Question 1125261

given system:

{{{4x-8=4y}}} ....eq.1
{{{4y+8=4x}}}....eq.2
-------------------- ----simplify, both sides in both equations divide by {{{4}}}

{{{x-2=y}}} ....eq.1
{{{y+2=x}}}....eq.2..->........solve or {{{y}}}
--------------------------

{{{x-2=y}}} ....eq.1
{{{x-2=y}}}....eq.2->........multiply by {{{-1}}}
---------------------------


{{{x-2=y}}} ....eq.1
{{{-x+2=-y}}}....eq.2
----------------------------add both eq.1 and eq.2

{{{x-2+(-x+2)=y+(-y)}}} 

{{{x-2-x+2=y-y}}} 

{{{0=0}}} 


 So we're left with {{{0=0}}} which means any {{{x}}} or {{{y}}} value will satisfy the system of equations. So there are an {{{infinite}}} number of {{{solutions}}} (we have coplanar lines, two lines that lie exactly on top of each other, actually it's same line), and this system is {{{dependent}}}.