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

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

Eliminatie (1)
003f - Stelsels oplossen - basis - 359ms - dynamic variables

a

Aftrekken geeft \(-2 x = -8 \text{,}\) dus \(x = 4 \text{.}\)

1p

○

\(\begin{rcases}x - 2 y = -4 \\ x = 4\end{rcases} \begin{matrix}4 - 2 y = -4 \\ -2 y = -8 \\ y = 4\end{matrix}\)

1p

○

De oplossing is \((x , y) = (4 , 4) \text{.}\)

1p

4p

b

\(\begin{cases}3 x - y = 5 \\ x + 5 y = -1\end{cases}\)

Eliminatie (2)
003g - Stelsels oplossen - basis - 11ms - dynamic variables

b

\(\begin{cases}3 x - y = 5 \\ x + 5 y = -1\end{cases}\) \(\begin{vmatrix}5 \\ 1\end{vmatrix}\) geeft \(\begin{cases}15 x - 5 y = 25 \\ x + 5 y = -1\end{cases}\)

1p

○

Optellen geeft \(16 x = 24 \text{,}\) dus \(x = 1\frac{1}{2} \text{.}\)

1p

○

\(\begin{rcases}3 x - y = 5 \\ x = 1\frac{1}{2}\end{rcases} \begin{matrix}3 ⋅ 1\frac{1}{2} - y = 5 \\ -y = \frac{1}{2} \\ y = -\frac{1}{2}\end{matrix}\)

1p

○

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

1p

4p

c

\(\begin{cases}3 a + 4 b = 2 \\ 4 a + 3 b = -2\end{cases}\)

Eliminatie (3)
003h - Stelsels oplossen - basis - 13ms - dynamic variables

c

\(\begin{cases}3 a + 4 b = 2 \\ 4 a + 3 b = -2\end{cases}\) \(\begin{vmatrix}3 \\ 4\end{vmatrix}\) geeft \(\begin{cases}9 a + 12 b = 6 \\ 16 a + 12 b = -8\end{cases}\)

1p

○

Aftrekken geeft \(-7 a = 14 \text{,}\) dus \(a = -2 \text{.}\)

1p

○

\(\begin{rcases}3 a + 4 b = 2 \\ a = -2\end{rcases} \begin{matrix}3 ⋅ -2 + 4 b = 2 \\ 4 b = 8 \\ b = 2\end{matrix}\)

1p

○

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

1p

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