Answer: \((x, y) = (-2, 4)\)
In the form \(a_1x + b_1y = c_1\), \(a_2x + b_2y = c_2\): rewrite \(4x + 3y - 4 = 0\) as \(4x + 3y = 4\) and \(6x = 8 - 5y\) as \(6x + 5y = 8\).\[4x + 3y = 4\]\[6x + 5y = 8\]
\(D = \begin{vmatrix} 4 & 3 \\ 6 & 5 \end{vmatrix}\)\[D = 4\times 5 - 3\times 6 = 2\]
\(D_x = \begin{vmatrix} 4 & 3 \\ 8 & 5 \end{vmatrix}\)\[D_x = 4\times 5 - 3\times 8 = -4\]
\(D_y = \begin{vmatrix} 4 & 4 \\ 6 & 8 \end{vmatrix}\)\[D_y = 4\times 8 - 4\times 6 = 8\]
By Cramer's rule:\[x = \frac{D_x}{D} = \frac{-4}{2} = -2\]\[y = \frac{D_y}{D} = \frac{8}{2} = 4\]