$3kx+9y = 1$
$x+3ky = 3$
$A = \begin{bmatrix} 3k & 9 |1 \\ 1 & 3 |3\end{bmatrix}$
and the answer is:
when $k \not =$ -1 or 1
$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{3k-27}{9k^2-9} \\ \frac{9k-1}{9k^2-9}\end{bmatrix} $
another question with 3 k's
$3kx+3ky = 5$
$12x+3ky = 1$
$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{12k}{9k^2-36} \\ \frac{3k-60}{9k^2-36}\end{bmatrix} $
how do they got the answer? I really need help. I don't even know what's the first step.