Getal & Ruimte (13e editie) - havo wiskunde B
'Stelsels oplossen'.
| havo wiskunde B | 1.4 Stelsels vergelijkingen |
opgave 1Los exact op. 3p a \(\begin{cases}4 x + 4 y = -6 \\ 5 x + 4 y = 2\end{cases}\) Eliminatie (1) 003f - Stelsels oplossen - basis - 318ms - dynamic variables a Aftrekken geeft \(-x = -8 \text{,}\) dus \(x = 8 \text{.}\) 1p ○ \(\begin{rcases}4 x + 4 y = -6 \\ x = 8\end{rcases} \begin{matrix}4 ⋅ 8 + 4 y = -6 \\ 4 y = -38 \\ y = -9\frac{1}{2}\end{matrix}\) 1p ○ De oplossing is \((x , y) = (8 , -9\frac{1}{2}) \text{.}\) 1p 4p b \(\begin{cases}3 p - 2 q = 1 \\ 4 p - 4 q = -2\end{cases}\) Eliminatie (2) 003g - Stelsels oplossen - basis - 10ms - dynamic variables b \(\begin{cases}3 p - 2 q = 1 \\ 4 p - 4 q = -2\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}6 p - 4 q = 2 \\ 4 p - 4 q = -2\end{cases}\) 1p ○ Aftrekken geeft \(2 p = 4 \text{,}\) dus \(p = 2 \text{.}\) 1p ○ \(\begin{rcases}3 p - 2 q = 1 \\ p = 2\end{rcases} \begin{matrix}3 ⋅ 2 - 2 q = 1 \\ -2 q = -5 \\ q = 2\frac{1}{2}\end{matrix}\) 1p ○ De oplossing is \((p , q) = (2 , 2\frac{1}{2}) \text{.}\) 1p 4p c \(\begin{cases}3 a + 5 b = 1 \\ 5 a - 3 b = -4\end{cases}\) Eliminatie (3) 003h - Stelsels oplossen - basis - 11ms - dynamic variables c \(\begin{cases}3 a + 5 b = 1 \\ 5 a - 3 b = -4\end{cases}\) \(\begin{vmatrix}3 \\ 5\end{vmatrix}\) geeft \(\begin{cases}9 a + 15 b = 3 \\ 25 a - 15 b = -20\end{cases}\) 1p ○ Optellen geeft \(34 a = -17 \text{,}\) dus \(a = -\frac{1}{2} \text{.}\) 1p ○ \(\begin{rcases}3 a + 5 b = 1 \\ a = -\frac{1}{2}\end{rcases} \begin{matrix}3 ⋅ -\frac{1}{2} + 5 b = 1 \\ 5 b = 2\frac{1}{2} \\ b = \frac{1}{2}\end{matrix}\) 1p ○ De oplossing is \((a , b) = (-\frac{1}{2} , \frac{1}{2}) \text{.}\) 1p 4p d \(\begin{cases}x = 7 y - 19 \\ x = 3 y - 7\end{cases}\) GelijkStellen 003i - Stelsels oplossen - basis - 1ms d Gelijk stellen geeft \(7 y - 19 = 3 y - 7\) 1p ○ \(4 y = 12\) dus \(y = 3\) 1p ○ \(\begin{rcases}x = 7 y - 19 \\ y = 3\end{rcases} \begin{matrix}x = 7 ⋅ 3 - 19 \\ x = 2\end{matrix}\) 1p ○ De oplossing is \((x , y) = (2 , 3) \text{.}\) 1p opgave 2Los exact op. 4p a \(\begin{cases}9 x + 2 y = -47 \\ x = 6 y + 1\end{cases}\) Substitutie (1) 003j - Stelsels oplossen - basis - 0ms - dynamic variables a Substitutie geeft \(9 (6 y + 1) + 2 y = -47\) 1p ○ Haakjes wegwerken geeft 1p ○ \(\begin{rcases}x = 6 y + 1 \\ y = -1\end{rcases} \begin{matrix}x = 6 ⋅ -1 + 1 \\ x = -5\end{matrix}\) 1p ○ De oplossing is \((x , y) = (-5 , -1) \text{.}\) 1p 4p b \(\begin{cases}a = 9 b - 15 \\ b = 3 a + 19\end{cases}\) Substitutie (2) 003k - Stelsels oplossen - basis - 0ms - dynamic variables b Substitutie geeft \(a = 9 (3 a + 19) - 15\) 1p ○ Haakjes wegwerken geeft 1p ○ \(\begin{rcases}b = 3 a + 19 \\ a = -6\end{rcases} \begin{matrix}b = 3 ⋅ -6 + 19 \\ b = 1\end{matrix}\) 1p ○ De oplossing is \((a , b) = (-6 , 1) \text{.}\) 1p |