Stelsels oplossen

Aftrekken geeft \(a=-10\text{.}\)

\(\begin{rcases}6a-3b=-6 \\ a=-10\end{rcases}\begin{matrix}6⋅-10-3b=-6 \\ -3b=54 \\ b=-18\end{matrix}\)

De oplossing is \((a, b)=(-10, -18)\text{.}\)

\(\begin{cases}4x+2y=-3 \\ 2x-6y=2\end{cases}\) \(\begin{vmatrix}3 \\ 1\end{vmatrix}\) geeft \(\begin{cases}12x+6y=-9 \\ 2x-6y=2\end{cases}\)

Optellen geeft \(14x=-7\text{,}\) dus \(x=-\frac{1}{2}\text{.}\)

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

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

\(\begin{cases}5p+6q=4 \\ 6p+5q=-4\end{cases}\) \(\begin{vmatrix}5 \\ 6\end{vmatrix}\) geeft \(\begin{cases}25p+30q=20 \\ 36p+30q=-24\end{cases}\)

Aftrekken geeft \(-11p=44\text{,}\) dus \(p=-4\text{.}\)

\(\begin{rcases}5p+6q=4 \\ p=-4\end{rcases}\begin{matrix}5⋅-4+6q=4 \\ 6q=24 \\ q=4\end{matrix}\)

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

Gelijk stellen geeft \(7x-17=5x-11\)

\(2x=6\) dus \(x=3\)

\(\begin{rcases}y=7x-17 \\ x=3\end{rcases}\begin{matrix}y=7⋅3-17 \\ y=4\end{matrix}\)

De oplossing is \((x, y)=(3, 4)\text{.}\)

Substitutie geeft \(9x+7(4x-30)=12\)

Haakjes wegwerken geeft
\(9x+28x-210=12\)
\(37x=222\)
\(x=6\)

\(\begin{rcases}y=4x-30 \\ x=6\end{rcases}\begin{matrix}y=4⋅6-30 \\ y=-6\end{matrix}\)

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

Substitutie geeft \(a=6(4a+21)-34\)

Haakjes wegwerken geeft
\(a=24a+126-34\)
\(-23a=92\)
\(a=-4\)

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

De oplossing is \((a, b)=(-4, 5)\text{.}\)