Joseph P. GCSE Maths tutor

Joseph P.

£18 /hr

Currently unavailable: for regular students

Studying: Economics BSc (Hons) (Bachelors) - Newcastle University

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About me

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.

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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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A
ChemistryA-level (A2)A
BiologyA-level (A2)A

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Subjects offered

SubjectQualificationPrices
MathsGCSE£18 /hr

Questions Joseph has answered

Solve the equations giving your answer in 2d.p (5 Marks).: x^2 + y^2 = 36 , x = 2y + 6

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

1 mark

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

1 mark

3.) Simplify the equation:

4y2 + 24y + 36 + y2 = 36

1 mark

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)

1 mark

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

1 mark

PS - Don't forget the 2 decimal places at the end

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

1 mark

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

1 mark

3.) Simplify the equation:

4y2 + 24y + 36 + y2 = 36

1 mark

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)

1 mark

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

1 mark

PS - Don't forget the 2 decimal places at the end

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2 years ago

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