Solve the following equations by Cramer's method: (x + y - 8)/2 = (x + 2y - 14)/3 = (3x - y)/4

Answer: \((x, y) = (2, 6)\)

From the first two parts: \(3(x + y - 8) = 2(x + 2y - 14)\), so \(3x + 3y - 24 = 2x + 4y - 28\), i.e. \(x - y = -4\).

From the first and third parts: \(4(x + y - 8) = 2(3x - y)\), so \(4x + 4y - 32 = 6x - 2y\), i.e. \(-2x + 6y = 32\), or \(-x + 3y = 16\).

In the form \(a_1x + b_1y = c_1\), \(a_2x + b_2y = c_2\): first form two equations (below).\[x - y = -4\]\[-x + 3y = 16\]

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

\(D_x = \begin{vmatrix} -4 & -1 \\ 16 & 3 \end{vmatrix}\)\[D_x = (-4)\times 3 - (-1)\times 16 = 4\]

\(D_y = \begin{vmatrix} 1 & -4 \\ -1 & 16 \end{vmatrix}\)\[D_y = 1\times 16 - (-4)\times (-1) = 12\]

By Cramer's rule:\[x = \frac{D_x}{D} = \frac{4}{2} = 2\]\[y = \frac{D_y}{D} = \frac{12}{2} = 6\]