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

\(\begin{rcases}p-2q=6 \\ q=\frac{1}{2}\end{rcases}\begin{matrix}p-2⋅\frac{1}{2}=6 \\ p=7\end{matrix}\)

De oplossing is \((p, q)=(7, \frac{1}{2})\text{.}\)

\(\begin{cases}x+y=-4 \\ 2x-6y=-4\end{cases}\) \(\begin{vmatrix}6 \\ 1\end{vmatrix}\) geeft \(\begin{cases}6x+6y=-24 \\ 2x-6y=-4\end{cases}\)

Optellen geeft \(8x=-28\text{,}\) dus \(x=-3\frac{1}{2}\text{.}\)

\(\begin{rcases}x+y=-4 \\ x=-3\frac{1}{2}\end{rcases}\begin{matrix}-3\frac{1}{2}+y=-4 \\ y=-\frac{1}{2}\end{matrix}\)

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

\(\begin{cases}3a-5b=5 \\ 4a-6b=3\end{cases}\) \(\begin{vmatrix}6 \\ 5\end{vmatrix}\) geeft \(\begin{cases}18a-30b=30 \\ 20a-30b=15\end{cases}\)

Aftrekken geeft \(-2a=15\text{,}\) dus \(a=-7\frac{1}{2}\text{.}\)

\(\begin{rcases}3a-5b=5 \\ a=-7\frac{1}{2}\end{rcases}\begin{matrix}3⋅-7\frac{1}{2}-5b=5 \\ -5b=27\frac{1}{2} \\ b=-5\frac{1}{2}\end{matrix}\)

De oplossing is \((a, b)=(-7\frac{1}{2}, -5\frac{1}{2})\text{.}\)