Stelsels oplossen

Eliminatie (1)
003f - Stelsels oplossen - 306ms - dynamic variables
code
003f
vaardigheid
Stelsels oplossen
start-up
306ms

Los exact op.

3p

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

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Aftrekken geeft \(x = -6 \text{.}\)

1p

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\(\begin{rcases}2 x - 2 y = -2 \\ x = -6\end{rcases} \begin{matrix}2 ⋅ -6 - 2 y = -2 \\ -2 y = 10 \\ y = -5\end{matrix}\)

1p

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De oplossing is \((x , y) = (-6 , -5) \text{.}\)

1p

Eliminatie (2)
003g - Stelsels oplossen - 12ms - dynamic variables
code
003g
vaardigheid
Stelsels oplossen
start-up
12ms

Los exact op.

4p

\(\begin{cases}a + b = 5 \\ 2 a - 2 b = -4\end{cases}\)

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\(\begin{cases}a + b = 5 \\ 2 a - 2 b = -4\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}2 a + 2 b = 10 \\ 2 a - 2 b = -4\end{cases}\)

1p

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Optellen geeft \(4 a = 6 \text{,}\) dus \(a = 1\frac{1}{2} \text{.}\)

1p

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

1p

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De oplossing is \((a , b) = (1\frac{1}{2} , 3\frac{1}{2}) \text{.}\)

1p

Eliminatie (3)
003h - Stelsels oplossen - 11ms - dynamic variables
code
003h
vaardigheid
Stelsels oplossen
start-up
11ms

Los exact op.

4p

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

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\(\begin{cases}2 p + 4 q = 1 \\ 5 p - 5 q = -5\end{cases}\) \(\begin{vmatrix}5 \\ 4\end{vmatrix}\) geeft \(\begin{cases}10 p + 20 q = 5 \\ 20 p - 20 q = -20\end{cases}\)

1p

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Optellen geeft \(30 p = -15 \text{,}\) dus \(p = -\frac{1}{2} \text{.}\)

1p

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\(\begin{rcases}2 p + 4 q = 1 \\ p = -\frac{1}{2}\end{rcases} \begin{matrix}2 ⋅ -\frac{1}{2} + 4 q = 1 \\ 4 q = 2 \\ q = \frac{1}{2}\end{matrix}\)

1p

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

1p

GelijkStellen
003i - Stelsels oplossen - 1ms
code
003i
vaardigheid
Stelsels oplossen
start-up
1ms

Los exact op.

4p

\(\begin{cases}y = 5 x + 22 \\ y = 9 x + 38\end{cases}\)

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Gelijk stellen geeft \(5 x + 22 = 9 x + 38\)

1p

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\(-4 x = 16\) dus \(x = -4\)

1p

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\(\begin{rcases}y = 5 x + 22 \\ x = -4\end{rcases} \begin{matrix}y = 5 ⋅ -4 + 22 \\ y = 2\end{matrix}\)

1p

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De oplossing is \((x , y) = (-4 , 2) \text{.}\)

1p

Substitutie (1)
003j - Stelsels oplossen - 0ms - dynamic variables
code
003j
vaardigheid
Stelsels oplossen
start-up
0ms

Los exact op.

4p

\(\begin{cases}9 x + 5 y = 69 \\ x = 7 y - 15\end{cases}\)

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Substitutie geeft \(9 (7 y - 15) + 5 y = 69\)

1p

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Haakjes wegwerken geeft
\(63 y - 135 + 5 y = 69\)
\(68 y = 204\)
\(y = 3\)

1p

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\(\begin{rcases}x = 7 y - 15 \\ y = 3\end{rcases} \begin{matrix}x = 7 ⋅ 3 - 15 \\ x = 6\end{matrix}\)

1p

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De oplossing is \((x , y) = (6 , 3) \text{.}\)

1p

Substitutie (2)
003k - Stelsels oplossen - 0ms - dynamic variables
code
003k
vaardigheid
Stelsels oplossen
start-up
0ms

Los exact op.

4p

\(\begin{cases}q = 3 p + 10 \\ p = 7 q + 10\end{cases}\)

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Substitutie geeft \(q = 3 (7 q + 10) + 10\)

1p

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Haakjes wegwerken geeft
\(q = 21 q + 30 + 10\)
\(-20 q = 40\)
\(q = -2\)

1p

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\(\begin{rcases}p = 7 q + 10 \\ q = -2\end{rcases} \begin{matrix}p = 7 ⋅ -2 + 10 \\ p = -4\end{matrix}\)

1p

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De oplossing is \((p , q) = (-4 , -2) \text{.}\)

1p