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 2^nd 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

1) 3x + y = 11 2) 2x + y = 8

Related Maths GCSE answers

Factorise 2x^3=10x+12x^2

Make F the subject of the formula: C= 5(F-32) / 9

How do I solve a simultaneous equation?

How do I solve simultaneous equations?