Stelsels oplossen

Optellen geeft \(4x=8\text{,}\) dus \(x=2\text{.}\)

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

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

\(\begin{cases}2p-4q=-5 \\ 4p-q=4\end{cases}\) \(\begin{vmatrix}1 \\ 4\end{vmatrix}\) geeft \(\begin{cases}2p-4q=-5 \\ 16p-4q=16\end{cases}\)

Aftrekken geeft \(-14p=-21\text{,}\) dus \(p=1\frac{1}{2}\text{.}\)

\(\begin{rcases}2p-4q=-5 \\ p=1\frac{1}{2}\end{rcases}\begin{matrix}2⋅1\frac{1}{2}-4q=-5 \\ -4q=-8 \\ q=2\end{matrix}\)

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

\(\begin{cases}6a+4b=-1 \\ 5a+3b=2\end{cases}\) \(\begin{vmatrix}3 \\ 4\end{vmatrix}\) geeft \(\begin{cases}18a+12b=-3 \\ 20a+12b=8\end{cases}\)

Aftrekken geeft \(-2a=-11\text{,}\) dus \(a=5\frac{1}{2}\text{.}\)

\(\begin{rcases}6a+4b=-1 \\ a=5\frac{1}{2}\end{rcases}\begin{matrix}6⋅5\frac{1}{2}+4b=-1 \\ 4b=-34 \\ b=-8\frac{1}{2}\end{matrix}\)

De oplossing is \((a, b)=(5\frac{1}{2}, -8\frac{1}{2})\text{.}\)

Gelijk stellen geeft \(8x+18=3x+8\)

\(5x=-10\) dus \(x=-2\)

\(\begin{rcases}y=8x+18 \\ x=-2\end{rcases}\begin{matrix}y=8⋅-2+18 \\ y=2\end{matrix}\)

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

Gelijk stellen geeft \(6y-26=3y-11\)

\(3y=15\) dus \(y=5\)

\(\begin{rcases}x=6y-26 \\ y=5\end{rcases}\begin{matrix}x=6⋅5-26 \\ x=4\end{matrix}\)

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

Substitutie geeft \(4x+8(6x-12)=-44\)

Haakjes wegwerken geeft \(4x+48x-96=-44\)
ofwel \(52x=52\) dus \(x=1\text{.}\)

\(\begin{rcases}y=6x-12 \\ x=1\end{rcases}\begin{matrix}y=6⋅1-12 \\ y=-6\end{matrix}\)

De oplossing is \((x, y)=(1, -6)\text{.}\)

Substitutie geeft \(8(6b-30)+3b=-36\)

Haakjes wegwerken geeft \(48b-240+3b=-36\)
ofwel \(51b=204\) dus \(b=4\text{.}\)

\(\begin{rcases}a=6b-30 \\ b=4\end{rcases}\begin{matrix}a=6⋅4-30 \\ a=-6\end{matrix}\)

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

Substitutie geeft \(q=6(3q+4)+27\)

Haakjes wegwerken geeft \(q=18q+24+27\)
ofwel \(-17q=51\) dus \(q=-3\text{.}\)

\(\begin{rcases}p=3q+4 \\ q=-3\end{rcases}\begin{matrix}p=3⋅-3+4 \\ p=-5\end{matrix}\)

De oplossing is \((p, q)=(-5, -3)\text{.}\)

Substitutie geeft \(a=5(7a+19)-27\)

Haakjes wegwerken geeft \(a=35a+95-27\)
ofwel \(-34a=68\) dus \(a=-2\text{.}\)

\(\begin{rcases}b=7a+19 \\ a=-2\end{rcases}\begin{matrix}b=7⋅-2+19 \\ b=5\end{matrix}\)

De oplossing is \((a, b)=(-2, 5)\text{.}\)