Answer: \((x, y) = (-\frac{2}{3}, 2)\)
In the form \(a_1x + b_1y = c_1\), \(a_2x + b_2y = c_2\): write \(3x + 5y - 8 = 0\) as \(3x + 5y = 8\).\[6x - 3y = -10\]\[3x + 5y = 8\]
\(D = \begin{vmatrix} 6 & -3 \\ 3 & 5 \end{vmatrix}\)\[D = 6\times 5 - (-3)\times 3 = 39\]
\(D_x = \begin{vmatrix} -10 & -3 \\ 8 & 5 \end{vmatrix}\)\[D_x = (-10)\times 5 - (-3)\times 8 = -26\]
\(D_y = \begin{vmatrix} 6 & -10 \\ 3 & 8 \end{vmatrix}\)\[D_y = 6\times 8 - (-10)\times 3 = 78\]
By Cramer's rule:\[x = \frac{D_x}{D} = \frac{-26}{39} = -\frac{2}{3}\]\[y = \frac{D_y}{D} = \frac{78}{39} = 2\]