Optellen geeft \(5 p = 10 \text{,}\) dus \(p = 2 \text{.}\)

\(\begin{rcases}4 p - q = 5 \\ p = 2\end{rcases} \begin{matrix}4 ⋅ 2 - q = 5 \\ -q = -3 \\ q = 3\end{matrix}\)

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

\(\begin{cases}4 x + 2 y = -2 \\ 2 x + 3 y = -5\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}4 x + 2 y = -2 \\ 4 x + 6 y = -10\end{cases}\)

Aftrekken geeft \(-4 y = 8 \text{,}\) dus \(y = -2 \text{.}\)

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

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

\(\begin{cases}5 a - 4 b = 5 \\ 6 a - 3 b = -3\end{cases}\) \(\begin{vmatrix}3 \\ 4\end{vmatrix}\) geeft \(\begin{cases}15 a - 12 b = 15 \\ 24 a - 12 b = -12\end{cases}\)

Aftrekken geeft \(-9 a = 27 \text{,}\) dus \(a = -3 \text{.}\)

\(\begin{rcases}5 a - 4 b = 5 \\ a = -3\end{rcases} \begin{matrix}5 ⋅ -3 - 4 b = 5 \\ -4 b = 20 \\ b = -5\end{matrix}\)

De oplossing is \((a , b) = (-3 , -5) \text{.}\)

Gelijk stellen geeft \(6 y + 14 = 2 y + 6\)

\(4 y = -8\) dus \(y = -2\)

\(\begin{rcases}x = 6 y + 14 \\ y = -2\end{rcases} \begin{matrix}x = 6 ⋅ -2 + 14 \\ x = 2\end{matrix}\)

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

Substitutie geeft \(2 (6 y + 19) + 3 y = -7\)

Haakjes wegwerken geeft
\(12 y + 38 + 3 y = -7\)
\(15 y = -45\)
\(y = -3\)

\(\begin{rcases}x = 6 y + 19 \\ y = -3\end{rcases} \begin{matrix}x = 6 ⋅ -3 + 19 \\ x = 1\end{matrix}\)

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

Substitutie geeft \(b = 5 (7 b - 31) - 15\)

Haakjes wegwerken geeft
\(b = 35 b - 155 - 15\)
\(-34 b = -170\)
\(b = 5\)

\(\begin{rcases}a = 7 b - 31 \\ b = 5\end{rcases} \begin{matrix}a = 7 ⋅ 5 - 31 \\ a = 4\end{matrix}\)

De oplossing is \((a , b) = (4 , 5) \text{.}\)