Solve the following equations by Cramer's method: 7x + 3y = 15 ; 12y - 5x = 39

Answer: \(x = \frac{7}{11},\ y = \frac{116}{33}\)

Write both equations in the form \(ax + by = c\):

\(7x + 3y = 15\) .....(1)

\(-5x + 12y = 39\) .....(2)

\(a_1 = 7, b_1 = 3, c_1 = 15\); \(a_2 = -5, b_2 = 12, c_2 = 39\)

\(D = \begin{vmatrix} 7 & 3 \\ -5 & 12 \end{vmatrix} = 7 \times 12 - 3 \times (-5) = 84 + 15 = 99\)

\(D_x = \begin{vmatrix} 15 & 3 \\ 39 & 12 \end{vmatrix} = 15 \times 12 - 3 \times 39 = 180 - 117 = 63\)

\(D_y = \begin{vmatrix} 7 & 15 \\ -5 & 39 \end{vmatrix} = 7 \times 39 - 15 \times (-5) = 273 + 75 = 348\)

By Cramer's rule:

\(x = \frac{D_x}{D} = \frac{63}{99} = \frac{7}{11}\), \(y = \frac{D_y}{D} = \frac{348}{99} = \frac{116}{33}\)

Check in (1): \(7 \times \frac{7}{11} + 3 \times \frac{116}{33} = \frac{49}{11} + \frac{116}{11} = \frac{165}{11} = 15\) ✓

Check in (2): \(12 \times \frac{116}{33} - 5 \times \frac{7}{11} = \frac{464}{11} - \frac{35}{11} = \frac{429}{11} = 39\) ✓

\(\therefore\) \(x = \frac{7}{11},\ y = \frac{116}{33}\)