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

\(\begin{cases}5 a + b = 1 \\ 3 a - b = -5\end{cases}\)

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

a

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

1p

\(\begin{rcases}5 a + b = 1 \\ a = -\frac{1}{2}\end{rcases} \begin{matrix}5 ⋅ -\frac{1}{2} + b = 1 \\ b = 3\frac{1}{2}\end{matrix}\)

1p

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

1p

4p

b

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

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

b

\(\begin{cases}6 x - y = -5 \\ 2 x - 2 y = -5\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}12 x - 2 y = -10 \\ 2 x - 2 y = -5\end{cases}\)

1p

Aftrekken geeft \(10 x = -5 \text{,}\) dus \(x = -\frac{1}{2} \text{.}\)

1p

\(\begin{rcases}6 x - y = -5 \\ x = -\frac{1}{2}\end{rcases} \begin{matrix}6 ⋅ -\frac{1}{2} - y = -5 \\ -y = -2 \\ y = 2\end{matrix}\)

1p

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

1p

4p

c

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

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

c

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

1p

Aftrekken geeft \(-3 x = 21 \text{,}\) dus \(x = -7 \text{.}\)

1p

\(\begin{rcases}4 x + 5 y = 2 \\ x = -7\end{rcases} \begin{matrix}4 ⋅ -7 + 5 y = 2 \\ 5 y = 30 \\ y = 6\end{matrix}\)

1p

De oplossing is \((x , y) = (-7 , 6) \text{.}\)

1p

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