Aftrekken geeft \(-2x=-3\text{,}\) dus \(x=1\frac{1}{2}\text{.}\)

\(\begin{rcases}3x-y=-2 \\ x=1\frac{1}{2}\end{rcases}\begin{matrix}3⋅1\frac{1}{2}-y=-2 \\ -y=-6\frac{1}{2} \\ y=6\frac{1}{2}\end{matrix}\)

De oplossing is \((x, y)=(1\frac{1}{2}, 6\frac{1}{2})\text{.}\)

\(\begin{cases}p+2q=-5 \\ 5p+5q=5\end{cases}\) \(\begin{vmatrix}5 \\ 1\end{vmatrix}\) geeft \(\begin{cases}5p+10q=-25 \\ 5p+5q=5\end{cases}\)

Aftrekken geeft \(5q=-30\text{,}\) dus \(q=-6\text{.}\)

\(\begin{rcases}p+2q=-5 \\ q=-6\end{rcases}\begin{matrix}p+2⋅-6=-5 \\ p=7\end{matrix}\)

De oplossing is \((p, q)=(7, -6)\text{.}\)

\(\begin{cases}3a-3b=-6 \\ 2a+2b=2\end{cases}\) \(\begin{vmatrix}2 \\ 3\end{vmatrix}\) geeft \(\begin{cases}6a-6b=-12 \\ 6a+6b=6\end{cases}\)

Optellen geeft \(12a=-6\text{,}\) dus \(a=-\frac{1}{2}\text{.}\)

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

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