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

'Stelsels oplossen'.

vwo wiskunde A k.1 Stelsels van lineaire vergelijkingen

Stelsels oplossen (3)

opgave 1

Los exact op.

3p

a

\(\begin{cases}6a-4b=-5 \\ 6a-2b=5\end{cases}\)

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

a

Aftrekken geeft \(-2b=-10\text{,}\) dus \(b=5\text{.}\)

1p

\(\begin{rcases}6a-4b=-5 \\ b=5\end{rcases}\begin{matrix}6a-4⋅5=-5 \\ 6a=15 \\ a=2\frac{1}{2}\end{matrix}\)

1p

De oplossing is \((a, b)=(2\frac{1}{2}, 5)\text{.}\)

1p

4p

b

\(\begin{cases}3a-2b=-4 \\ 2a-b=-5\end{cases}\)

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

b

\(\begin{cases}3a-2b=-4 \\ 2a-b=-5\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}3a-2b=-4 \\ 4a-2b=-10\end{cases}\)

1p

Aftrekken geeft \(-a=6\text{,}\) dus \(a=-6\text{.}\)

1p

\(\begin{rcases}3a-2b=-4 \\ a=-6\end{rcases}\begin{matrix}3⋅-6-2b=-4 \\ -2b=14 \\ b=-7\end{matrix}\)

1p

De oplossing is \((a, b)=(-6, -7)\text{.}\)

1p

4p

c

\(\begin{cases}4x-2y=3 \\ 5x+3y=1\end{cases}\)

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

c

\(\begin{cases}4x-2y=3 \\ 5x+3y=1\end{cases}\) \(\begin{vmatrix}3 \\ 2\end{vmatrix}\) geeft \(\begin{cases}12x-6y=9 \\ 10x+6y=2\end{cases}\)

1p

Optellen geeft \(22x=11\text{,}\) dus \(x=\frac{1}{2}\text{.}\)

1p

\(\begin{rcases}4x-2y=3 \\ x=\frac{1}{2}\end{rcases}\begin{matrix}4⋅\frac{1}{2}-2y=3 \\ -2y=1 \\ y=-\frac{1}{2}\end{matrix}\)

1p

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

1p

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