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

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

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

\(\begin{cases}4 p - 3 q = -5 \\ p - q = 5\end{cases}\) \(\begin{vmatrix}1 \\ 3\end{vmatrix}\) geeft \(\begin{cases}4 p - 3 q = -5 \\ 3 p - 3 q = 15\end{cases}\)

Aftrekken geeft \(p = -20 \text{.}\)

\(\begin{rcases}4 p - 3 q = -5 \\ p = -20\end{rcases} \begin{matrix}4 ⋅ -20 - 3 q = -5 \\ -3 q = 75 \\ q = -25\end{matrix}\)

De oplossing is \((p , q) = (-20 , -25) \text{.}\)

\(\begin{cases}5 a + 3 b = 3 \\ 6 a + 4 b = 2\end{cases}\) \(\begin{vmatrix}4 \\ 3\end{vmatrix}\) geeft \(\begin{cases}20 a + 12 b = 12 \\ 18 a + 12 b = 6\end{cases}\)

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

\(\begin{rcases}5 a + 3 b = 3 \\ a = 3\end{rcases} \begin{matrix}5 ⋅ 3 + 3 b = 3 \\ 3 b = -12 \\ b = -4\end{matrix}\)

De oplossing is \((a , b) = (3 , -4) \text{.}\)