Answer: \((m, n) = (6, 5)\)
In the form \(a_1m + b_1n = c_1\), \(a_2m + b_2n = c_2\):\[4m + 6n = 54\]\[3m + 2n = 28\]
\(D = \begin{vmatrix} 4 & 6 \\ 3 & 2 \end{vmatrix}\)\[D = 4\times 2 - 6\times 3 = -10\]
\(D_m = \begin{vmatrix} 54 & 6 \\ 28 & 2 \end{vmatrix}\)\[D_m = 54\times 2 - 6\times 28 = -60\]
\(D_n = \begin{vmatrix} 4 & 54 \\ 3 & 28 \end{vmatrix}\)\[D_n = 4\times 28 - 54\times 3 = -50\]
By Cramer's rule:\[m = \frac{D_m}{D} = \frac{-60}{-10} = 6\]\[n = \frac{D_n}{D} = \frac{-50}{-10} = 5\]