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

Answered by Roxane Z. Maths tutor

4382 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

In a class there are 12 girls and 18 boys. what proportion of the girls are there in the class? Simplify


A and B are on the line 3x+2y=6. At A x=0, what is y? At B y=0, what is x?


If a student wishes to have a ratio of 2:7 red pens to yellow pens in their pencil case: a) if they have 50 pens total what is the maximum amount they can carry with them b) if they have 18 red and 31 yellow what is the maximum amount they can carry


How can you differentiate when to use SohCahToa and when to use the sine/cosine rules?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences