Optellen geeft \(5 p = 10 \text{,}\) dus \(p = 2 \text{.}\)

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

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

\(\begin{cases}5 a - 6 b = -4 \\ 2 a - b = 4\end{cases}\) \(\begin{vmatrix}1 \\ 6\end{vmatrix}\) geeft \(\begin{cases}5 a - 6 b = -4 \\ 12 a - 6 b = 24\end{cases}\)

Aftrekken geeft \(-7 a = -28 \text{,}\) dus \(a = 4 \text{.}\)

\(\begin{rcases}5 a - 6 b = -4 \\ a = 4\end{rcases} \begin{matrix}5 ⋅ 4 - 6 b = -4 \\ -6 b = -24 \\ b = 4\end{matrix}\)

De oplossing is \((a , b) = (4 , 4) \text{.}\)

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

Aftrekken geeft \(x = 3 \text{.}\)

\(\begin{rcases}3 x + 4 y = -3 \\ x = 3\end{rcases} \begin{matrix}3 ⋅ 3 + 4 y = -3 \\ 4 y = -12 \\ y = -3\end{matrix}\)

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