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.
in the second equation we multiply through by 3 like before
now both the coefficients of x are 15 so we can subtract one equation from the other
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.
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