Stelsels oplossen

Aftrekken geeft \(2 a = 9 \text{,}\) dus \(a = 4\frac{1}{2} \text{.}\)

\(\begin{rcases}3 a + b = 5 \\ a = 4\frac{1}{2}\end{rcases} \begin{matrix}3 ⋅ 4\frac{1}{2} + b = 5 \\ b = -8\frac{1}{2}\end{matrix}\)

De oplossing is \((a , b) = (4\frac{1}{2} , -8\frac{1}{2}) \text{.}\)

\(\begin{cases}3 p - 4 q = 5 \\ 2 p - q = -5\end{cases}\) \(\begin{vmatrix}1 \\ 4\end{vmatrix}\) geeft \(\begin{cases}3 p - 4 q = 5 \\ 8 p - 4 q = -20\end{cases}\)

Aftrekken geeft \(-5 p = 25 \text{,}\) dus \(p = -5 \text{.}\)

\(\begin{rcases}3 p - 4 q = 5 \\ p = -5\end{rcases} \begin{matrix}3 ⋅ -5 - 4 q = 5 \\ -4 q = 20 \\ q = -5\end{matrix}\)

De oplossing is \((p , q) = (-5 , -5) \text{.}\)

\(\begin{cases}5 a - 3 b = -4 \\ 2 a + 4 b = 1\end{cases}\) \(\begin{vmatrix}4 \\ 3\end{vmatrix}\) geeft \(\begin{cases}20 a - 12 b = -16 \\ 6 a + 12 b = 3\end{cases}\)

Optellen geeft \(26 a = -13 \text{,}\) dus \(a = -\frac{1}{2} \text{.}\)

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

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

Gelijk stellen geeft \(5 x - 1 = 3 x + 1\)

\(2 x = 2\) dus \(x = 1\)

\(\begin{rcases}y = 5 x - 1 \\ x = 1\end{rcases} \begin{matrix}y = 5 ⋅ 1 - 1 \\ y = 4\end{matrix}\)

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

Substitutie geeft \(5 (9 y - 48) + 2 y = -5\)

Haakjes wegwerken geeft
\(45 y - 240 + 2 y = -5\)
\(47 y = 235\)
\(y = 5\)

\(\begin{rcases}x = 9 y - 48 \\ y = 5\end{rcases} \begin{matrix}x = 9 ⋅ 5 - 48 \\ x = -3\end{matrix}\)

De oplossing is \((x , y) = (-3 , 5) \text{.}\)

Substitutie geeft \(y = 3 (5 y + 10) + 12\)

Haakjes wegwerken geeft
\(y = 15 y + 30 + 12\)
\(-14 y = 42\)
\(y = -3\)

\(\begin{rcases}x = 5 y + 10 \\ y = -3\end{rcases} \begin{matrix}x = 5 ⋅ -3 + 10 \\ x = -5\end{matrix}\)

De oplossing is \((x , y) = (-5 , -3) \text{.}\)