Step 1) Because the second equation is already solved for #y# we can substitute #(4x + 4)# for #y# in the first equation and solve for #x#:
#3x + 2y = 12# becomes:
#3x + 2(4x + 4) = 12#
#3x + (2 xx 4x) + (2 xx 4) = 12#
#3x + 8x + 8 = 12#
#(3 + 8)x + 8 = 12#
#11x + 8 = 12#
#11x + 8 - color(red)(8) = 12 - color(red)(8)#
#11x + 0 = 4#
#11x = 4#
#(11x)/color(red)(11) = 4/color(red)(11)#
#(color(red)(cancel(color(black)(11)))x)/cancel(color(red)(11)) = 4/11#
#x = 4/11#
Step 2) Substitute #4/11# for #x# in the second equation and solve for #y#:
#y = 4x + 4# becomes:
#y = (4 xx 4/11) + 4#
#y = 16/11 + 4#
#y = 16/11 + (4 xx 11/11)#
#y = 16/11 + 44/11#
#y = (16 + 44)/11#
#y = 60/11#
The Solution Is:
#x = 4/11# and #y = 60/11#
Or
#(4/11, 60/11)#
