#color(red)y=color(white)(aa)3x+11#
#color(red)y=-2x+1#
Because #3x+11# and #-2x+1# both equal #color(red)y#, they must be equal to each other.
#color(white)(a^1)3x+11=-2x+1#
#+2xcolor(white)(aaaaaaa)+2xcolor(white)(aaaaaaa)# Add #2x# to both sides
#color(white)(a^1)5x+11 =color(white)(aaa)1#
#color(white)(aaaa)-11color(white)(aa)-11color(white)(aaaa)# Subtract #11# from both sides
#color(white)(aa)(5x)/5color(white)(aaaa)=(-10)/5color(white)(aaa)# Divide both sides by #5#
#color(white)(aa)x=-2#
To find #y#, substitute #x=color(blue)(-2)# into either of the original equations.
#y=3color(blue)x+11#
#y=3(color(blue)(-2))+11#
#y=-6+11#
#y=5#
The solution #(color(blue)x,color(red)y)# is #(color(blue)(-2),color(red)5)#
