Optellen geeft \(6 x = 9 \text{,}\) dus \(x = 1\frac{1}{2} \text{.}\)

\(\begin{rcases}2 x + y = 6 \\ x = 1\frac{1}{2}\end{rcases} \begin{matrix}2 ⋅ 1\frac{1}{2} + y = 6 \\ y = 3\end{matrix}\)

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

\(\begin{cases}a - 3 b = 2 \\ 2 a - 2 b = 6\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}2 a - 6 b = 4 \\ 2 a - 2 b = 6\end{cases}\)

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

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

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

\(\begin{cases}2 a - 4 b = 1 \\ 3 a - 5 b = 2\end{cases}\) \(\begin{vmatrix}5 \\ 4\end{vmatrix}\) geeft \(\begin{cases}10 a - 20 b = 5 \\ 12 a - 20 b = 8\end{cases}\)

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

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

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