Aftrekken geeft \(2 y = 3 \text{,}\) dus \(y = 1\frac{1}{2} \text{.}\)

\(\begin{rcases}4 x - 4 y = 2 \\ y = 1\frac{1}{2}\end{rcases} \begin{matrix}4 x - 4 ⋅ 1\frac{1}{2} = 2 \\ 4 x = 8 \\ x = 2\end{matrix}\)

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

\(\begin{cases}x - y = -4 \\ 3 x - 4 y = 3\end{cases}\) \(\begin{vmatrix}4 \\ 1\end{vmatrix}\) geeft \(\begin{cases}4 x - 4 y = -16 \\ 3 x - 4 y = 3\end{cases}\)

Aftrekken geeft \(x = -19 \text{.}\)

\(\begin{rcases}x - y = -4 \\ x = -19\end{rcases} \begin{matrix}-19 - y = -4 \\ -y = 15 \\ y = -15\end{matrix}\)

De oplossing is \((x , y) = (-19 , -15) \text{.}\)

\(\begin{cases}3 p + 3 q = 6 \\ 5 p - 5 q = -5\end{cases}\) \(\begin{vmatrix}5 \\ 3\end{vmatrix}\) geeft \(\begin{cases}15 p + 15 q = 30 \\ 15 p - 15 q = -15\end{cases}\)

Optellen geeft \(30 p = 15 \text{,}\) dus \(p = \frac{1}{2} \text{.}\)

\(\begin{rcases}3 p + 3 q = 6 \\ p = \frac{1}{2}\end{rcases} \begin{matrix}3 ⋅ \frac{1}{2} + 3 q = 6 \\ 3 q = 4\frac{1}{2} \\ q = 1\frac{1}{2}\end{matrix}\)

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

Gelijk stellen geeft \(8 x + 9 = 4 x + 5\)

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

\(\begin{rcases}y = 8 x + 9 \\ x = -1\end{rcases} \begin{matrix}y = 8 ⋅ -1 + 9 \\ y = 1\end{matrix}\)

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

Substitutie geeft \(5 a + 4 (2 a + 4) = 29\)

Haakjes wegwerken geeft
\(5 a + 8 a + 16 = 29\)
\(13 a = 13\)
\(a = 1\)

\(\begin{rcases}b = 2 a + 4 \\ a = 1\end{rcases} \begin{matrix}b = 2 ⋅ 1 + 4 \\ b = 6\end{matrix}\)

De oplossing is \((a , b) = (1 , 6) \text{.}\)

Substitutie geeft \(a = 7 (2 a - 9) - 2\)

Haakjes wegwerken geeft
\(a = 14 a - 63 - 2\)
\(-13 a = -65\)
\(a = 5\)

\(\begin{rcases}b = 2 a - 9 \\ a = 5\end{rcases} \begin{matrix}b = 2 ⋅ 5 - 9 \\ b = 1\end{matrix}\)

De oplossing is \((a , b) = (5 , 1) \text{.}\)