Dividing the first equation by -3 we get:
[tex]\begin{gathered} \frac{15y-3}{-3}=\frac{-3x}{-3}, \\ \frac{15y}{-3}-\frac{3}{-3}=\frac{-3x}{-3}, \\ -5y+1=x\text{.} \end{gathered}[/tex]Substituting x=-5y+1 in the second equation we get:
[tex]6(-5y+1)-2y=6.[/tex]Applying the distributive property we get:
[tex]\begin{gathered} 6\times(-5y)+6\times1-2y=6, \\ -30y+6-2y=6. \end{gathered}[/tex]Adding like terms we get:
[tex]-32y+6=6.[/tex]Subtracting 6 from the above equation we get:
[tex]\begin{gathered} -32y+6-6=6-6, \\ -32y=0. \end{gathered}[/tex]Dividing the above equation by -32 we get:
[tex]\begin{gathered} \frac{-32y}{-32}=\frac{0}{-32}, \\ y=0. \end{gathered}[/tex]Finally, substituting y=0 in x=-5y+1 we get:
[tex]\begin{gathered} x=-5\cdot0+1 \\ =0+1=1. \end{gathered}[/tex]Answer: The solution to the given system of equations is:
[tex]x=1\text{ and y=0.}[/tex]