Aftrekken geeft \(x=3\text{.}\)

\(\begin{rcases}5x-2y=4 \\ x=3\end{rcases}\begin{matrix}5⋅3-2y=4 \\ -2y=-11 \\ y=5\frac{1}{2}\end{matrix}\)

De oplossing is \((x, y)=(3, 5\frac{1}{2})\text{.}\)

\(\begin{cases}3p+4q=-2 \\ p+3q=6\end{cases}\) \(\begin{vmatrix}1 \\ 3\end{vmatrix}\) geeft \(\begin{cases}3p+4q=-2 \\ 3p+9q=18\end{cases}\)

Aftrekken geeft \(-5q=-20\text{,}\) dus \(q=4\text{.}\)

\(\begin{rcases}3p+4q=-2 \\ q=4\end{rcases}\begin{matrix}3p+4⋅4=-2 \\ 3p=-18 \\ p=-6\end{matrix}\)

De oplossing is \((p, q)=(-6, 4)\text{.}\)

\(\begin{cases}6a-4b=-1 \\ 5a+5b=-5\end{cases}\) \(\begin{vmatrix}5 \\ 4\end{vmatrix}\) 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 \(7x-31=3x-11\)

\(4x=20\) dus \(x=5\)

\(\begin{rcases}y=7x-31 \\ x=5\end{rcases}\begin{matrix}y=7⋅5-31 \\ y=4\end{matrix}\)

De oplossing is \((x, y)=(5, 4)\text{.}\)

Substitutie geeft \(9(8y-26)+3y=66\)

Haakjes wegwerken geeft
\(72y-234+3y=66\)
\(75y=300\)
\(y=4\)

\(\begin{rcases}x=8y-26 \\ y=4\end{rcases}\begin{matrix}x=8⋅4-26 \\ x=6\end{matrix}\)

De oplossing is \((x, y)=(6, 4)\text{.}\)

Substitutie geeft \(a=3(5a-24)+2\)

Haakjes wegwerken geeft
\(a=15a-72+2\)
\(-14a=-70\)
\(a=5\)

\(\begin{rcases}b=5a-24 \\ a=5\end{rcases}\begin{matrix}b=5⋅5-24 \\ b=1\end{matrix}\)

De oplossing is \((a, b)=(5, 1)\text{.}\)