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

\(\begin{cases}4 x + 6 y = 6 \\ 4 x + 5 y = -1\end{cases}\)

Eliminatie (1)
003f - Stelsels oplossen - basis - 318ms - dynamic variables

a

Aftrekken geeft \(y = 7 \text{.}\)

1p

\(\begin{rcases}4 x + 6 y = 6 \\ y = 7\end{rcases} \begin{matrix}4 x + 6 ⋅ 7 = 6 \\ 4 x = -36 \\ x = -9\end{matrix}\)

1p

De oplossing is \((x , y) = (-9 , 7) \text{.}\)

1p

4p

b

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

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

b

\(\begin{cases}5 x - 4 y = -4 \\ 2 x - 2 y = 6\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}5 x - 4 y = -4 \\ 4 x - 4 y = 12\end{cases}\)

1p

Aftrekken geeft \(x = -16 \text{.}\)

1p

\(\begin{rcases}5 x - 4 y = -4 \\ x = -16\end{rcases} \begin{matrix}5 ⋅ -16 - 4 y = -4 \\ -4 y = 76 \\ y = -19\end{matrix}\)

1p

De oplossing is \((x , y) = (-16 , -19) \text{.}\)

1p

4p

c

\(\begin{cases}3 a + 5 b = -1 \\ 4 a - 4 b = 4\end{cases}\)

Eliminatie (3)
003h - Stelsels oplossen - basis - 11ms - dynamic variables

c

\(\begin{cases}3 a + 5 b = -1 \\ 4 a - 4 b = 4\end{cases}\) \(\begin{vmatrix}4 \\ 5\end{vmatrix}\) geeft \(\begin{cases}12 a + 20 b = -4 \\ 20 a - 20 b = 20\end{cases}\)

1p

Optellen geeft \(32 a = 16 \text{,}\) dus \(a = \frac{1}{2} \text{.}\)

1p

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

1p

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

1p

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