Answer
\[\left\{ {\left( { - 2,5} \right)} \right\}\]
Work Step by Step
\[\begin{gathered}
\left\{ {\begin{array}{*{20}{c}}
{3x + y = - 1} \\
{5x + 4y = 10}
\end{array}} \right. \hfill \\
{\text{Given the system}} \hfill \\
\left\{ {\begin{array}{*{20}{c}}
{{a_1}x + {b_1}y = {c_1}} \\
{{a_2}x + {b_2}y = {c_2}}
\end{array}} \right. \hfill \\
D = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{b_1}} \\
{{a_2}}&{{b_2}}
\end{array}} \right|,\,\,\,{D_x} = \left| {\begin{array}{*{20}{c}}
{{c_1}}&{{b_1}} \\
{{c_2}}&{{b_2}}
\end{array}} \right|,\,\,{D_y} = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{c_1}} \\
{{a_2}}&{{c_2}}
\end{array}} \right| \hfill \\
{\text{First find }}D,{\text{ If }}D \ne 0,\,{\text{ then find }}{D_x}{\text{ and }}{D_y} \hfill \\
D = \left| {\begin{array}{*{20}{c}}
3&1 \\
5&4
\end{array}} \right| = 12 - 5 = 7 \hfill \\
{D_x} = \left| {\begin{array}{*{20}{c}}
{ - 1}&1 \\
{10}&4
\end{array}} \right| = - 4 - 10 = - 14 \hfill \\
{D_y} = \left| {\begin{array}{*{20}{c}}
3&{ - 1} \\
5&{10}
\end{array}} \right| = 30 + 5 = 35 \hfill \\
{\text{Using the Cramer's rule}} \hfill \\
x = \frac{{{D_x}}}{D} = \frac{{ - 14}}{7} = - 2,\,\,\,\,\,\,\,\,\,y = \frac{{{D_y}}}{D} = \frac{{35}}{7} = 5 \hfill \\
{\text{The solution set is}} \hfill \\
\left\{ {\left( { - 2,5} \right)} \right\} \hfill \\
\end{gathered} \]
