Answer
\[\left\{ {\left( { - 2,1} \right)} \right\}\]
Work Step by Step
\[\begin{gathered}
\left\{ {\begin{array}{*{20}{c}}
{3x + 2y = - 4} \\
{2x - y = - 5}
\end{array}} \right. \hfill \\
{\text{Given the system}} \hfill \\
\left\{ {\begin{array}{*{20}{c}}
{{a_1}x + {b_1}y = {c_1}} \\
{{a_2}x + {b_2}y = {c_2}}
\end{array}} \right. \hfill \\
D = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{b_1}} \\
{{a_2}}&{{b_2}}
\end{array}} \right|,\,\,\,{D_x} = \left| {\begin{array}{*{20}{c}}
{{c_1}}&{{b_1}} \\
{{c_2}}&{{b_2}}
\end{array}} \right|,\,\,{D_y} = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{c_1}} \\
{{a_2}}&{{c_2}}
\end{array}} \right| \hfill \\
{\text{First find }}D,{\text{ If }}D \ne 0,\,{\text{ then find }}{D_x}{\text{ and }}{D_y} \hfill \\
D = \left| {\begin{array}{*{20}{c}}
3&2 \\
2&{ - 1}
\end{array}} \right| = - 3 - 4 = - 7 \hfill \\
{D_x} = \left| {\begin{array}{*{20}{c}}
{ - 4}&2 \\
{ - 5}&{ - 1}
\end{array}} \right| = 4 + 10 = 14 \hfill \\
{D_y} = \left| {\begin{array}{*{20}{c}}
3&{ - 4} \\
2&{ - 5}
\end{array}} \right| = - 15 + 8 = - 7 \hfill \\
{\text{Using the Cramer's rule}} \hfill \\
x = \frac{{{D_x}}}{D} = \frac{{14}}{{ - 7}} = - 2,\,\,\,\,\,\,\,\,\,y = \frac{{{D_y}}}{D} = \frac{{ - 7}}{{ - 7}} = 1 \hfill \\
{\text{The solution set is}} \hfill \\
\left\{ {\left( { - 2,1} \right)} \right\} \hfill \\
\end{gathered} \]
