Getal & Ruimte (13e editie) - vwo wiskunde C
'Stelsels oplossen'.
| vwo wiskunde A | k.1 Stelsels van lineaire vergelijkingen |
opgave 1Los exact op. 3p a \(\begin{cases}x + y = -5 \\ 3 x - y = -5\end{cases}\) Eliminatie (1) 003f - Stelsels oplossen - basis - 318ms - dynamic variables a Optellen geeft \(4 x = -10 \text{,}\) dus \(x = -2\frac{1}{2} \text{.}\) 1p ○ \(\begin{rcases}x + y = -5 \\ x = -2\frac{1}{2}\end{rcases} \begin{matrix}-2\frac{1}{2} + y = -5 \\ y = -2\frac{1}{2}\end{matrix}\) 1p ○ De oplossing is \((x , y) = (-2\frac{1}{2} , -2\frac{1}{2}) \text{.}\) 1p 4p b \(\begin{cases}6 x + 4 y = 1 \\ 2 x - 6 y = 4\end{cases}\) Eliminatie (2) 003g - Stelsels oplossen - basis - 10ms - dynamic variables b \(\begin{cases}6 x + 4 y = 1 \\ 2 x - 6 y = 4\end{cases}\) \(\begin{vmatrix}1 \\ 3\end{vmatrix}\) geeft \(\begin{cases}6 x + 4 y = 1 \\ 6 x - 18 y = 12\end{cases}\) 1p ○ Aftrekken geeft \(22 y = -11 \text{,}\) dus \(y = -\frac{1}{2} \text{.}\) 1p ○ \(\begin{rcases}6 x + 4 y = 1 \\ y = -\frac{1}{2}\end{rcases} \begin{matrix}6 x + 4 ⋅ -\frac{1}{2} = 1 \\ 6 x = 3 \\ x = \frac{1}{2}\end{matrix}\) 1p ○ De oplossing is \((x , y) = (\frac{1}{2} , -\frac{1}{2}) \text{.}\) 1p 4p c \(\begin{cases}2 p - 3 q = 2 \\ 3 p - 4 q = 5\end{cases}\) Eliminatie (3) 003h - Stelsels oplossen - basis - 11ms - dynamic variables c \(\begin{cases}2 p - 3 q = 2 \\ 3 p - 4 q = 5\end{cases}\) \(\begin{vmatrix}4 \\ 3\end{vmatrix}\) geeft \(\begin{cases}8 p - 12 q = 8 \\ 9 p - 12 q = 15\end{cases}\) 1p ○ Aftrekken geeft \(-p = -7 \text{,}\) dus \(p = 7 \text{.}\) 1p ○ \(\begin{rcases}2 p - 3 q = 2 \\ p = 7\end{rcases} \begin{matrix}2 ⋅ 7 - 3 q = 2 \\ -3 q = -12 \\ q = 4\end{matrix}\) 1p ○ De oplossing is \((p , q) = (7 , 4) \text{.}\) 1p |