Getal & Ruimte (12e editie) - vwo wiskunde C
'Stelsels oplossen'.
| vwo wiskunde A | k.1 Stelsels van lineaire vergelijkingen |
opgave 1Los exact op. 3p a \(\begin{cases}5 a + b = 1 \\ 3 a - b = -5\end{cases}\) Eliminatie (1) 003f - Stelsels oplossen - basis - 317ms - dynamic variables a Optellen geeft \(8 a = -4 \text{,}\) dus \(a = -\frac{1}{2} \text{.}\) 1p ○ \(\begin{rcases}5 a + b = 1 \\ a = -\frac{1}{2}\end{rcases} \begin{matrix}5 ⋅ -\frac{1}{2} + b = 1 \\ b = 3\frac{1}{2}\end{matrix}\) 1p ○ De oplossing is \((a , b) = (-\frac{1}{2} , 3\frac{1}{2}) \text{.}\) 1p 4p b \(\begin{cases}6 x - y = -5 \\ 2 x - 2 y = -5\end{cases}\) Eliminatie (2) 003g - Stelsels oplossen - basis - 10ms - dynamic variables b \(\begin{cases}6 x - y = -5 \\ 2 x - 2 y = -5\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}12 x - 2 y = -10 \\ 2 x - 2 y = -5\end{cases}\) 1p ○ Aftrekken geeft \(10 x = -5 \text{,}\) dus \(x = -\frac{1}{2} \text{.}\) 1p ○ \(\begin{rcases}6 x - y = -5 \\ x = -\frac{1}{2}\end{rcases} \begin{matrix}6 ⋅ -\frac{1}{2} - y = -5 \\ -y = -2 \\ y = 2\end{matrix}\) 1p ○ De oplossing is \((x , y) = (-\frac{1}{2} , 2) \text{.}\) 1p 4p c \(\begin{cases}4 x + 5 y = 2 \\ 3 x + 3 y = -3\end{cases}\) Eliminatie (3) 003h - Stelsels oplossen - basis - 7ms - dynamic variables c \(\begin{cases}4 x + 5 y = 2 \\ 3 x + 3 y = -3\end{cases}\) \(\begin{vmatrix}3 \\ 5\end{vmatrix}\) geeft \(\begin{cases}12 x + 15 y = 6 \\ 15 x + 15 y = -15\end{cases}\) 1p ○ Aftrekken geeft \(-3 x = 21 \text{,}\) dus \(x = -7 \text{.}\) 1p ○ \(\begin{rcases}4 x + 5 y = 2 \\ x = -7\end{rcases} \begin{matrix}4 ⋅ -7 + 5 y = 2 \\ 5 y = 30 \\ y = 6\end{matrix}\) 1p ○ De oplossing is \((x , y) = (-7 , 6) \text{.}\) 1p |