If #y=4-4x#, then we can replace the #color(blue)(y)# in #4x+color(blue)(y)=4# with #4-4x#:
#4x+(4-4x)=4#
Solve for #x#
#4x+4-4x=4#
subtract #4# on both sides
#4x-4x=0#
#0x=0#
#x=oo#
#color(white)(0)#
Let's try solving for #y# now.
We need to isolate #x#:
#4x+y=4#
#4x=4-y#
#x=(4-y)/4#
Now, let's replace #color(blue)(x)# with #(4-y)/4# in #y=4-4color(blue)(x)#
#y=4-cancel(4)(color(black)((4-y)/cancel(4)))#
#y=4-4-y#
#0y=0#
#y=oo#
The reason #x# and #y# both equal #oo# is that any number, multiplied by #0#, equals #0#. So, #x# can equal #-500#, #8/19#, or #9.01xx10^13#, and still equal #0#, thus making the equation true. The same fact also applies for #y#.
Both #x# and #y# equal all real numbers
