Stelsels oplossen

Aftrekken geeft \(2 x = -7 \text{,}\) dus \(x = -3\frac{1}{2} \text{.}\)

\(\begin{rcases}4 x - 2 y = -2 \\ x = -3\frac{1}{2}\end{rcases} \begin{matrix}4 ⋅ -3\frac{1}{2} - 2 y = -2 \\ -2 y = 12 \\ y = -6\end{matrix}\)

De oplossing is \((x , y) = (-3\frac{1}{2} , -6) \text{.}\)

\(\begin{cases}5 x + 6 y = -2 \\ x + 2 y = -4\end{cases}\) \(\begin{vmatrix}1 \\ 3\end{vmatrix}\) geeft \(\begin{cases}5 x + 6 y = -2 \\ 3 x + 6 y = -12\end{cases}\)

Aftrekken geeft \(2 x = 10 \text{,}\) dus \(x = 5 \text{.}\)

\(\begin{rcases}5 x + 6 y = -2 \\ x = 5\end{rcases} \begin{matrix}5 ⋅ 5 + 6 y = -2 \\ 6 y = -27 \\ y = -4\frac{1}{2}\end{matrix}\)

De oplossing is \((x , y) = (5 , -4\frac{1}{2}) \text{.}\)

\(\begin{cases}2 p + 2 q = 4 \\ 3 p - 5 q = 2\end{cases}\) \(\begin{vmatrix}5 \\ 2\end{vmatrix}\) geeft \(\begin{cases}10 p + 10 q = 20 \\ 6 p - 10 q = 4\end{cases}\)

Optellen geeft \(16 p = 24 \text{,}\) dus \(p = 1\frac{1}{2} \text{.}\)

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

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

Gelijk stellen geeft \(8 y - 27 = 4 y - 15\)

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

\(\begin{rcases}x = 8 y - 27 \\ y = 3\end{rcases} \begin{matrix}x = 8 ⋅ 3 - 27 \\ x = -3\end{matrix}\)

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

Substitutie geeft \(9 (8 b + 4) + 6 b = -42\)

Haakjes wegwerken geeft
\(72 b + 36 + 6 b = -42\)
\(78 b = -78\)
\(b = -1\)

\(\begin{rcases}a = 8 b + 4 \\ b = -1\end{rcases} \begin{matrix}a = 8 ⋅ -1 + 4 \\ a = -4\end{matrix}\)

De oplossing is \((a , b) = (-4 , -1) \text{.}\)

Substitutie geeft \(a = 4 (2 a - 9) + 8\)

Haakjes wegwerken geeft
\(a = 8 a - 36 + 8\)
\(-7 a = -28\)
\(a = 4\)

\(\begin{rcases}b = 2 a - 9 \\ a = 4\end{rcases} \begin{matrix}b = 2 ⋅ 4 - 9 \\ b = -1\end{matrix}\)

De oplossing is \((a , b) = (4 , -1) \text{.}\)