Given:
[tex]\begin{gathered} 9x-3y=-6 \\ 5y=15x+10 \end{gathered}[/tex]To determine the solutions of the given system of equations, we first solve for x in 9x-3y=-6:
[tex]\begin{gathered} 9x-3y=-6 \\ \text{Simplify and rearrange} \\ 9x=-6+3y \\ x=\frac{3y-6}{9} \\ \\ x=\frac{y-2}{3} \end{gathered}[/tex]Next, we substitute x =(y-2)/3 into 5y=15x+10. So,
[tex]\begin{gathered} 5y=15x+10 \\ 5y=15(\frac{y-2}{3})+10 \\ \text{Simplify} \\ 5y=5(y-2)+10 \\ 5y=5y-10+10 \\ 5y=5y \end{gathered}[/tex]Since 5y=5y is redundant information, this is a dependent system. Therefore, the given system of equations have infinitely many solutions.
