Aftrekken geeft \(p=-6\text{.}\)

\(\begin{rcases}2p+2q=-2 \\ p=-6\end{rcases}\begin{matrix}2⋅-6+2q=-2 \\ 2q=10 \\ q=5\end{matrix}\)

De oplossing is \((p, q)=(-6, 5)\text{.}\)

\(\begin{cases}5a-6b=-1 \\ 2a-2b=2\end{cases}\) \(\begin{vmatrix}1 \\ 3\end{vmatrix}\) geeft \(\begin{cases}5a-6b=-1 \\ 6a-6b=6\end{cases}\)

Aftrekken geeft \(-a=-7\text{,}\) dus \(a=7\text{.}\)

\(\begin{rcases}5a-6b=-1 \\ a=7\end{rcases}\begin{matrix}5⋅7-6b=-1 \\ -6b=-36 \\ b=6\end{matrix}\)

De oplossing is \((a, b)=(7, 6)\text{.}\)

\(\begin{cases}5x+5y=-5 \\ 6x-2y=-2\end{cases}\) \(\begin{vmatrix}2 \\ 5\end{vmatrix}\) geeft \(\begin{cases}10x+10y=-10 \\ 30x-10y=-10\end{cases}\)

Optellen geeft \(40x=-20\text{,}\) dus \(x=-\frac{1}{2}\text{.}\)

\(\begin{rcases}5x+5y=-5 \\ x=-\frac{1}{2}\end{rcases}\begin{matrix}5⋅-\frac{1}{2}+5y=-5 \\ 5y=-2\frac{1}{2} \\ y=-\frac{1}{2}\end{matrix}\)

De oplossing is \((x, y)=(-\frac{1}{2}, -\frac{1}{2})\text{.}\)