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

'Stelsels oplossen'.

vwo wiskunde A k.1 Stelsels van lineaire vergelijkingen

Stelsels oplossen (4)

opgave 1

Los exact op.

3p

a

\(\begin{cases}p-2q=5 \\ 2p+2q=1\end{cases}\)

Eliminatie (1)
003f - Stelsels oplossen - basis - 416ms - dynamic variables

a

Optellen geeft \(3p=6\text{,}\) dus \(p=2\text{.}\)

1p

○

\(\begin{rcases}p-2q=5 \\ p=2\end{rcases}\begin{matrix}2-2q=5 \\ -2q=3 \\ q=-1\frac{1}{2}\end{matrix}\)

1p

○

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

1p

4p

b

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

Eliminatie (2)
003g - Stelsels oplossen - basis - 8ms - dynamic variables

b

\(\begin{cases}4a-6b=-6 \\ 2a-2b=3\end{cases}\) \(\begin{vmatrix}1 \\ 3\end{vmatrix}\) geeft \(\begin{cases}4a-6b=-6 \\ 6a-6b=9\end{cases}\)

1p

○

Aftrekken geeft \(-2a=-15\text{,}\) dus \(a=7\frac{1}{2}\text{.}\)

1p

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\(\begin{rcases}4a-6b=-6 \\ a=7\frac{1}{2}\end{rcases}\begin{matrix}4⋅7\frac{1}{2}-6b=-6 \\ -6b=-36 \\ b=6\end{matrix}\)

1p

○

De oplossing is \((a, b)=(7\frac{1}{2}, 6)\text{.}\)

1p

4p

c

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

Eliminatie (3)
003h - Stelsels oplossen - basis - 10ms - dynamic variables

c

\(\begin{cases}4a+2b=-5 \\ 5a+3b=-3\end{cases}\) \(\begin{vmatrix}3 \\ 2\end{vmatrix}\) geeft \(\begin{cases}12a+6b=-15 \\ 10a+6b=-6\end{cases}\)

1p

○

Aftrekken geeft \(2a=-9\text{,}\) dus \(a=-4\frac{1}{2}\text{.}\)

1p

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\(\begin{rcases}4a+2b=-5 \\ a=-4\frac{1}{2}\end{rcases}\begin{matrix}4⋅-4\frac{1}{2}+2b=-5 \\ 2b=13 \\ b=6\frac{1}{2}\end{matrix}\)

1p

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De oplossing is \((a, b)=(-4\frac{1}{2}, 6\frac{1}{2})\text{.}\)

1p

opgave 2

De lijnen \(k{:}\,4x-3y=2\) en \(l{:}\,2x-y=-1\) snijden elkaar in het punt \(S\text{.}\)

4p

Bereken de coördinaten van \(S\text{.}\)

SnijpuntVanTweeLijnen (1)
00bs - Stelsels oplossen - basis - 470ms - data pool: #928 (470ms)

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\(\begin{cases}4x-3y=2 \\ 2x-y=-1\end{cases}\) \(\begin{vmatrix}1 \\ 3\end{vmatrix}\) geeft \(\begin{cases}4x-3y=2 \\ 6x-3y=-3\end{cases}\)

1p

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Aftrekken geeft \(-2x=5\) dus \(x=-2\frac{1}{2}\text{.}\)

1p

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\(\begin{rcases}4x-3y=2 \\ x=-2\frac{1}{2}\end{rcases}\begin{matrix}4⋅-2\frac{1}{2}-3y=2 \\ -3y=12 \\ y=-4\end{matrix}\)

1p

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Dus \(S(-2\frac{1}{2}, -4)\text{.}\)

1p
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