Solve the following simultaneous equations using Cramer's rule: 2x + 3y = 2 ; x - y/2 = 1/2

Answer: \((x, y) = (\frac{5}{8}, \frac{1}{4})\)

In the form \(a_1x + b_1y = c_1\), \(a_2x + b_2y = c_2\): multiply \(x - \frac{y}{2} = \frac{1}{2}\) by 2 to get \(2x - y = 1\).\[2x + 3y = 2\]\[2x - y = 1\]

\(D = \begin{vmatrix} 2 & 3 \\ 2 & -1 \end{vmatrix}\)\[D = 2\times (-1) - 3\times 2 = -8\]

\(D_x = \begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix}\)\[D_x = 2\times (-1) - 3\times 1 = -5\]

\(D_y = \begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix}\)\[D_y = 2\times 1 - 2\times 2 = -2\]

By Cramer's rule:\[x = \frac{D_x}{D} = \frac{-5}{-8} = \frac{5}{8}\]\[y = \frac{D_y}{D} = \frac{-2}{-8} = \frac{1}{4}\]