Aftrekken geeft \(2 b = 10 \text{,}\) dus \(b = 5 \text{.}\)

\(\begin{rcases}2 a - b = 4 \\ b = 5\end{rcases} \begin{matrix}2 a - 1 ⋅ 5 = 4 \\ 2 a = 9 \\ a = 4\frac{1}{2}\end{matrix}\)

De oplossing is \((a , b) = (4\frac{1}{2} , 5) \text{.}\)

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

Aftrekken geeft \(2 a = -14 \text{,}\) dus \(a = -7 \text{.}\)

\(\begin{rcases}a - b = -2 \\ a = -7\end{rcases} \begin{matrix}-7 - b = -2 \\ -b = 5 \\ b = -5\end{matrix}\)

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

\(\begin{cases}5 p + 3 q = -1 \\ 6 p - 2 q = -4\end{cases}\) \(\begin{vmatrix}2 \\ 3\end{vmatrix}\) geeft \(\begin{cases}10 p + 6 q = -2 \\ 18 p - 6 q = -12\end{cases}\)

Optellen geeft \(28 p = -14 \text{,}\) dus \(p = -\frac{1}{2} \text{.}\)

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

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