Aftrekken geeft \(p=-8\text{.}\)

\(\begin{rcases}2p+2q=-6 \\ p=-8\end{rcases}\begin{matrix}2⋅-8+2q=-6 \\ 2q=10 \\ q=5\end{matrix}\)

De oplossing is \((p, q)=(-8, 5)\text{.}\)

\(\begin{cases}4a+2b=3 \\ 3a-b=-4\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}4a+2b=3 \\ 6a-2b=-8\end{cases}\)

Optellen geeft \(10a=-5\text{,}\) dus \(a=-\frac{1}{2}\text{.}\)

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

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

\(\begin{cases}5a-2b=-6 \\ 4a-3b=-2\end{cases}\) \(\begin{vmatrix}3 \\ 2\end{vmatrix}\) geeft \(\begin{cases}15a-6b=-18 \\ 8a-6b=-4\end{cases}\)

Aftrekken geeft \(7a=-14\text{,}\) dus \(a=-2\text{.}\)

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

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

Gelijk stellen geeft \(8x-37=3x-12\)

\(5x=25\) dus \(x=5\)

\(\begin{rcases}y=8x-37 \\ x=5\end{rcases}\begin{matrix}y=8⋅5-37 \\ y=3\end{matrix}\)

De oplossing is \((x, y)=(5, 3)\text{.}\)

Substitutie geeft \(6x+5(3x+3)=36\)

Haakjes wegwerken geeft
\(6x+15x+15=36\)
\(21x=21\)
\(x=1\)

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

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

Substitutie geeft \(y=9(3y+12)-30\)

Haakjes wegwerken geeft
\(y=27y+108-30\)
\(-26y=78\)
\(y=-3\)

\(\begin{rcases}x=3y+12 \\ y=-3\end{rcases}\begin{matrix}x=3⋅-3+12 \\ x=3\end{matrix}\)

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