Stelsels oplossen

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

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

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

\(\begin{cases}4 x - 6 y = -1 \\ x + 2 y = 5\end{cases}\) \(\begin{vmatrix}1 \\ 3\end{vmatrix}\) geeft \(\begin{cases}4 x - 6 y = -1 \\ 3 x + 6 y = 15\end{cases}\)

Optellen geeft \(7 x = 14 \text{,}\) dus \(x = 2 \text{.}\)

\(\begin{rcases}4 x - 6 y = -1 \\ x = 2\end{rcases} \begin{matrix}4 ⋅ 2 - 6 y = -1 \\ -6 y = -9 \\ y = 1\frac{1}{2}\end{matrix}\)

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

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

Optellen geeft \(12 a = -18 \text{,}\) dus \(a = -1\frac{1}{2} \text{.}\)

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

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

Gelijk stellen geeft \(9 y - 28 = 7 y - 22\)

\(2 y = 6\) dus \(y = 3\)

\(\begin{rcases}x = 9 y - 28 \\ y = 3\end{rcases} \begin{matrix}x = 9 ⋅ 3 - 28 \\ x = -1\end{matrix}\)

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

Substitutie geeft \(5 (7 y + 13) + 4 y = 26\)

Haakjes wegwerken geeft
\(35 y + 65 + 4 y = 26\)
\(39 y = -39\)
\(y = -1\)

\(\begin{rcases}x = 7 y + 13 \\ y = -1\end{rcases} \begin{matrix}x = 7 ⋅ -1 + 13 \\ x = 6\end{matrix}\)

De oplossing is \((x , y) = (6 , -1) \text{.}\)

Substitutie geeft \(q = 7 (4 q + 8) + 25\)

Haakjes wegwerken geeft
\(q = 28 q + 56 + 25\)
\(-27 q = 81\)
\(q = -3\)

\(\begin{rcases}p = 4 q + 8 \\ q = -3\end{rcases} \begin{matrix}p = 4 ⋅ -3 + 8 \\ p = -4\end{matrix}\)

De oplossing is \((p , q) = (-4 , -3) \text{.}\)