Aftrekken geeft \(b = -2 \text{.}\)

\(\begin{rcases}3 a + 6 b = -3 \\ b = -2\end{rcases} \begin{matrix}3 a + 6 ⋅ -2 = -3 \\ 3 a = 9 \\ a = 3\end{matrix}\)

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

\(\begin{cases}4 x + 6 y = 1 \\ x - y = -1\end{cases}\) \(\begin{vmatrix}1 \\ 6\end{vmatrix}\) geeft \(\begin{cases}4 x + 6 y = 1 \\ 6 x - 6 y = -6\end{cases}\)

Optellen geeft \(10 x = -5 \text{,}\) dus \(x = -\frac{1}{2} \text{.}\)

\(\begin{rcases}4 x + 6 y = 1 \\ x = -\frac{1}{2}\end{rcases} \begin{matrix}4 ⋅ -\frac{1}{2} + 6 y = 1 \\ 6 y = 3 \\ y = \frac{1}{2}\end{matrix}\)

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

\(\begin{cases}5 a - 5 b = -5 \\ 3 a - 4 b = 2\end{cases}\) \(\begin{vmatrix}4 \\ 5\end{vmatrix}\) geeft \(\begin{cases}20 a - 20 b = -20 \\ 15 a - 20 b = 10\end{cases}\)

Aftrekken geeft \(5 a = -30 \text{,}\) dus \(a = -6 \text{.}\)

\(\begin{rcases}5 a - 5 b = -5 \\ a = -6\end{rcases} \begin{matrix}5 ⋅ -6 - 5 b = -5 \\ -5 b = 25 \\ b = -5\end{matrix}\)

De oplossing is \((a , b) = (-6 , -5) \text{.}\)

Gelijk stellen geeft \(6 y + 7 = 8 y + 11\)

\(-2 y = 4\) dus \(y = -2\)

\(\begin{rcases}x = 6 y + 7 \\ y = -2\end{rcases} \begin{matrix}x = 6 ⋅ -2 + 7 \\ x = -5\end{matrix}\)

De oplossing is \((x , y) = (-5 , -2) \text{.}\)

Substitutie geeft \(6 (9 y - 25) + 2 y = 18\)

Haakjes wegwerken geeft
\(54 y - 150 + 2 y = 18\)
\(56 y = 168\)
\(y = 3\)

\(\begin{rcases}x = 9 y - 25 \\ y = 3\end{rcases} \begin{matrix}x = 9 ⋅ 3 - 25 \\ x = 2\end{matrix}\)

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

Substitutie geeft \(p = 9 (5 p - 18) + 30\)

Haakjes wegwerken geeft
\(p = 45 p - 162 + 30\)
\(-44 p = -132\)
\(p = 3\)

\(\begin{rcases}q = 5 p - 18 \\ p = 3\end{rcases} \begin{matrix}q = 5 ⋅ 3 - 18 \\ q = -3\end{matrix}\)

De oplossing is \((p , q) = (3 , -3) \text{.}\)