Stelsels oplossen

Optellen geeft \(6x=-3\text{,}\) dus \(x=-\frac{1}{2}\text{.}\)

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

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

\(\begin{cases}2p-q=-6 \\ 5p-3q=3\end{cases}\) \(\begin{vmatrix}3 \\ 1\end{vmatrix}\) geeft \(\begin{cases}6p-3q=-18 \\ 5p-3q=3\end{cases}\)

Aftrekken geeft \(p=-21\text{.}\)

\(\begin{rcases}2p-q=-6 \\ p=-21\end{rcases}\begin{matrix}2⋅-21-q=-6 \\ -q=36 \\ q=-36\end{matrix}\)

De oplossing is \((p, q)=(-21, -36)\text{.}\)

\(\begin{cases}3a+4b=6 \\ 4a+5b=6\end{cases}\) \(\begin{vmatrix}5 \\ 4\end{vmatrix}\) geeft \(\begin{cases}15a+20b=30 \\ 16a+20b=24\end{cases}\)

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

\(\begin{rcases}3a+4b=6 \\ a=-6\end{rcases}\begin{matrix}3⋅-6+4b=6 \\ 4b=24 \\ b=6\end{matrix}\)

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

Gelijk stellen geeft \(8x+12=4x+8\)

\(4x=-4\) dus \(x=-1\)

\(\begin{rcases}y=8x+12 \\ x=-1\end{rcases}\begin{matrix}y=8⋅-1+12 \\ y=4\end{matrix}\)

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

Gelijk stellen geeft \(3y-5=6y-11\)

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

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

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

Substitutie geeft \(2x+6(4x-13)=-26\)

Haakjes wegwerken geeft \(2x+24x-78=-26\)
ofwel \(26x=52\) dus \(x=2\text{.}\)

\(\begin{rcases}y=4x-13 \\ x=2\end{rcases}\begin{matrix}y=4⋅2-13 \\ y=-5\end{matrix}\)

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

Substitutie geeft \(3(6b+34)+4b=-30\)

Haakjes wegwerken geeft \(18b+102+4b=-30\)
ofwel \(22b=-132\) dus \(b=-6\text{.}\)

\(\begin{rcases}a=6b+34 \\ b=-6\end{rcases}\begin{matrix}a=6⋅-6+34 \\ a=-2\end{matrix}\)

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

Substitutie geeft \(y=6(2y+8)-26\)

Haakjes wegwerken geeft \(y=12y+48-26\)
ofwel \(-11y=22\) dus \(y=-2\text{.}\)

\(\begin{rcases}x=2y+8 \\ y=-2\end{rcases}\begin{matrix}x=2⋅-2+8 \\ x=4\end{matrix}\)

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

Substitutie geeft \(a=5(8a-18)+12\)

Haakjes wegwerken geeft \(a=40a-90+12\)
ofwel \(-39a=-78\) dus \(a=2\text{.}\)

\(\begin{rcases}b=8a-18 \\ a=2\end{rcases}\begin{matrix}b=8⋅2-18 \\ b=-2\end{matrix}\)

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