Answer: \((m, n) = (1, 4)\)
In the form \(a_1m + b_1n = c_1\), \(a_2m + b_2n = c_2\):\[4m - 2n = -4\]\[4m + 3n = 16\]
\(D = \begin{vmatrix} 4 & -2 \\ 4 & 3 \end{vmatrix}\)\[D = 4\times 3 - (-2)\times 4 = 20\]
\(D_m = \begin{vmatrix} -4 & -2 \\ 16 & 3 \end{vmatrix}\)\[D_m = (-4)\times 3 - (-2)\times 16 = 20\]
\(D_n = \begin{vmatrix} 4 & -4 \\ 4 & 16 \end{vmatrix}\)\[D_n = 4\times 16 - (-4)\times 4 = 80\]
By Cramer's rule:\[m = \frac{D_m}{D} = \frac{20}{20} = 1\]\[n = \frac{D_n}{D} = \frac{80}{20} = 4\]