Getal & Ruimte (12e editie) - vwo wiskunde C
'Stelsels oplossen'.
| vwo wiskunde A | k.1 Stelsels van lineaire vergelijkingen |
opgave 1Los exact op. 3p a \(\begin{cases}x - 2 y = -4 \\ 3 x - 2 y = 4\end{cases}\) Eliminatie (1) 003f - Stelsels oplossen - basis - 359ms - dynamic variables a Aftrekken geeft \(-2 x = -8 \text{,}\) dus \(x = 4 \text{.}\) 1p ○ \(\begin{rcases}x - 2 y = -4 \\ x = 4\end{rcases} \begin{matrix}4 - 2 y = -4 \\ -2 y = -8 \\ y = 4\end{matrix}\) 1p ○ De oplossing is \((x , y) = (4 , 4) \text{.}\) 1p 4p b \(\begin{cases}3 x - y = 5 \\ x + 5 y = -1\end{cases}\) Eliminatie (2) 003g - Stelsels oplossen - basis - 11ms - dynamic variables b \(\begin{cases}3 x - y = 5 \\ x + 5 y = -1\end{cases}\) \(\begin{vmatrix}5 \\ 1\end{vmatrix}\) geeft \(\begin{cases}15 x - 5 y = 25 \\ x + 5 y = -1\end{cases}\) 1p ○ Optellen geeft \(16 x = 24 \text{,}\) dus \(x = 1\frac{1}{2} \text{.}\) 1p ○ \(\begin{rcases}3 x - y = 5 \\ x = 1\frac{1}{2}\end{rcases} \begin{matrix}3 ⋅ 1\frac{1}{2} - y = 5 \\ -y = \frac{1}{2} \\ y = -\frac{1}{2}\end{matrix}\) 1p ○ De oplossing is \((x , y) = (1\frac{1}{2} , -\frac{1}{2}) \text{.}\) 1p 4p c \(\begin{cases}3 a + 4 b = 2 \\ 4 a + 3 b = -2\end{cases}\) Eliminatie (3) 003h - Stelsels oplossen - basis - 13ms - dynamic variables c \(\begin{cases}3 a + 4 b = 2 \\ 4 a + 3 b = -2\end{cases}\) \(\begin{vmatrix}3 \\ 4\end{vmatrix}\) geeft \(\begin{cases}9 a + 12 b = 6 \\ 16 a + 12 b = -8\end{cases}\) 1p ○ Aftrekken geeft \(-7 a = 14 \text{,}\) dus \(a = -2 \text{.}\) 1p ○ \(\begin{rcases}3 a + 4 b = 2 \\ a = -2\end{rcases} \begin{matrix}3 ⋅ -2 + 4 b = 2 \\ 4 b = 8 \\ b = 2\end{matrix}\) 1p ○ De oplossing is \((a , b) = (-2 , 2) \text{.}\) 1p |