Answer
\[\left\{ {\left( {0, - 2} \right)} \right\}\]
Work Step by Step
\[\begin{gathered}
\left\{ {\begin{array}{*{20}{c}}
{3x + 2y = - 4} \\
{5x - y = 2}
\end{array}} \right. \hfill \\
{\text{Given the system}} \hfill \\
\left\{ {\begin{array}{*{20}{c}}
{{a_1}x + {b_1}y = {c_1}} \\
{{a_2}x + {b_2}y = {c_2}}
\end{array}} \right. \hfill \\
D = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{b_1}} \\
{{a_2}}&{{b_2}}
\end{array}} \right|,\,\,\,{D_x} = \left| {\begin{array}{*{20}{c}}
{{c_1}}&{{b_1}} \\
{{c_2}}&{{b_2}}
\end{array}} \right|,\,\,{D_y} = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{c_1}} \\
{{a_2}}&{{c_2}}
\end{array}} \right| \hfill \\
{\text{First find }}D,{\text{ If }}D \ne 0,\,{\text{ then find }}{D_x}{\text{ and }}{D_y} \hfill \\
D = \left| {\begin{array}{*{20}{c}}
3&2 \\
5&{ - 1}
\end{array}} \right| = - 3 - 10 = - 13 \hfill \\
{D_x} = \left| {\begin{array}{*{20}{c}}
{ - 4}&2 \\
2&{ - 1}
\end{array}} \right| = 4 - 4 = 0 \hfill \\
{D_y} = \left| {\begin{array}{*{20}{c}}
3&{ - 4} \\
5&2
\end{array}} \right| = 6 + 20 = 26 \hfill \\
{\text{Using the Cramer's rule}} \hfill \\
x = \frac{{{D_x}}}{D} = \frac{0}{{ - 13}} = 0,\,\,\,\,\,\,\,\,\,y = \frac{{{D_y}}}{D} = \frac{{26}}{{ - 13}} = - 2 \hfill \\
{\text{The solution set is}} \hfill \\
\left\{ {\left( {0, - 2} \right)} \right\} \hfill \\
\end{gathered} \]
