Getal & Ruimte (12e editie) - vwo wiskunde C
'Stelsels oplossen'.
| vwo wiskunde A | k.1 Stelsels van lineaire vergelijkingen |
opgave 1Los exact op. 3p a \(\begin{cases}x+y=-4 \\ 3x-y=-6\end{cases}\) Eliminatie (1) 003f - Stelsels oplossen - basis - 439ms - dynamic variables a Optellen geeft \(4x=-10\text{,}\) dus \(x=-2\frac{1}{2}\text{.}\) 1p ○ \(\begin{rcases}x+y=-4 \\ x=-2\frac{1}{2}\end{rcases}\begin{matrix}-2\frac{1}{2}+y=-4 \\ y=-1\frac{1}{2}\end{matrix}\) 1p ○ De oplossing is \((x, y)=(-2\frac{1}{2}, -1\frac{1}{2})\text{.}\) 1p 4p b \(\begin{cases}5a-4b=-4 \\ a-b=-2\end{cases}\) Eliminatie (2) 003g - Stelsels oplossen - basis - 21ms - dynamic variables b \(\begin{cases}5a-4b=-4 \\ a-b=-2\end{cases}\) \(\begin{vmatrix}1 \\ 4\end{vmatrix}\) geeft \(\begin{cases}5a-4b=-4 \\ 4a-4b=-8\end{cases}\) 1p ○ Aftrekken geeft \(a=4\text{.}\) 1p ○ \(\begin{rcases}5a-4b=-4 \\ a=4\end{rcases}\begin{matrix}5⋅4-4b=-4 \\ -4b=-24 \\ b=6\end{matrix}\) 1p ○ De oplossing is \((a, b)=(4, 6)\text{.}\) 1p 4p c \(\begin{cases}4a+5b=4 \\ 3a+3b=6\end{cases}\) Eliminatie (3) 003h - Stelsels oplossen - basis - 17ms - dynamic variables c \(\begin{cases}4a+5b=4 \\ 3a+3b=6\end{cases}\) \(\begin{vmatrix}3 \\ 5\end{vmatrix}\) geeft \(\begin{cases}12a+15b=12 \\ 15a+15b=30\end{cases}\) 1p ○ Aftrekken geeft \(-3a=-18\text{,}\) dus \(a=6\text{.}\) 1p ○ \(\begin{rcases}4a+5b=4 \\ a=6\end{rcases}\begin{matrix}4⋅6+5b=4 \\ 5b=-20 \\ b=-4\end{matrix}\) 1p ○ De oplossing is \((a, b)=(6, -4)\text{.}\) 1p |