Answer
$$\eqalign{
& {\text{Infinitely many solutions}}{\text{.}} \cr
& \left( {4y + 2\,,y} \right) \cr} $$
Work Step by Step
$$\eqalign{
& \,2x - 8y\, = 4\,\,\,\left( {\bf{1}} \right)\,\, \cr
& x - 4y = 2\,\,\,\,\,\,\left( {\bf{2}} \right)\, \cr
& \cr
& {\text{Multiply the equation }}\left( {\bf{2}} \right){\text{ by }} - 2 \cr
& \,\,\,2x - 8y\, = 4 \cr
& - 2x + 8y = - 4 \cr
& {\text{Add both equations}} \cr
& \,\,\,2x - 8y\, = 4 \cr
& \underline { - 2x + 8y = - 4} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 = 0\,\,\,\,\,\,\,\,\,\,\, \cr
& {\text{The result}},{\text{ }}0 = 0,{\text{ is a true statement}},{\text{ which indicates that the }} \cr
& {\text{equations are equivalent}}{\text{. Therefore,}} \cr
& {\text{The system has infinitely many solutions}}{\text{.}} \cr
& \cr
& {\text{Solve the equation }}\left( {\bf{2}} \right){\text{ for }}x \cr
& x - 4y = 2\, \cr
& x = 4y + 2\, \cr
& \cr
& {\text{The solutions of the system can be written in the form of a set }} \cr
& {\text{of ordered pairs }}\left( {4y + 2\,,y} \right) \cr} $$
