Stelsels oplossen

Optellen geeft \(6 a = 3 \text{,}\) dus \(a = \frac{1}{2} \text{.}\)

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

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

\(\begin{cases}3 a + 4 b = 4 \\ a - b = 6\end{cases}\) \(\begin{vmatrix}1 \\ 4\end{vmatrix}\) geeft \(\begin{cases}3 a + 4 b = 4 \\ 4 a - 4 b = 24\end{cases}\)

Optellen geeft \(7 a = 28 \text{,}\) dus \(a = 4 \text{.}\)

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

De oplossing is \((a , b) = (4 , -2) \text{.}\)

\(\begin{cases}4 x - 4 y = 4 \\ 5 x - 6 y = -2\end{cases}\) \(\begin{vmatrix}3 \\ 2\end{vmatrix}\) geeft \(\begin{cases}12 x - 12 y = 12 \\ 10 x - 12 y = -4\end{cases}\)

Aftrekken geeft \(2 x = 16 \text{,}\) dus \(x = 8 \text{.}\)

\(\begin{rcases}4 x - 4 y = 4 \\ x = 8\end{rcases} \begin{matrix}4 ⋅ 8 - 4 y = 4 \\ -4 y = -28 \\ y = 7\end{matrix}\)

De oplossing is \((x , y) = (8 , 7) \text{.}\)

Gelijk stellen geeft \(5 y - 2 = 2 y + 1\)

\(3 y = 3\) dus \(y = 1\)

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

De oplossing is \((x , y) = (3 , 1) \text{.}\)

Substitutie geeft \(8 p + 4 (6 p - 19) = 20\)

Haakjes wegwerken geeft
\(8 p + 24 p - 76 = 20\)
\(32 p = 96\)
\(p = 3\)

\(\begin{rcases}q = 6 p - 19 \\ p = 3\end{rcases} \begin{matrix}q = 6 ⋅ 3 - 19 \\ q = -1\end{matrix}\)

De oplossing is \((p , q) = (3 , -1) \text{.}\)

Substitutie geeft \(b = 8 (6 b - 17) + 42\)

Haakjes wegwerken geeft
\(b = 48 b - 136 + 42\)
\(-47 b = -94\)
\(b = 2\)

\(\begin{rcases}a = 6 b - 17 \\ b = 2\end{rcases} \begin{matrix}a = 6 ⋅ 2 - 17 \\ a = -5\end{matrix}\)

De oplossing is \((a , b) = (-5 , 2) \text{.}\)