Getal & Ruimte (12e editie) - vwo wiskunde C

'Stelsels oplossen'.

vwo wiskunde A k.1 Stelsels van lineaire vergelijkingen

Stelsels oplossen (3)

opgave 1

Los exact op.

3p

a

\(\begin{cases}4x-y=3 \\ 5x-y=-2\end{cases}\)

Eliminatie (1)
003f - Stelsels oplossen - basis - 361ms - dynamic variables

a

Aftrekken geeft \(-x=5\text{,}\) dus \(x=-5\text{.}\)

1p

\(\begin{rcases}4x-y=3 \\ x=-5\end{rcases}\begin{matrix}4⋅-5-y=3 \\ -y=23 \\ y=-23\end{matrix}\)

1p

De oplossing is \((x, y)=(-5, -23)\text{.}\)

1p

4p

b

\(\begin{cases}2a+b=2 \\ a-2b=6\end{cases}\)

Eliminatie (2)
003g - Stelsels oplossen - basis - 15ms - dynamic variables

b

\(\begin{cases}2a+b=2 \\ a-2b=6\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}4a+2b=4 \\ a-2b=6\end{cases}\)

1p

Optellen geeft \(5a=10\text{,}\) dus \(a=2\text{.}\)

1p

\(\begin{rcases}2a+b=2 \\ a=2\end{rcases}\begin{matrix}2⋅2+b=2 \\ b=-2\end{matrix}\)

1p

De oplossing is \((a, b)=(2, -2)\text{.}\)

1p

4p

c

\(\begin{cases}4a+4b=2 \\ 6a+3b=-3\end{cases}\)

Eliminatie (3)
003h - Stelsels oplossen - basis - 12ms - dynamic variables

c

\(\begin{cases}4a+4b=2 \\ 6a+3b=-3\end{cases}\) \(\begin{vmatrix}3 \\ 4\end{vmatrix}\) geeft \(\begin{cases}12a+12b=6 \\ 24a+12b=-12\end{cases}\)

1p

Aftrekken geeft \(-12a=18\text{,}\) dus \(a=-1\frac{1}{2}\text{.}\)

1p

\(\begin{rcases}4a+4b=2 \\ a=-1\frac{1}{2}\end{rcases}\begin{matrix}4⋅-1\frac{1}{2}+4b=2 \\ 4b=8 \\ b=2\end{matrix}\)

1p

De oplossing is \((a, b)=(-1\frac{1}{2}, 2)\text{.}\)

1p

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