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

RZ
Answered by Roxane Z. Maths tutor

5093 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Triangle PQR is a right angled triangle. PQ is the hypotenuse and is 5cm long and QR has length 3cm. What is the length of side PR? Show your working.


Draw the graph y=4x+2


Show x^2 + 8x +15 = 0 in the form of (x+b)^2 +c (complete the square) and then solve the equation


Whats the inverse of y = 2x+1/x-1 ?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning