Optellen geeft \(6a=12\text{,}\) 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 \((a, b)=(2, -2)\text{.}\)

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

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

\(\begin{rcases}2x-2y=-5 \\ x=-16\end{rcases}\begin{matrix}2⋅-16-2y=-5 \\ -2y=27 \\ y=-13\frac{1}{2}\end{matrix}\)

De oplossing is \((x, y)=(-16, -13\frac{1}{2})\text{.}\)

\(\begin{cases}3x+5y=-6 \\ 2x+4y=-5\end{cases}\) \(\begin{vmatrix}4 \\ 5\end{vmatrix}\) geeft \(\begin{cases}12x+20y=-24 \\ 10x+20y=-25\end{cases}\)

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

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

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