y + 6)2 + y2 = 36
2.) Expand the brackets:
(2y + 6)(2y + 6) + y2 = 36
4y2 + (2y x 6) + (2y x 6) + 36 + y2 = 36
4y2 + 12y + 12y + 36 + y2 = 36
3.) Simplify the equation:
4y2 + 24y + 36 + y2 = 36
4.) Subtract 36 from both sides and add the 4y2 and y2:
5y2 + 24y = 0
This is now a quadratic equation. It can either be solved using the quadratic formula or through factorising. I shall do the factorising method.
5.) Take out y from both 5y2 and 24y:
y(5y + 24) = 0 y = (y + 0)
6.) Find the values of y that would make inside the bracket become zero:
(y + 0) = 0 (5y + 24) = 0
y = 0 5y = -24
y = -24/5 = -4.8
7.) Substitute the y values into the equation to find the corresponding x values:
y = 0 y = -4.80
x = (2 x 0) + 6 x = (2 x (-4.8)) +6
x = 6.00 x = (-9.6) + 6
x = -3.60
PS - Don't forget the 2 decimal places at the endsee more