Stelsels oplossen

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

\(\begin{rcases}4 x - 2 y = -3 \\ x = -3\end{rcases} \begin{matrix}4 ⋅ -3 - 2 y = -3 \\ -2 y = 9 \\ y = -4\frac{1}{2}\end{matrix}\)

De oplossing is \((x , y) = (-3 , -4\frac{1}{2}) \text{.}\)

\(\begin{cases}2 a + 6 b = -5 \\ a + 4 b = -1\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}2 a + 6 b = -5 \\ 2 a + 8 b = -2\end{cases}\)

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

\(\begin{rcases}2 a + 6 b = -5 \\ b = 1\frac{1}{2}\end{rcases} \begin{matrix}2 a + 6 ⋅ 1\frac{1}{2} = -5 \\ 2 a = -14 \\ a = -7\end{matrix}\)

De oplossing is \((a , b) = (-7 , 1\frac{1}{2}) \text{.}\)

\(\begin{cases}2 a + 2 b = 4 \\ 3 a - 5 b = 2\end{cases}\) \(\begin{vmatrix}5 \\ 2\end{vmatrix}\) geeft \(\begin{cases}10 a + 10 b = 20 \\ 6 a - 10 b = 4\end{cases}\)

Optellen geeft \(16 a = 24 \text{,}\) 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 \(9 y + 13 = 7 y + 11\)

\(2 y = -2\) dus \(y = -1\)

\(\begin{rcases}x = 9 y + 13 \\ y = -1\end{rcases} \begin{matrix}x = 9 ⋅ -1 + 13 \\ x = 4\end{matrix}\)

De oplossing is \((x , y) = (4 , -1) \text{.}\)

Substitutie geeft \(8 p + 5 (2 p + 8) = 22\)

Haakjes wegwerken geeft
\(8 p + 10 p + 40 = 22\)
\(18 p = -18\)
\(p = -1\)

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

De oplossing is \((p , q) = (-1 , 6) \text{.}\)

Substitutie geeft \(x = 8 (2 x - 18) + 54\)

Haakjes wegwerken geeft
\(x = 16 x - 144 + 54\)
\(-15 x = -90\)
\(x = 6\)

\(\begin{rcases}y = 2 x - 18 \\ x = 6\end{rcases} \begin{matrix}y = 2 ⋅ 6 - 18 \\ y = -6\end{matrix}\)

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