Stelsels oplossen

Aftrekken geeft \(2y=-3\text{,}\) dus \(y=-1\frac{1}{2}\text{.}\)

\(\begin{rcases}2x+4y=3 \\ y=-1\frac{1}{2}\end{rcases}\begin{matrix}2x+4⋅-1\frac{1}{2}=3 \\ 2x=9 \\ x=4\frac{1}{2}\end{matrix}\)

De oplossing is \((x, y)=(4\frac{1}{2}, -1\frac{1}{2})\text{.}\)

\(\begin{cases}2x-y=4 \\ 6x+2y=-3\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}4x-2y=8 \\ 6x+2y=-3\end{cases}\)

Optellen geeft \(10x=5\text{,}\) dus \(x=\frac{1}{2}\text{.}\)

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

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

\(\begin{cases}3p-5q=6 \\ 4p-6q=-5\end{cases}\) \(\begin{vmatrix}6 \\ 5\end{vmatrix}\) geeft \(\begin{cases}18p-30q=36 \\ 20p-30q=-25\end{cases}\)

Aftrekken geeft \(-2p=61\text{,}\) dus \(p=-30\frac{1}{2}\text{.}\)

\(\begin{rcases}3p-5q=6 \\ p=-30\frac{1}{2}\end{rcases}\begin{matrix}3⋅-30\frac{1}{2}-5q=6 \\ -5q=97\frac{1}{2} \\ q=-19\frac{1}{2}\end{matrix}\)

De oplossing is \((p, q)=(-30\frac{1}{2}, -19\frac{1}{2})\text{.}\)

Gelijk stellen geeft \(2y+7=7y+32\)

\(-5y=25\) dus \(y=-5\)

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

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

Substitutie geeft \(9a+3(2a-14)=48\)

Haakjes wegwerken geeft
\(9a+6a-42=48\)
\(15a=90\)
\(a=6\)

\(\begin{rcases}b=2a-14 \\ a=6\end{rcases}\begin{matrix}b=2⋅6-14 \\ b=-2\end{matrix}\)

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

Substitutie geeft \(b=9(4b-21)+14\)

Haakjes wegwerken geeft
\(b=36b-189+14\)
\(-35b=-175\)
\(b=5\)

\(\begin{rcases}a=4b-21 \\ b=5\end{rcases}\begin{matrix}a=4⋅5-21 \\ a=-1\end{matrix}\)

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