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

\(\begin{cases}x+y=-4 \\ 3x-y=-6\end{cases}\)

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

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