Getal & Ruimte (13e 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

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

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

a

Optellen geeft \(6 x = 9 \text{,}\) dus \(x = 1\frac{1}{2} \text{.}\)

1p

\(\begin{rcases}2 x - 4 y = 5 \\ x = 1\frac{1}{2}\end{rcases} \begin{matrix}2 ⋅ 1\frac{1}{2} - 4 y = 5 \\ -4 y = 2 \\ y = -\frac{1}{2}\end{matrix}\)

1p

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

1p

4p

b

\(\begin{cases}4 p + 2 q = -6 \\ 6 p + 4 q = -1\end{cases}\)

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

b

\(\begin{cases}4 p + 2 q = -6 \\ 6 p + 4 q = -1\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}8 p + 4 q = -12 \\ 6 p + 4 q = -1\end{cases}\)

1p

Aftrekken geeft \(2 p = -11 \text{,}\) dus \(p = -5\frac{1}{2} \text{.}\)

1p

\(\begin{rcases}4 p + 2 q = -6 \\ p = -5\frac{1}{2}\end{rcases} \begin{matrix}4 ⋅ -5\frac{1}{2} + 2 q = -6 \\ 2 q = 16 \\ q = 8\end{matrix}\)

1p

De oplossing is \((p , q) = (-5\frac{1}{2} , 8) \text{.}\)

1p

4p

c

\(\begin{cases}3 x - 5 y = -2 \\ 2 x - 4 y = 6\end{cases}\)

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

c

\(\begin{cases}3 x - 5 y = -2 \\ 2 x - 4 y = 6\end{cases}\) \(\begin{vmatrix}4 \\ 5\end{vmatrix}\) geeft \(\begin{cases}12 x - 20 y = -8 \\ 10 x - 20 y = 30\end{cases}\)

1p

Aftrekken geeft \(2 x = -38 \text{,}\) dus \(x = -19 \text{.}\)

1p

\(\begin{rcases}3 x - 5 y = -2 \\ x = -19\end{rcases} \begin{matrix}3 ⋅ -19 - 5 y = -2 \\ -5 y = 55 \\ y = -11\end{matrix}\)

1p

De oplossing is \((x , y) = (-19 , -11) \text{.}\)

1p

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