Answer :
[tex]x=6y+30 \\
x=\frac{3}{2}y \\ \\
\hbox{substitute } \frac{3}{2}y \hbox{ for x in the 1st equation:} \\
\frac{3}{2}y=6y+30 \\
\frac{3}{2}y-6y=30 \\
3y-12y=60 \\
-9y=60 \\
y=-\frac{60}{9}=-\frac{20}{3}=-6\frac{2}{3} \\ \\
x=\frac{3}{2} \cdot (-\frac{20}{3})=-\frac{20}{2}=-10 \\ \\
\left \{ {{x=-10} \atop {y=-6\frac{2}{3}}} \right.[/tex]