Aftrekken geeft \(2x=-6\text{,}\) dus \(x=-3\text{.}\)

\(\begin{rcases}6x-y=-3 \\ x=-3\end{rcases}\begin{matrix}6⋅-3-y=-3 \\ -y=15 \\ y=-15\end{matrix}\)

De oplossing is \((x, y)=(-3, -15)\text{.}\)

\(\begin{cases}6a+5b=2 \\ 3a+2b=-1\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}6a+5b=2 \\ 6a+4b=-2\end{cases}\)

Aftrekken geeft \(b=4\text{.}\)

\(\begin{rcases}6a+5b=2 \\ b=4\end{rcases}\begin{matrix}6a+5⋅4=2 \\ 6a=-18 \\ a=-3\end{matrix}\)

De oplossing is \((a, b)=(-3, 4)\text{.}\)

\(\begin{cases}3p-3q=-6 \\ 5p+5q=5\end{cases}\) \(\begin{vmatrix}5 \\ 3\end{vmatrix}\) geeft \(\begin{cases}15p-15q=-30 \\ 15p+15q=15\end{cases}\)

Optellen geeft \(30p=-15\text{,}\) dus \(p=-\frac{1}{2}\text{.}\)

\(\begin{rcases}3p-3q=-6 \\ p=-\frac{1}{2}\end{rcases}\begin{matrix}3⋅-\frac{1}{2}-3q=-6 \\ -3q=-4\frac{1}{2} \\ q=1\frac{1}{2}\end{matrix}\)

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