Answer
\[\left\{ {\left( { - 16,6} \right)} \right\}\]
Work Step by Step
\[\begin{gathered}
\left\{ {\begin{array}{*{20}{c}}
{ - \frac{3}{4}x + \frac{2}{3}y = 16} \\
{\frac{5}{2}x + \frac{1}{2}y = - 37}
\end{array}} \right. \hfill \\
{\text{Given the system}} \hfill \\
\left\{ {\begin{array}{*{20}{c}}
{{a_1}x + {b_1}y = {c_1}} \\
{{a_2}x + {b_2}y = {c_2}}
\end{array}} \right. \hfill \\
D = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{b_1}} \\
{{a_2}}&{{b_2}}
\end{array}} \right|,\,\,\,{D_x} = \left| {\begin{array}{*{20}{c}}
{{c_1}}&{{b_1}} \\
{{c_2}}&{{b_2}}
\end{array}} \right|,\,\,{D_y} = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{c_1}} \\
{{a_2}}&{{c_2}}
\end{array}} \right| \hfill \\
{\text{First find }}D,{\text{ If }}D \ne 0,\,{\text{ then find }}{D_x}{\text{ and }}{D_y} \hfill \\
D = \left| {\begin{array}{*{20}{c}}
{ - 3/4}&{2/3} \\
{5/2}&{1/2}
\end{array}} \right| = - \frac{3}{8} - \frac{5}{3} = - \frac{{49}}{{24}} \hfill \\
{D_x} = \left| {\begin{array}{*{20}{c}}
{16}&{2/3} \\
{ - 37}&{1/2}
\end{array}} \right| = 8 + \frac{{74}}{3} = \frac{{98}}{3} \hfill \\
{D_y} = \left| {\begin{array}{*{20}{c}}
{ - 3/4}&{16} \\
{5/2}&{ - 37}
\end{array}} \right| = \frac{{111}}{4} - 40 = - \frac{{49}}{4} \hfill \\
{\text{Using the Cramer's rule}} \hfill \\
x = \frac{{{D_x}}}{D} = \frac{{98/3}}{{ - 49/24}} = - 16,\,\,\,\,\,\,\,\,\,y = \frac{{{D_y}}}{D} = \frac{{ - 49/4}}{{ - 49/24}} = 6 \hfill \\
{\text{The solution set is}} \hfill \\
\left\{ {\left( { - 16,6} \right)} \right\} \hfill \\
\end{gathered} \]
