Optellen geeft \(4a=6\text{,}\) dus \(a=1\frac{1}{2}\text{.}\)

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

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

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

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

\(\begin{rcases}3a+4b=5 \\ a=-7\end{rcases}\begin{matrix}3⋅-7+4b=5 \\ 4b=26 \\ b=6\frac{1}{2}\end{matrix}\)

De oplossing is \((a, b)=(-7, 6\frac{1}{2})\text{.}\)

\(\begin{cases}5p-5q=5 \\ 6p-4q=-5\end{cases}\) \(\begin{vmatrix}4 \\ 5\end{vmatrix}\) geeft \(\begin{cases}20p-20q=20 \\ 30p-20q=-25\end{cases}\)

Aftrekken geeft \(-10p=45\text{,}\) dus \(p=-4\frac{1}{2}\text{.}\)

\(\begin{rcases}5p-5q=5 \\ p=-4\frac{1}{2}\end{rcases}\begin{matrix}5⋅-4\frac{1}{2}-5q=5 \\ -5q=27\frac{1}{2} \\ q=-5\frac{1}{2}\end{matrix}\)

De oplossing is \((p, q)=(-4\frac{1}{2}, -5\frac{1}{2})\text{.}\)