Getal & Ruimte (12e editie) - vwo wiskunde A
'Stelsels oplossen'.
| vwo wiskunde A | k.1 Stelsels van lineaire vergelijkingen |
opgave 1Los exact op. 3p a \(\begin{cases}4 x - 6 y = -6 \\ 3 x - 6 y = 3\end{cases}\) Eliminatie (1) 003f - Stelsels oplossen - basis - 317ms - dynamic variables a Aftrekken geeft \(x = -9 \text{.}\) 1p ○ \(\begin{rcases}4 x - 6 y = -6 \\ x = -9\end{rcases} \begin{matrix}4 ⋅ -9 - 6 y = -6 \\ -6 y = 30 \\ y = -5\end{matrix}\) 1p ○ De oplossing is \((x , y) = (-9 , -5) \text{.}\) 1p 4p b \(\begin{cases}x + y = -6 \\ 4 x - 4 y = 4\end{cases}\) Eliminatie (2) 003g - Stelsels oplossen - basis - 10ms - dynamic variables b \(\begin{cases}x + y = -6 \\ 4 x - 4 y = 4\end{cases}\) \(\begin{vmatrix}4 \\ 1\end{vmatrix}\) geeft \(\begin{cases}4 x + 4 y = -24 \\ 4 x - 4 y = 4\end{cases}\) 1p ○ Optellen geeft \(8 x = -20 \text{,}\) dus \(x = -2\frac{1}{2} \text{.}\) 1p ○ \(\begin{rcases}x + y = -6 \\ x = -2\frac{1}{2}\end{rcases} \begin{matrix}-2\frac{1}{2} + y = -6 \\ y = -3\frac{1}{2}\end{matrix}\) 1p ○ De oplossing is \((x , y) = (-2\frac{1}{2} , -3\frac{1}{2}) \text{.}\) 1p 4p c \(\begin{cases}2 p + 4 q = 6 \\ 3 p + 3 q = -3\end{cases}\) Eliminatie (3) 003h - Stelsels oplossen - basis - 7ms - dynamic variables c \(\begin{cases}2 p + 4 q = 6 \\ 3 p + 3 q = -3\end{cases}\) \(\begin{vmatrix}3 \\ 4\end{vmatrix}\) geeft \(\begin{cases}6 p + 12 q = 18 \\ 12 p + 12 q = -12\end{cases}\) 1p ○ Aftrekken geeft \(-6 p = 30 \text{,}\) dus \(p = -5 \text{.}\) 1p ○ \(\begin{rcases}2 p + 4 q = 6 \\ p = -5\end{rcases} \begin{matrix}2 ⋅ -5 + 4 q = 6 \\ 4 q = 16 \\ q = 4\end{matrix}\) 1p ○ De oplossing is \((p , q) = (-5 , 4) \text{.}\) 1p |