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

\(\begin{rcases}2 x + 3 y = -6 \\ y = 4\end{rcases} \begin{matrix}2 x + 3 ⋅ 4 = -6 \\ 2 x = -18 \\ x = -9\end{matrix}\)

De oplossing is \((x , y) = (-9 , 4) \text{.}\)

\(\begin{cases}3 p - 3 q = 3 \\ 2 p + 6 q = 6\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}6 p - 6 q = 6 \\ 2 p + 6 q = 6\end{cases}\)

Optellen geeft \(8 p = 12 \text{,}\) dus \(p = 1\frac{1}{2} \text{.}\)

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

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

\(\begin{cases}3 x - 4 y = -3 \\ 4 x - 6 y = 3\end{cases}\) \(\begin{vmatrix}3 \\ 2\end{vmatrix}\) geeft \(\begin{cases}9 x - 12 y = -9 \\ 8 x - 12 y = 6\end{cases}\)

Aftrekken geeft \(x = -15 \text{.}\)

\(\begin{rcases}3 x - 4 y = -3 \\ x = -15\end{rcases} \begin{matrix}3 ⋅ -15 - 4 y = -3 \\ -4 y = 42 \\ y = -10\frac{1}{2}\end{matrix}\)

De oplossing is \((x , y) = (-15 , -10\frac{1}{2}) \text{.}\)