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

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

Eliminatie (1)
003f - Stelsels oplossen - basis - 317ms - dynamic variables

a

Aftrekken geeft \(x = -9 \text{.}\)

1p

\(\begin{rcases}4 x - 6 y = -6 \\ x = -9\end{rcases} \begin{matrix}4 ⋅ -9 - 6 y = -6 \\ -6 y = 30 \\ y = -5\end{matrix}\)

1p

De oplossing is \((x , y) = (-9 , -5) \text{.}\)

1p

4p

b

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

Eliminatie (2)
003g - Stelsels oplossen - basis - 10ms - dynamic variables

b

\(\begin{cases}x + y = -6 \\ 4 x - 4 y = 4\end{cases}\) \(\begin{vmatrix}4 \\ 1\end{vmatrix}\) geeft \(\begin{cases}4 x + 4 y = -24 \\ 4 x - 4 y = 4\end{cases}\)

1p

Optellen geeft \(8 x = -20 \text{,}\) dus \(x = -2\frac{1}{2} \text{.}\)

1p

\(\begin{rcases}x + y = -6 \\ x = -2\frac{1}{2}\end{rcases} \begin{matrix}-2\frac{1}{2} + y = -6 \\ y = -3\frac{1}{2}\end{matrix}\)

1p

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

1p

4p

c

\(\begin{cases}2 p + 4 q = 6 \\ 3 p + 3 q = -3\end{cases}\)

Eliminatie (3)
003h - Stelsels oplossen - basis - 7ms - dynamic variables

c

\(\begin{cases}2 p + 4 q = 6 \\ 3 p + 3 q = -3\end{cases}\) \(\begin{vmatrix}3 \\ 4\end{vmatrix}\) geeft \(\begin{cases}6 p + 12 q = 18 \\ 12 p + 12 q = -12\end{cases}\)

1p

Aftrekken geeft \(-6 p = 30 \text{,}\) dus \(p = -5 \text{.}\)

1p

\(\begin{rcases}2 p + 4 q = 6 \\ p = -5\end{rcases} \begin{matrix}2 ⋅ -5 + 4 q = 6 \\ 4 q = 16 \\ q = 4\end{matrix}\)

1p

De oplossing is \((p , q) = (-5 , 4) \text{.}\)

1p

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