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 .
