Stelsels oplossen

Aftrekken geeft \(y = -7 \text{.}\)

\(\begin{rcases}6 x + 4 y = -1 \\ y = -7\end{rcases} \begin{matrix}6 x + 4 ⋅ -7 = -1 \\ 6 x = 27 \\ x = 4\frac{1}{2}\end{matrix}\)

De oplossing is \((x , y) = (4\frac{1}{2} , -7) \text{.}\)

\(\begin{cases}a - 3 b = 5 \\ 6 a + 6 b = -6\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}2 a - 6 b = 10 \\ 6 a + 6 b = -6\end{cases}\)

Optellen geeft \(8 a = 4 \text{,}\) dus \(a = \frac{1}{2} \text{.}\)

\(\begin{rcases}a - 3 b = 5 \\ a = \frac{1}{2}\end{rcases} \begin{matrix}\frac{1}{2} - 3 b = 5 \\ -3 b = 4\frac{1}{2} \\ b = -1\frac{1}{2}\end{matrix}\)

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

\(\begin{cases}2 x - 6 y = -6 \\ 3 x - 5 y = -3\end{cases}\) \(\begin{vmatrix}5 \\ 6\end{vmatrix}\) geeft \(\begin{cases}10 x - 30 y = -30 \\ 18 x - 30 y = -18\end{cases}\)

Aftrekken geeft \(-8 x = -12 \text{,}\) dus \(x = 1\frac{1}{2} \text{.}\)

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

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

Gelijk stellen geeft \(6 x - 34 = 2 x - 14\)

\(4 x = 20\) dus \(x = 5\)

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

De oplossing is \((x , y) = (5 , -4) \text{.}\)

Substitutie geeft \(4 a + 7 (2 a + 7) = 13\)

Haakjes wegwerken geeft
\(4 a + 14 a + 49 = 13\)
\(18 a = -36\)
\(a = -2\)

\(\begin{rcases}b = 2 a + 7 \\ a = -2\end{rcases} \begin{matrix}b = 2 ⋅ -2 + 7 \\ b = 3\end{matrix}\)

De oplossing is \((a , b) = (-2 , 3) \text{.}\)

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

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

\(\begin{rcases}q = 2 p + 18 \\ p = -6\end{rcases} \begin{matrix}q = 2 ⋅ -6 + 18 \\ q = 6\end{matrix}\)

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