Solve the following equations by Cramer's method: 3x - 2y = 5/2 ; (1/3)x + 3y = -4/3

Answer: \((x, y) = (\frac{1}{2}, -\frac{1}{2})\)

In the form \(a_1x + b_1y = c_1\), \(a_2x + b_2y = c_2\): multiply \(3x - 2y = \frac{5}{2}\) by 2 to get \(6x - 4y = 5\), and \(\frac{1}{3}x + 3y = -\frac{4}{3}\) by 3 to get \(x + 9y = -4\).\[6x - 4y = 5\]\[x + 9y = -4\]

\(D = \begin{vmatrix} 6 & -4 \\ 1 & 9 \end{vmatrix}\)\[D = 6\times 9 - (-4)\times 1 = 58\]

\(D_x = \begin{vmatrix} 5 & -4 \\ -4 & 9 \end{vmatrix}\)\[D_x = 5\times 9 - (-4)\times (-4) = 29\]

\(D_y = \begin{vmatrix} 6 & 5 \\ 1 & -4 \end{vmatrix}\)\[D_y = 6\times (-4) - 5\times 1 = -29\]

By Cramer's rule:\[x = \frac{D_x}{D} = \frac{29}{58} = \frac{1}{2}\]\[y = \frac{D_y}{D} = \frac{-29}{58} = -\frac{1}{2}\]