Roxane Z.

Currently unavailable: for new students

Mathematics and Philosophy (Bachelors) - Warwick University

5.0

185 completed lessons

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Ratings & Reviews

5from 80 customer reviews

Dan (Student)

January 23 2018

Another great lesson as always thanks !!

Dan (Student)

June 27 2017

Roxane is a brilliant french tutor she is very patient and understanding,she is also great at simplifying things that confuse me so i can understand them. Thank you !!

Dan (Student)

April 18 2018

Dan (Student)

April 10 2018

Qualifications

MathematicsInternational Baccalaureate (IB) (HL)7
FrenchInternational Baccalaureate (IB) (HL)6
PhysicsInternational Baccalaureate (IB) (HL)7
ChemistryInternational Baccalaureate (IB) (SL)7
EnglishInternational Baccalaureate (IB) (SL)6
GermanInternational Baccalaureate (IB) (SL)7

Subjects offered

SubjectQualificationPrices
FrenchA Level£24 /hr
FrenchGCSE£22 /hr
MathsGCSE£22 /hr

How do I solve simultaneous equations?

Always follow the same three steps : equating – substracting – substituting but you can have different cases :

Case 1: we only need to multiply one equation

Example:
2x+5y=9
6x-4y=8

Note : Always check first that the equation is solvable (= there are as many or more equations than there are unknowns.)
We notice that 6x is a multiple of 2x. So we only need to multiply the second equation to equate the coefficients.
Substracting the 2nd equation from 3 times the first one gives 19y=19 so y=1
Now we substitute into the first equation to get
2x+5*1=9
2x=4
x=2

The solution is x=2, y=1

Case 2 : we need to multiply both equations

Example:
3x+5y = 14
4x+3y=15

We proceed the same way as in the first case but to equate coefficients we need to multiply both equations, because the coefficients are not multiples of each other.
Multiplying the first equation by 3 and substracting 5 times the second gives 11x=33 so x=3.
Now we substitute back in to get
9+5y=14
5y=5
y=1

The solution is x=3, y=1

Always follow the same three steps : equating – substracting – substituting but you can have different cases :

Case 1: we only need to multiply one equation

Example:
2x+5y=9
6x-4y=8

Note : Always check first that the equation is solvable (= there are as many or more equations than there are unknowns.)
We notice that 6x is a multiple of 2x. So we only need to multiply the second equation to equate the coefficients.
Substracting the 2nd equation from 3 times the first one gives 19y=19 so y=1
Now we substitute into the first equation to get
2x+5*1=9
2x=4
x=2

The solution is x=2, y=1

Case 2 : we need to multiply both equations

Example:
3x+5y = 14
4x+3y=15

We proceed the same way as in the first case but to equate coefficients we need to multiply both equations, because the coefficients are not multiples of each other.
Multiplying the first equation by 3 and substracting 5 times the second gives 11x=33 so x=3.
Now we substitute back in to get
9+5y=14
5y=5
y=1

The solution is x=3, y=1

3 years ago

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