Stelsels oplossen

Aftrekken geeft \(-2y=-7\text{,}\) dus \(y=3\frac{1}{2}\text{.}\)

\(\begin{rcases}4x-6y=-5 \\ y=3\frac{1}{2}\end{rcases}\begin{matrix}4x-6⋅3\frac{1}{2}=-5 \\ 4x=16 \\ x=4\end{matrix}\)

De oplossing is \((x, y)=(4, 3\frac{1}{2})\text{.}\)

\(\begin{cases}4a+2b=-6 \\ 2a-b=5\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}4a+2b=-6 \\ 4a-2b=10\end{cases}\)

Optellen geeft \(8a=4\text{,}\) dus \(a=\frac{1}{2}\text{.}\)

\(\begin{rcases}4a+2b=-6 \\ a=\frac{1}{2}\end{rcases}\begin{matrix}4⋅\frac{1}{2}+2b=-6 \\ 2b=-8 \\ b=-4\end{matrix}\)

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

\(\begin{cases}3a-5b=2 \\ 2a-4b=4\end{cases}\) \(\begin{vmatrix}4 \\ 5\end{vmatrix}\) geeft \(\begin{cases}12a-20b=8 \\ 10a-20b=20\end{cases}\)

Aftrekken geeft \(2a=-12\text{,}\) dus \(a=-6\text{.}\)

\(\begin{rcases}3a-5b=2 \\ a=-6\end{rcases}\begin{matrix}3⋅-6-5b=2 \\ -5b=20 \\ b=-4\end{matrix}\)

De oplossing is \((a, b)=(-6, -4)\text{.}\)

Gelijk stellen geeft \(7y+15=4y+9\)

\(3y=-6\) dus \(y=-2\)

\(\begin{rcases}x=7y+15 \\ y=-2\end{rcases}\begin{matrix}x=7⋅-2+15 \\ x=1\end{matrix}\)

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

Substitutie geeft \(6x+3(2x)=-24\)

Haakjes wegwerken geeft
\(6x+6x+0=-24\)
\(12x=-24\)
\(x=-2\)

\(\begin{rcases}y=2x+0 \\ x=-2\end{rcases}\begin{matrix}y=2⋅-2+0 \\ y=-4\end{matrix}\)

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

Substitutie geeft \(p=8(4p+10)+44\)

Haakjes wegwerken geeft
\(p=32p+80+44\)
\(-31p=124\)
\(p=-4\)

\(\begin{rcases}q=4p+10 \\ p=-4\end{rcases}\begin{matrix}q=4⋅-4+10 \\ q=-6\end{matrix}\)

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