Answer
\[\left\{ {\left( { - 4,12} \right)} \right\}\]
Work Step by Step
\[\begin{gathered}
\left\{ {\begin{array}{*{20}{c}}
{\frac{1}{2}x + \frac{1}{3}y = 2} \\
{\frac{3}{2}x - \frac{1}{2}y = - 12}
\end{array}} \right. \hfill \\
{\text{Given the system}} \hfill \\
\left\{ {\begin{array}{*{20}{c}}
{{a_1}x + {b_1}y = {c_1}} \\
{{a_2}x + {b_2}y = {c_2}}
\end{array}} \right. \hfill \\
D = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{b_1}} \\
{{a_2}}&{{b_2}}
\end{array}} \right|,\,\,\,{D_x} = \left| {\begin{array}{*{20}{c}}
{{c_1}}&{{b_1}} \\
{{c_2}}&{{b_2}}
\end{array}} \right|,\,\,{D_y} = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{c_1}} \\
{{a_2}}&{{c_2}}
\end{array}} \right| \hfill \\
{\text{First find }}D,{\text{ If }}D \ne 0,\,{\text{ then find }}{D_x}{\text{ and }}{D_y} \hfill \\
D = \left| {\begin{array}{*{20}{c}}
{1/2}&{1/3} \\
{3/2}&{ - 1/2}
\end{array}} \right| = - \frac{1}{4} - \frac{1}{2} = - \frac{3}{4} \hfill \\
{D_x} = \left| {\begin{array}{*{20}{c}}
2&{1/3} \\
{ - 12}&{ - 1/2}
\end{array}} \right| = - 1 + 4 = 3 \hfill \\
{D_y} = \left| {\begin{array}{*{20}{c}}
{1/2}&2 \\
{3/2}&{ - 12}
\end{array}} \right| = - 6 - 3 = - 9 \hfill \\
{\text{Using the Cramer's rule}} \hfill \\
x = \frac{{{D_x}}}{D} = \frac{3}{{ - 3/4}} = - 4,\,\,\,\,\,\,\,\,\,y = \frac{{{D_y}}}{D} = \frac{{ - 9}}{{ - 3/4}} = 12 \hfill \\
{\text{The solution set is}} \hfill \\
\left\{ {\left( { - 4,12} \right)} \right\} \hfill \\
\end{gathered} \]
