Stelsels oplossen

Aftrekken geeft \(b = -10 \text{.}\)

\(\begin{rcases}6 a + 4 b = -4 \\ b = -10\end{rcases} \begin{matrix}6 a + 4 ⋅ -10 = -4 \\ 6 a = 36 \\ a = 6\end{matrix}\)

De oplossing is \((a , b) = (6 , -10) \text{.}\)

\(\begin{cases}x + y = -6 \\ 3 x - 3 y = 6\end{cases}\) \(\begin{vmatrix}3 \\ 1\end{vmatrix}\) geeft \(\begin{cases}3 x + 3 y = -18 \\ 3 x - 3 y = 6\end{cases}\)

Optellen geeft \(6 x = -12 \text{,}\) dus \(x = -2 \text{.}\)

\(\begin{rcases}x + y = -6 \\ x = -2\end{rcases} \begin{matrix}-2 + y = -6 \\ y = -4\end{matrix}\)

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

\(\begin{cases}4 a - 5 b = -3 \\ 6 a - 4 b = 6\end{cases}\) \(\begin{vmatrix}4 \\ 5\end{vmatrix}\) geeft \(\begin{cases}16 a - 20 b = -12 \\ 30 a - 20 b = 30\end{cases}\)

Aftrekken geeft \(-14 a = -42 \text{,}\) dus \(a = 3 \text{.}\)

\(\begin{rcases}4 a - 5 b = -3 \\ a = 3\end{rcases} \begin{matrix}4 ⋅ 3 - 5 b = -3 \\ -5 b = -15 \\ b = 3\end{matrix}\)

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

Gelijk stellen geeft \(7 y + 2 = 4 y - 1\)

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

\(\begin{rcases}x = 7 y + 2 \\ y = -1\end{rcases} \begin{matrix}x = 7 ⋅ -1 + 2 \\ x = -5\end{matrix}\)

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

Substitutie geeft \(3 p + 4 (5 p + 8) = 9\)

Haakjes wegwerken geeft
\(3 p + 20 p + 32 = 9\)
\(23 p = -23\)
\(p = -1\)

\(\begin{rcases}q = 5 p + 8 \\ p = -1\end{rcases} \begin{matrix}q = 5 ⋅ -1 + 8 \\ q = 3\end{matrix}\)

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

Substitutie geeft \(y = 5 (3 y + 10) - 22\)

Haakjes wegwerken geeft
\(y = 15 y + 50 - 22\)
\(-14 y = 28\)
\(y = -2\)

\(\begin{rcases}x = 3 y + 10 \\ y = -2\end{rcases} \begin{matrix}x = 3 ⋅ -2 + 10 \\ x = 4\end{matrix}\)

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