Optellen geeft \(4x=8\text{,}\) dus \(x=2\text{.}\)

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

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

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

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

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

De oplossing is \((a, b)=(4, 2\frac{1}{2})\text{.}\)

\(\begin{cases}5x+6y=-4 \\ 4x+5y=-2\end{cases}\) \(\begin{vmatrix}5 \\ 6\end{vmatrix}\) geeft \(\begin{cases}25x+30y=-20 \\ 24x+30y=-12\end{cases}\)

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

\(\begin{rcases}5x+6y=-4 \\ x=-8\end{rcases}\begin{matrix}5⋅-8+6y=-4 \\ 6y=36 \\ y=6\end{matrix}\)

De oplossing is \((x, y)=(-8, 6)\text{.}\)

Gelijk stellen geeft \(2y+8=9y+29\)

\(-7y=21\) dus \(y=-3\)

\(\begin{rcases}x=2y+8 \\ y=-3\end{rcases}\begin{matrix}x=2⋅-3+8 \\ x=2\end{matrix}\)

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

Substitutie geeft \(6(7q-8)+5q=46\)

Haakjes wegwerken geeft
\(42q-48+5q=46\)
\(47q=94\)
\(q=2\)

\(\begin{rcases}p=7q-8 \\ q=2\end{rcases}\begin{matrix}p=7⋅2-8 \\ p=6\end{matrix}\)

De oplossing is \((p, q)=(6, 2)\text{.}\)

Substitutie geeft \(a=6(4a+1)+17\)

Haakjes wegwerken geeft
\(a=24a+6+17\)
\(-23a=23\)
\(a=-1\)

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

De oplossing is \((a, b)=(-1, -3)\text{.}\)