Optellen geeft \(4 p = 10 \text{,}\) dus \(p = 2\frac{1}{2} \text{.}\)

\(\begin{rcases}3 p + q = 4 \\ p = 2\frac{1}{2}\end{rcases} \begin{matrix}3 ⋅ 2\frac{1}{2} + q = 4 \\ q = -3\frac{1}{2}\end{matrix}\)

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

\(\begin{cases}2 x - 6 y = -4 \\ x - 4 y = 3\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}2 x - 6 y = -4 \\ 2 x - 8 y = 6\end{cases}\)

Aftrekken geeft \(2 y = -10 \text{,}\) dus \(y = -5 \text{.}\)

\(\begin{rcases}2 x - 6 y = -4 \\ y = -5\end{rcases} \begin{matrix}2 x - 6 ⋅ -5 = -4 \\ 2 x = -34 \\ x = -17\end{matrix}\)

De oplossing is \((x , y) = (-17 , -5) \text{.}\)

\(\begin{cases}3 a - 5 b = -1 \\ 2 a - 3 b = -3\end{cases}\) \(\begin{vmatrix}3 \\ 5\end{vmatrix}\) geeft \(\begin{cases}9 a - 15 b = -3 \\ 10 a - 15 b = -15\end{cases}\)

Aftrekken geeft \(-a = 12 \text{,}\) dus \(a = -12 \text{.}\)

\(\begin{rcases}3 a - 5 b = -1 \\ a = -12\end{rcases} \begin{matrix}3 ⋅ -12 - 5 b = -1 \\ -5 b = 35 \\ b = -7\end{matrix}\)

De oplossing is \((a , b) = (-12 , -7) \text{.}\)

Gelijk stellen geeft \(2 y - 5 = 6 y - 9\)

\(-4 y = -4\) dus \(y = 1\)

\(\begin{rcases}x = 2 y - 5 \\ y = 1\end{rcases} \begin{matrix}x = 2 ⋅ 1 - 5 \\ x = -3\end{matrix}\)

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

Substitutie geeft \(7 (9 y - 34) + 4 y = 30\)

Haakjes wegwerken geeft
\(63 y - 238 + 4 y = 30\)
\(67 y = 268\)
\(y = 4\)

\(\begin{rcases}x = 9 y - 34 \\ y = 4\end{rcases} \begin{matrix}x = 9 ⋅ 4 - 34 \\ x = 2\end{matrix}\)

De oplossing is \((x , y) = (2 , 4) \text{.}\)

Substitutie geeft \(b = 6 (4 b - 9) + 8\)

Haakjes wegwerken geeft
\(b = 24 b - 54 + 8\)
\(-23 b = -46\)
\(b = 2\)

\(\begin{rcases}a = 4 b - 9 \\ b = 2\end{rcases} \begin{matrix}a = 4 ⋅ 2 - 9 \\ a = -1\end{matrix}\)

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