Optellen geeft \(8a=4\text{,}\) dus \(a=\frac{1}{2}\text{.}\)

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

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

\(\begin{cases}3x-4y=1 \\ x-2y=-3\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}3x-4y=1 \\ 2x-4y=-6\end{cases}\)

Aftrekken geeft \(x=7\text{.}\)

\(\begin{rcases}3x-4y=1 \\ x=7\end{rcases}\begin{matrix}3⋅7-4y=1 \\ -4y=-20 \\ y=5\end{matrix}\)

De oplossing is \((x, y)=(7, 5)\text{.}\)

\(\begin{cases}2p+2q=4 \\ 5p+3q=5\end{cases}\) \(\begin{vmatrix}3 \\ 2\end{vmatrix}\) geeft \(\begin{cases}6p+6q=12 \\ 10p+6q=10\end{cases}\)

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

\(\begin{rcases}2p+2q=4 \\ p=-\frac{1}{2}\end{rcases}\begin{matrix}2⋅-\frac{1}{2}+2q=4 \\ 2q=5 \\ q=2\frac{1}{2}\end{matrix}\)

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