Optellen geeft \(6x=3\text{,}\) dus \(x=\frac{1}{2}\text{.}\)

\(\begin{rcases}5x+y=-1 \\ x=\frac{1}{2}\end{rcases}\begin{matrix}5⋅\frac{1}{2}+y=-1 \\ y=-3\frac{1}{2}\end{matrix}\)

De oplossing is \((x, y)=(\frac{1}{2}, -3\frac{1}{2})\text{.}\)

\(\begin{cases}4p-4q=4 \\ p-2q=4\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}4p-4q=4 \\ 2p-4q=8\end{cases}\)

Aftrekken geeft \(2p=-4\text{,}\) dus \(p=-2\text{.}\)

\(\begin{rcases}4p-4q=4 \\ p=-2\end{rcases}\begin{matrix}4⋅-2-4q=4 \\ -4q=12 \\ q=-3\end{matrix}\)

De oplossing is \((p, q)=(-2, -3)\text{.}\)

\(\begin{cases}3a+5b=3 \\ 2a+2b=4\end{cases}\) \(\begin{vmatrix}2 \\ 5\end{vmatrix}\) geeft \(\begin{cases}6a+10b=6 \\ 10a+10b=20\end{cases}\)

Aftrekken geeft \(-4a=-14\text{,}\) dus \(a=3\frac{1}{2}\text{.}\)

\(\begin{rcases}3a+5b=3 \\ a=3\frac{1}{2}\end{rcases}\begin{matrix}3⋅3\frac{1}{2}+5b=3 \\ 5b=-7\frac{1}{2} \\ b=-1\frac{1}{2}\end{matrix}\)

De oplossing is \((a, b)=(3\frac{1}{2}, -1\frac{1}{2})\text{.}\)