Answer
\[\left\{ {\left( {2,2} \right)} \right\}\]
Work Step by Step
\[\begin{gathered}
\left\{ {\begin{array}{*{20}{c}}
{x + y = 4} \\
{2x - y = 2}
\end{array}} \right. \hfill \\
{\text{Given the system}} \hfill \\
\left\{ {\begin{array}{*{20}{c}}
{{a_1}x + {b_1}y = {c_1}} \\
{{a_2}x + {b_2}y = {c_2}}
\end{array}} \right. \hfill \\
D = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{b_1}} \\
{{a_2}}&{{b_2}}
\end{array}} \right|,\,\,\,{D_x} = \left| {\begin{array}{*{20}{c}}
{{c_1}}&{{b_1}} \\
{{c_2}}&{{b_2}}
\end{array}} \right|,\,\,{D_y} = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{c_1}} \\
{{a_2}}&{{c_2}}
\end{array}} \right| \hfill \\
{\text{First find }}D,{\text{ If }}D \ne 0,\,{\text{ then find }}{D_x}{\text{ and }}{D_y} \hfill \\
D = \left| {\begin{array}{*{20}{c}}
1&1 \\
2&{ - 1}
\end{array}} \right| = - 1 - 2 = - 3 \hfill \\
{D_x} = \left| {\begin{array}{*{20}{c}}
4&1 \\
2&{ - 1}
\end{array}} \right| = - 4 - 2 = - 6 \hfill \\
{D_y} = \left| {\begin{array}{*{20}{c}}
1&4 \\
2&2
\end{array}} \right| = 2 - 8 = - 6 \hfill \\
{\text{Using the Cramer's rule}} \hfill \\
x = \frac{{{D_x}}}{D} = \frac{{ - 6}}{{ - 3}} = 2,\,\,\,\,\,\,\,\,\,y = \frac{{{D_y}}}{D} = \frac{{ - 6}}{{ - 3}} = 2 \hfill \\
{\text{The solution set is}} \hfill \\
\left\{ {\left( {2,2} \right)} \right\} \hfill \\
\end{gathered} \]
