Step 1) Because each equation is already solved for #y# we can equate the right sides of each equation and solve for #x#:
#-3x - 12 = x + 4#
#color(red)(3x) - 3x - 12 - color(blue)(4) = color(red)(3x) + x + 4 - color(blue)(4)#
#-16 = color(red)(3x) + 1x + 0#
#-16 = (color(red)(3) + 1)x#
#-16 = 4x#
#-16/color(red)(4) = (4x)/color(red)(4)#
#-4 = (color(red)(cancel(color(black)(4)))x)/cancel(color(red)(4))#
#-4 = x#
#x = -4#
Step 2) Substitute #-4# for #x# into either equation and calculate #y#. I will use the second equation because it has less operands:
#y = x + 4# becomes:
#y = -4 + 4#
#y = 0#
The solution is: #x = -4# and #y = 0# or #(-4, 0)#
