Answer
$$\eqalign{
& {\text{Infinitely many solutions}}{\text{.}} \cr
& \left( {\frac{{6 - 2y}}{7},y} \right) \cr} $$
Work Step by Step
$$\eqalign{
& 7x + 2y\,\,\,\, = \,6\,\,\,\,\,\,\,\left( {\bf{1}} \right) \cr
& 14x + 4y = 12\,\,\,\,\left( {\bf{2}} \right) \cr
& {\text{Multiply the equation }}\left( {\bf{1}} \right){\text{ by }} - {\text{2}} \cr
& - 14x - 4y\,\,\,\, = \, - 12 \cr
& \,\,\,\,14x + 4y\,\,\, = \,\,12 \cr
& {\text{Add both equations}} \cr
& - 14x - 4y\,\,\,\, = \, - 12 \cr
& \underline {\,\,\,\,14x + 4y\,\,\, = \,\,12} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 = 0\,\,\,\,\,\,\,\,\,\,\, \cr
& {\text{The result}},{\text{ }}0 = 0,{\text{ is a true statement}},{\text{ which indicates that the }} \cr
& {\text{equations are equivalent}}{\text{. Therefore,}} \cr
& {\text{The system has infinitely many solutions}}{\text{.}} \cr
& \cr
& {\text{Solve the equation }}\left( {\bf{1}} \right){\text{ for }}x \cr
& 7x + 2y = \,6\,\, \cr
& y = \frac{{6 - 2y}}{7} \cr
& \cr
& {\text{The solutions of the system can be written in the form of a set }} \cr
& {\text{of ordered pairs }}\left( {\frac{{6 - 2y}}{7},y} \right) \cr} $$
