Stelsels oplossen

Aftrekken geeft \(2 x = 9 \text{,}\) dus \(x = 4\frac{1}{2} \text{.}\)

\(\begin{rcases}3 x + 5 y = 6 \\ x = 4\frac{1}{2}\end{rcases} \begin{matrix}3 ⋅ 4\frac{1}{2} + 5 y = 6 \\ 5 y = -7\frac{1}{2} \\ y = -1\frac{1}{2}\end{matrix}\)

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

\(\begin{cases}2 x + 4 y = 1 \\ x - y = 5\end{cases}\) \(\begin{vmatrix}1 \\ 4\end{vmatrix}\) geeft \(\begin{cases}2 x + 4 y = 1 \\ 4 x - 4 y = 20\end{cases}\)

Optellen geeft \(6 x = 21 \text{,}\) dus \(x = 3\frac{1}{2} \text{.}\)

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

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

\(\begin{cases}3 a + 5 b = 4 \\ 2 a - 4 b = -1\end{cases}\) \(\begin{vmatrix}4 \\ 5\end{vmatrix}\) geeft \(\begin{cases}12 a + 20 b = 16 \\ 10 a - 20 b = -5\end{cases}\)

Optellen geeft \(22 a = 11 \text{,}\) dus \(a = \frac{1}{2} \text{.}\)

\(\begin{rcases}3 a + 5 b = 4 \\ a = \frac{1}{2}\end{rcases} \begin{matrix}3 ⋅ \frac{1}{2} + 5 b = 4 \\ 5 b = 2\frac{1}{2} \\ b = \frac{1}{2}\end{matrix}\)

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

Gelijk stellen geeft \(8 x - 38 = 6 x - 28\)

\(2 x = 10\) dus \(x = 5\)

\(\begin{rcases}y = 8 x - 38 \\ x = 5\end{rcases} \begin{matrix}y = 8 ⋅ 5 - 38 \\ y = 2\end{matrix}\)

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

Substitutie geeft \(3 p + 9 (5 p - 6) = 42\)

Haakjes wegwerken geeft
\(3 p + 45 p - 54 = 42\)
\(48 p = 96\)
\(p = 2\)

\(\begin{rcases}q = 5 p - 6 \\ p = 2\end{rcases} \begin{matrix}q = 5 ⋅ 2 - 6 \\ q = 4\end{matrix}\)

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

Substitutie geeft \(x = 8 (2 x - 3) - 36\)

Haakjes wegwerken geeft
\(x = 16 x - 24 - 36\)
\(-15 x = -60\)
\(x = 4\)

\(\begin{rcases}y = 2 x - 3 \\ x = 4\end{rcases} \begin{matrix}y = 2 ⋅ 4 - 3 \\ y = 5\end{matrix}\)

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