simulatneous equations are solved by eliminating one of the unknowns such as x or y. there are a number of ways to do this such as substitution or subtraction/addition. For this example we will use subtraction addition as i always found that the easier method at GCSE since it often avoids fractions.

e.g. 2y=3x+12 and 3y=5x+14

we need to choose one of the two unknowns to eliminate and then find their lowest common denominator. I will choose x, where the LCD is 15 (3*5=15). now we multiply the entire equation through in order that the number in front of x (called the coefficient) is 15. in the first equation, we multiply through by 5.

(2*5)y=(3*5)x+(12*5)

10y=15x+60

in the second equation we multiply through by 3 like before

(3*3)y=(5*3)x+(14*3)

9y=15x+42

now both the coefficients of x are 15 so we can subtract one equation from the other

(10-9)y=(15-15)x+(60-42)

y=0x+18

y=18

we now have the y value, since x is now zero. to find the x value all we need to do is go back to one of the first equations and put the value y=18 into the equation to resolve for x.

(2*18)=3x+12

36-12=3x

24=3x

x=8

and thats everything. By making sure we can subtract one equation from another to eliminate one of the unkowns we can subsequently obtain both values. in this case, y=18 and x=8