Getal & Ruimte (13e 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

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

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

a

Optellen geeft \(4 x = -10 \text{,}\) dus \(x = -2\frac{1}{2} \text{.}\)

1p

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

1p

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

1p

4p

b

\(\begin{cases}6 x + 4 y = 1 \\ 2 x - 6 y = 4\end{cases}\)

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

b

\(\begin{cases}6 x + 4 y = 1 \\ 2 x - 6 y = 4\end{cases}\) \(\begin{vmatrix}1 \\ 3\end{vmatrix}\) geeft \(\begin{cases}6 x + 4 y = 1 \\ 6 x - 18 y = 12\end{cases}\)

1p

Aftrekken geeft \(22 y = -11 \text{,}\) dus \(y = -\frac{1}{2} \text{.}\)

1p

\(\begin{rcases}6 x + 4 y = 1 \\ y = -\frac{1}{2}\end{rcases} \begin{matrix}6 x + 4 ⋅ -\frac{1}{2} = 1 \\ 6 x = 3 \\ x = \frac{1}{2}\end{matrix}\)

1p

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

1p

4p

c

\(\begin{cases}2 p - 3 q = 2 \\ 3 p - 4 q = 5\end{cases}\)

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

c

\(\begin{cases}2 p - 3 q = 2 \\ 3 p - 4 q = 5\end{cases}\) \(\begin{vmatrix}4 \\ 3\end{vmatrix}\) geeft \(\begin{cases}8 p - 12 q = 8 \\ 9 p - 12 q = 15\end{cases}\)

1p

Aftrekken geeft \(-p = -7 \text{,}\) dus \(p = 7 \text{.}\)

1p

\(\begin{rcases}2 p - 3 q = 2 \\ p = 7\end{rcases} \begin{matrix}2 ⋅ 7 - 3 q = 2 \\ -3 q = -12 \\ q = 4\end{matrix}\)

1p

De oplossing is \((p , q) = (7 , 4) \text{.}\)

1p

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