Answer
\[{\text{the solution set is }}\emptyset \]
Work Step by Step
\[\begin{gathered}
\left\{ {\begin{array}{*{20}{c}}
{6x + y = - 3} \\
{12x + 2y = 1}
\end{array}} \right. \hfill \\
{\text{Given the system}} \hfill \\
\left\{ {\begin{array}{*{20}{c}}
{{a_1}x + {b_1}y = {c_1}} \\
{{a_2}x + {b_2}y = {c_2}}
\end{array}} \right. \hfill \\
D = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{b_1}} \\
{{a_2}}&{{b_2}}
\end{array}} \right|,\,\,\,{D_x} = \left| {\begin{array}{*{20}{c}}
{{c_1}}&{{b_1}} \\
{{c_2}}&{{b_2}}
\end{array}} \right|,\,\,{D_y} = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{c_1}} \\
{{a_2}}&{{c_2}}
\end{array}} \right| \hfill \\
{\text{First find }}D,{\text{ If }}D \ne 0,\,{\text{ then find }}{D_x}{\text{ and }}{D_y} \hfill \\
D = \left| {\begin{array}{*{20}{c}}
6&1 \\
{12}&2
\end{array}} \right| = 12 - 12 = 0 \hfill \\
{\text{Using another method to determine the solution set}} \hfill \\
\left\{ {\begin{array}{*{20}{c}}
{6x + y = - 3\,\,\left( {\mathbf{1}} \right)} \\
{12x + 2y = 1\,\,\left( {\mathbf{2}} \right)}
\end{array}} \right. \hfill \\
{\text{Multiply the equation }}\left( {\mathbf{1}} \right){\text{ by }} - 2 \hfill \\
\left\{ {\begin{array}{*{20}{c}}
{ - 12x - 2y = 6\,} \\
{12x + 2y = 1\,}
\end{array}} \right. \hfill \\
{\text{Add both equations}} \hfill \\
\underline {\left\{ {\begin{array}{*{20}{c}}
{ - 12x - 2y = 6\,} \\
{12x + 2y = 1\,}
\end{array}} \right.} \hfill \\
0 + 0 = 7 \hfill \\
0 = 7 \hfill \\
{\text{Then,}} \hfill \\
{\text{The graphs are parallel lines}},{\text{ so there is no solution}} \hfill \\
{\text{and the solution set is }}\emptyset \hfill \\
\end{gathered} \]
