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

\(\begin{cases}6p-2q=-2 \\ 2p-2q=4\end{cases}\)

Eliminatie (1)
003f - Stelsels oplossen - basis - 304ms - dynamic variables

a

Aftrekken geeft \(4p=-6\text{,}\) dus \(p=-1\frac{1}{2}\text{.}\)

1p

○

\(\begin{rcases}6p-2q=-2 \\ p=-1\frac{1}{2}\end{rcases}\begin{matrix}6⋅-1\frac{1}{2}-2q=-2 \\ -2q=7 \\ q=-3\frac{1}{2}\end{matrix}\)

1p

○

De oplossing is \((p, q)=(-1\frac{1}{2}, -3\frac{1}{2})\text{.}\)

1p

4p

b

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

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

b

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

1p

○

Optellen geeft \(14x=-7\text{,}\) dus \(x=-\frac{1}{2}\text{.}\)

1p

○

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

1p

○

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

1p

4p

c

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

Eliminatie (3)
003h - Stelsels oplossen - basis - 8ms - dynamic variables

c

\(\begin{cases}6a+4b=3 \\ 4a+6b=-3\end{cases}\) \(\begin{vmatrix}3 \\ 2\end{vmatrix}\) geeft \(\begin{cases}18a+12b=9 \\ 8a+12b=-6\end{cases}\)

1p

○

Aftrekken geeft \(10a=15\text{,}\) dus \(a=1\frac{1}{2}\text{.}\)

1p

○

\(\begin{rcases}6a+4b=3 \\ a=1\frac{1}{2}\end{rcases}\begin{matrix}6⋅1\frac{1}{2}+4b=3 \\ 4b=-6 \\ b=-1\frac{1}{2}\end{matrix}\)

1p

○

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

1p

opgave 2

De lijnen \(k{:}\,4x+2y=4\) en \(l{:}\,2x-y=0\) snijden elkaar in het punt \(S\text{.}\)

4p

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

SnijpuntVanTweeLijnen (1)
00bs - Stelsels oplossen - basis - 218ms - data pool: #928 (218ms)

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\(\begin{cases}4x+2y=4 \\ 2x-y=0\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}4x+2y=4 \\ 4x-2y=0\end{cases}\)

1p

○

Optellen geeft \(8x=4\) dus \(x=\frac{1}{2}\text{.}\)

1p

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

1p

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

1p
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