How do you solve #4x + y = 4# and #y = 4 - 4x# using substitution?

1 Answer
May 12, 2017

Answer:

Both #x# and #y# equal all real numbers

Explanation:

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