Question

Solve the following equation by Cramer’s method.

4m – 2n = –4 ; 4m + 3n = 16

Answer 1

4m – 2n = –4 ; 4m + 3n = 16

\[D = \begin{vmatrix}4 & - 2 \\ 4 & 3\end{vmatrix} = 4 \times 3 - \left( - 2 \right) \times 4 = 12 + 8 = 20\]

\[ D_m = \begin{vmatrix}- 4 & - 2 \\ 16 & 3\end{vmatrix} = - 4 \times 3 - \left( - 2 \right) \times 16 = - 12 + 32 = 20\]

\[ D_n = \begin{vmatrix}4 & - 4 \\ 4 & 16\end{vmatrix} = 4 \times 16 - \left( - 4 \right) \times 4 = 64 + 16 = 80\]

\[m = \frac{D_m}{D} = \frac{20}{20} = 1\]

\[n = \frac{D_n}{D} = \frac{80}{20} = 4\]

(m, n) = (1, 4)