Aftrekken geeft \(-x = -3 \text{,}\) dus \(x = 3 \text{.}\)

\(\begin{rcases}4 x - 3 y = 3 \\ x = 3\end{rcases} \begin{matrix}4 ⋅ 3 - 3 y = 3 \\ -3 y = -9 \\ y = 3\end{matrix}\)

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

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

Optellen geeft \(6 p = -12 \text{,}\) dus \(p = -2 \text{.}\)

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

De oplossing is \((p , q) = (-2 , 3) \text{.}\)

\(\begin{cases}6 a + 3 b = 3 \\ 5 a + 2 b = 4\end{cases}\) \(\begin{vmatrix}2 \\ 3\end{vmatrix}\) geeft \(\begin{cases}12 a + 6 b = 6 \\ 15 a + 6 b = 12\end{cases}\)

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

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

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

Gelijk stellen geeft \(2 x - 15 = 6 x - 35\)

\(-4 x = -20\) dus \(x = 5\)

\(\begin{rcases}y = 2 x - 15 \\ x = 5\end{rcases} \begin{matrix}y = 2 ⋅ 5 - 15 \\ y = -5\end{matrix}\)

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

Substitutie geeft \(4 (6 b + 8) + 2 b = 6\)

Haakjes wegwerken geeft
\(24 b + 32 + 2 b = 6\)
\(26 b = -26\)
\(b = -1\)

\(\begin{rcases}a = 6 b + 8 \\ b = -1\end{rcases} \begin{matrix}a = 6 ⋅ -1 + 8 \\ a = 2\end{matrix}\)

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

Substitutie geeft \(y = 3 (6 y - 4) - 5\)

Haakjes wegwerken geeft
\(y = 18 y - 12 - 5\)
\(-17 y = -17\)
\(y = 1\)

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

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