Writing this using Latex, your matrix equation is
\begin{bmatrix}2 & 4 \\ -1 & -1 \end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}= \begin{bmatrix}-9 \\ 2\end{bmatrix}
You give your solution as x= 0.03, y= 0.16666.
It's easy enough to check that your self isn't it?
2x+ 4y= 2(0.03)+ 4(0.1666)= 0.06+ 0.66664= 0.72664, not -9! (Not even close.)
You seem to be under the impression that the inverse matrix to \begin{bmatrix}-2 & 4 \\ -1 & -1\end{bmatrix} is (1/6)\begin{bmatrix}-1 & -4 \\ 1 & 2\end{bmatrix}. It isn't as, again, you could have checked:
\begin{bmatrix}2 & 4 \\ -1 & -1 \end{bmatrix}\begin{bmatrix}-\frac{1}{6} & -\frac{2}{3} \\ \frac{1}{6} & \frac{1}{3}\end{bmatrix}= \begin{bmatrix}-\frac{2}{6}+ \frac{4}{6} & -\frac{4}{3}+ \frac{4}{3} \\ \frac{1}{6}- \frac{1}{6} & \frac{2}{3}- \frac{1}{3}\end{bmatrix}= \begin{bmatrix}-\frac{1}{3} & 0 \\ 0 & \frac{1}{3} \end{bmatrix}
NOT the identity matrix (though pretty close)!
Looks to me like you calculated the determinant wrong. Be careful with the signs.
While your method of finding the determinant and cofactors will work to find the inverse matrix (it was the first method I learned), using the "augmented matrix" and row operations, as Hilbert2 suggests, is much easier and less error prone.
