Answer
$$\eqalign{
& {\text{Infinitely many solutions}}{\text{.}} \cr
& \left( {\frac{{ - 5y - 2}}{3},y} \right) \cr} $$
Work Step by Step
$$\eqalign{
& 3x\,\, + \,\,\,\,5y\,\, = - 2\,\,\,\,\,\left( {\bf{1}} \right) \cr
& 9x + \,\,15y\,\, = - 6\,\,\,\,\,\left( {\bf{2}} \right) \cr
& {\text{Multiply the equation }}\left( {\bf{1}} \right){\text{ by }} - 3 \cr
& - 9x - 15\,\,y\,\, = 6 \cr
& \,\,\,\,9x + \,\,15y\,\, = - 6 \cr
& {\text{Add both equations}} \cr
& - 9x - 15\,\,y\,\, = 6 \cr
& \underline {\,\,\,\,9x + \,\,15y\,\, = - 6} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 = 0\,\,\,\,\,\,\,\,\,\,\, \cr
& {\text{The result}},{\text{ }}0 = 0,{\text{ is a true statement}},{\text{ which indicates that the }} \cr
& {\text{equations are equivalent}}{\text{. Therefore,}} \cr
& {\text{The system has infinitely many solutions}}{\text{.}} \cr
& {\text{Solve the equation }}\left( {\bf{1}} \right){\text{ for }}x \cr
& 3x\,\, + \,\,\,\,5y\,\, = - 2 \cr
& x = \frac{{ - 5y - 2}}{3} \cr
& \cr
& {\text{The solutions of the system can be written in the form of a set }} \cr
& {\text{of ordered pairs }}\left( {\frac{{ - 5y - 2}}{3},y} \right) \cr
& \cr
& {\text{Infinitely many solutions}}{\text{.}} \cr
& \left( {\frac{{ - 5y - 2}}{3},y} \right) \cr} $$
