Answer
\[\left\{ {\left( {1,4} \right)} \right\}\]
Work Step by Step
\[\begin{gathered}
\left\{ {\begin{array}{*{20}{c}}
{4x - y = 0} \\
{2x + 3y = 14}
\end{array}} \right. \hfill \\
{\text{Given the system}} \hfill \\
\left\{ {\begin{array}{*{20}{c}}
{{a_1}x + {b_1}y = {c_1}} \\
{{a_2}x + {b_2}y = {c_2}}
\end{array}} \right. \hfill \\
D = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{b_1}} \\
{{a_2}}&{{b_2}}
\end{array}} \right|,\,\,\,{D_x} = \left| {\begin{array}{*{20}{c}}
{{c_1}}&{{b_1}} \\
{{c_2}}&{{b_2}}
\end{array}} \right|,\,\,{D_y} = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{c_1}} \\
{{a_2}}&{{c_2}}
\end{array}} \right| \hfill \\
{\text{First find }}D,{\text{ If }}D \ne 0,\,{\text{ then find }}{D_x}{\text{ and }}{D_y} \hfill \\
D = \left| {\begin{array}{*{20}{c}}
4&{ - 1} \\
2&3
\end{array}} \right| = 12 + 2 = 14 \hfill \\
{D_x} = \left| {\begin{array}{*{20}{c}}
0&{ - 1} \\
{14}&3
\end{array}} \right| = 0 + 14 = 14 \hfill \\
{D_y} = \left| {\begin{array}{*{20}{c}}
4&0 \\
2&{14}
\end{array}} \right| = 56 \hfill \\
{\text{Using the Cramer's rule}} \hfill \\
x = \frac{{{D_x}}}{D} = \frac{{14}}{{14}} = 1,\,\,\,\,\,\,\,\,\,y = \frac{{{D_y}}}{D} = \frac{{56}}{{14}} = 4 \hfill \\
{\text{The solution set is}} \hfill \\
\left\{ {\left( {1,4} \right)} \right\} \hfill \\
\end{gathered} \]
