We start with:

#12x + 7y = 3#,

#4x + 3y = -2#.

There are (as far as I know) three methods of solving simultaneous equations. I'll choose substitution. Therefore we must leave either #x# or #y# alone. I'll choose #x# in the second equation (you can choose any of the 2 variables in any of the 2 equations):

#4x + 3y = -2#,

#4x = -3y -2#,

#x = (-3y - 2)/4#.

We now substitute #x# in the other equation such that:

#12x + 7y = 3#,

#12((-3y - 2)/4) + 7y = 3#, 12 and 4 cancel out,

#3(-3y - 2) + 7y = 3#,

#-9y -6 +7y = 3#, we now add #y#s and pass the #6# to the other side,

#-2y = 9#,

#2y = -9#,

#y = -4.5#.

We now substitute #y# by this value in any of the 2 equations. I'll choose the second one:

#4x + 3(-4.5) = -2#,

#4x - 13.5 = -2#, we now pass #-13.5# to the other side,

#4x = 11.5#,

#x = 2.875#.

Hope it Helps! :D .