Optellen geeft \(6 x = 3 \text{,}\) dus \(x = \frac{1}{2} \text{.}\)

\(\begin{rcases}5 x + y = 4 \\ x = \frac{1}{2}\end{rcases} \begin{matrix}5 ⋅ \frac{1}{2} + y = 4 \\ y = 1\frac{1}{2}\end{matrix}\)

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

\(\begin{cases}6 p - 3 q = 6 \\ p - q = 4\end{cases}\) \(\begin{vmatrix}1 \\ 3\end{vmatrix}\) geeft \(\begin{cases}6 p - 3 q = 6 \\ 3 p - 3 q = 12\end{cases}\)

Aftrekken geeft \(3 p = -6 \text{,}\) dus \(p = -2 \text{.}\)

\(\begin{rcases}6 p - 3 q = 6 \\ p = -2\end{rcases} \begin{matrix}6 ⋅ -2 - 3 q = 6 \\ -3 q = 18 \\ q = -6\end{matrix}\)

De oplossing is \((p , q) = (-2 , -6) \text{.}\)

\(\begin{cases}6 a + 5 b = -6 \\ 4 a + 4 b = -2\end{cases}\) \(\begin{vmatrix}4 \\ 5\end{vmatrix}\) geeft \(\begin{cases}24 a + 20 b = -24 \\ 20 a + 20 b = -10\end{cases}\)

Aftrekken geeft \(4 a = -14 \text{,}\) dus \(a = -3\frac{1}{2} \text{.}\)

\(\begin{rcases}6 a + 5 b = -6 \\ a = -3\frac{1}{2}\end{rcases} \begin{matrix}6 ⋅ -3\frac{1}{2} + 5 b = -6 \\ 5 b = 15 \\ b = 3\end{matrix}\)

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