\(2x+3y=1 \\
-3x+y=15 \\ \\
\hbox{from the 2nd equation:} \\
-3x+y=15 \\
y=15+3x \\ \\
\hbox{substitute 15+3x for y in the 1st equation:} \\
2x+3(15+3x)=1 \\
2x+45+9x=1 \\
9x+2x=1-45 \\
11x=-44 \\
x=-4 \\ \\
\hbox{solve one of the equations for y}: \\
2x+3y=1 \\
2 \cdot (-4) + 3y=1 \\
-8+3y=1 \\
3y=1+8 \\
3y=9 \\
y=3 \\ \\
\hbox{the answer:} \\
x=-4 \\
y=3\)
