Aftrekken geeft \(-4b=-8\text{,}\) dus \(b=2\text{.}\)

\(\begin{rcases}2a-6b=-2 \\ b=2\end{rcases}\begin{matrix}2a-6⋅2=-2 \\ 2a=10 \\ a=5\end{matrix}\)

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

\(\begin{cases}x-2y=5 \\ 3x-4y=-6\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}2x-4y=10 \\ 3x-4y=-6\end{cases}\)

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

\(\begin{rcases}x-2y=5 \\ x=-16\end{rcases}\begin{matrix}-16-2y=5 \\ -2y=21 \\ y=-10\frac{1}{2}\end{matrix}\)

De oplossing is \((x, y)=(-16, -10\frac{1}{2})\text{.}\)

\(\begin{cases}6a+4b=1 \\ 5a-5b=5\end{cases}\) \(\begin{vmatrix}5 \\ 4\end{vmatrix}\) geeft \(\begin{cases}30a+20b=5 \\ 20a-20b=20\end{cases}\)

Optellen geeft \(50a=25\text{,}\) dus \(a=\frac{1}{2}\text{.}\)

\(\begin{rcases}6a+4b=1 \\ a=\frac{1}{2}\end{rcases}\begin{matrix}6⋅\frac{1}{2}+4b=1 \\ 4b=-2 \\ b=-\frac{1}{2}\end{matrix}\)

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

Gelijk stellen geeft \(5x+23=8x+38\)

\(-3x=15\) dus \(x=-5\)

\(\begin{rcases}y=5x+23 \\ x=-5\end{rcases}\begin{matrix}y=5⋅-5+23 \\ y=-2\end{matrix}\)

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

Substitutie geeft \(2p+5(6p-24)=40\)

Haakjes wegwerken geeft
\(2p+30p-120=40\)
\(32p=160\)
\(p=5\)

\(\begin{rcases}q=6p-24 \\ p=5\end{rcases}\begin{matrix}q=6⋅5-24 \\ q=6\end{matrix}\)

De oplossing is \((p, q)=(5, 6)\text{.}\)

Substitutie geeft \(x=7(3x+12)-24\)

Haakjes wegwerken geeft
\(x=21x+84-24\)
\(-20x=60\)
\(x=-3\)

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

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