Stelsels oplossen

Aftrekken geeft \(x = -6 \text{.}\)

\(\begin{rcases}2 x - 2 y = -2 \\ x = -6\end{rcases} \begin{matrix}2 ⋅ -6 - 2 y = -2 \\ -2 y = 10 \\ y = -5\end{matrix}\)

De oplossing is \((x , y) = (-6 , -5) \text{.}\)

\(\begin{cases}a + b = 5 \\ 2 a - 2 b = -4\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}2 a + 2 b = 10 \\ 2 a - 2 b = -4\end{cases}\)

Optellen geeft \(4 a = 6 \text{,}\) dus \(a = 1\frac{1}{2} \text{.}\)

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

De oplossing is \((a , b) = (1\frac{1}{2} , 3\frac{1}{2}) \text{.}\)

\(\begin{cases}2 p + 4 q = 1 \\ 5 p - 5 q = -5\end{cases}\) \(\begin{vmatrix}5 \\ 4\end{vmatrix}\) geeft \(\begin{cases}10 p + 20 q = 5 \\ 20 p - 20 q = -20\end{cases}\)

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

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

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

Gelijk stellen geeft \(5 x + 22 = 9 x + 38\)

\(-4 x = 16\) dus \(x = -4\)

\(\begin{rcases}y = 5 x + 22 \\ x = -4\end{rcases} \begin{matrix}y = 5 ⋅ -4 + 22 \\ y = 2\end{matrix}\)

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

Substitutie geeft \(9 (7 y - 15) + 5 y = 69\)

Haakjes wegwerken geeft
\(63 y - 135 + 5 y = 69\)
\(68 y = 204\)
\(y = 3\)

\(\begin{rcases}x = 7 y - 15 \\ y = 3\end{rcases} \begin{matrix}x = 7 ⋅ 3 - 15 \\ x = 6\end{matrix}\)

De oplossing is \((x , y) = (6 , 3) \text{.}\)

Substitutie geeft \(q = 3 (7 q + 10) + 10\)

Haakjes wegwerken geeft
\(q = 21 q + 30 + 10\)
\(-20 q = 40\)
\(q = -2\)

\(\begin{rcases}p = 7 q + 10 \\ q = -2\end{rcases} \begin{matrix}p = 7 ⋅ -2 + 10 \\ p = -4\end{matrix}\)

De oplossing is \((p , q) = (-4 , -2) \text{.}\)