Getal & Ruimte (13e editie) - havo wiskunde B

'Stelsels oplossen'.

havo wiskunde B 1.4 Stelsels vergelijkingen

Stelsels oplossen (6)

opgave 1

Los exact op.

3p

a

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

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

a

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

1p

\(\begin{rcases}4 x + 4 y = -6 \\ x = 8\end{rcases} \begin{matrix}4 ⋅ 8 + 4 y = -6 \\ 4 y = -38 \\ y = -9\frac{1}{2}\end{matrix}\)

1p

De oplossing is \((x , y) = (8 , -9\frac{1}{2}) \text{.}\)

1p

4p

b

\(\begin{cases}3 p - 2 q = 1 \\ 4 p - 4 q = -2\end{cases}\)

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

b

\(\begin{cases}3 p - 2 q = 1 \\ 4 p - 4 q = -2\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}6 p - 4 q = 2 \\ 4 p - 4 q = -2\end{cases}\)

1p

Aftrekken geeft \(2 p = 4 \text{,}\) dus \(p = 2 \text{.}\)

1p

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

1p

De oplossing is \((p , q) = (2 , 2\frac{1}{2}) \text{.}\)

1p

4p

c

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

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

c

\(\begin{cases}3 a + 5 b = 1 \\ 5 a - 3 b = -4\end{cases}\) \(\begin{vmatrix}3 \\ 5\end{vmatrix}\) geeft \(\begin{cases}9 a + 15 b = 3 \\ 25 a - 15 b = -20\end{cases}\)

1p

Optellen geeft \(34 a = -17 \text{,}\) 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

4p

d

\(\begin{cases}x = 7 y - 19 \\ x = 3 y - 7\end{cases}\)

GelijkStellen
003i - Stelsels oplossen - basis - 1ms

d

Gelijk stellen geeft \(7 y - 19 = 3 y - 7\)

1p

\(4 y = 12\) dus \(y = 3\)

1p

\(\begin{rcases}x = 7 y - 19 \\ y = 3\end{rcases} \begin{matrix}x = 7 ⋅ 3 - 19 \\ x = 2\end{matrix}\)

1p

De oplossing is \((x , y) = (2 , 3) \text{.}\)

1p

opgave 2

Los exact op.

4p

a

\(\begin{cases}9 x + 2 y = -47 \\ x = 6 y + 1\end{cases}\)

Substitutie (1)
003j - Stelsels oplossen - basis - 0ms - dynamic variables

a

Substitutie geeft \(9 (6 y + 1) + 2 y = -47\)

1p

Haakjes wegwerken geeft
\(54 y + 9 + 2 y = -47\)
\(56 y = -56\)
\(y = -1\)

1p

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

1p

De oplossing is \((x , y) = (-5 , -1) \text{.}\)

1p

4p

b

\(\begin{cases}a = 9 b - 15 \\ b = 3 a + 19\end{cases}\)

Substitutie (2)
003k - Stelsels oplossen - basis - 0ms - dynamic variables

b

Substitutie geeft \(a = 9 (3 a + 19) - 15\)

1p

Haakjes wegwerken geeft
\(a = 27 a + 171 - 15\)
\(-26 a = 156\)
\(a = -6\)

1p

\(\begin{rcases}b = 3 a + 19 \\ a = -6\end{rcases} \begin{matrix}b = 3 ⋅ -6 + 19 \\ b = 1\end{matrix}\)

1p

De oplossing is \((a , b) = (-6 , 1) \text{.}\)

1p

"