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

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

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

a

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

1p

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

1p

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

1p

4p

b

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

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

b

\(\begin{cases}4 p + 2 q = 4 \\ 5 p + q = -4\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}4 p + 2 q = 4 \\ 10 p + 2 q = -8\end{cases}\)

1p

Aftrekken geeft \(-6 p = 12 \text{,}\) dus \(p = -2 \text{.}\)

1p

\(\begin{rcases}4 p + 2 q = 4 \\ p = -2\end{rcases} \begin{matrix}4 ⋅ -2 + 2 q = 4 \\ 2 q = 12 \\ q = 6\end{matrix}\)

1p

De oplossing is \((p , q) = (-2 , 6) \text{.}\)

1p

4p

c

\(\begin{cases}4 a - 6 b = -3 \\ 3 a - 5 b = 4\end{cases}\)

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

c

\(\begin{cases}4 a - 6 b = -3 \\ 3 a - 5 b = 4\end{cases}\) \(\begin{vmatrix}5 \\ 6\end{vmatrix}\) geeft \(\begin{cases}20 a - 30 b = -15 \\ 18 a - 30 b = 24\end{cases}\)

1p

Aftrekken geeft \(2 a = -39 \text{,}\) dus \(a = -19\frac{1}{2} \text{.}\)

1p

\(\begin{rcases}4 a - 6 b = -3 \\ a = -19\frac{1}{2}\end{rcases} \begin{matrix}4 ⋅ -19\frac{1}{2} - 6 b = -3 \\ -6 b = 75 \\ b = -12\frac{1}{2}\end{matrix}\)

1p

De oplossing is \((a , b) = (-19\frac{1}{2} , -12\frac{1}{2}) \text{.}\)

1p

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