Stelsels vergelijkingen

Aftrekken geeft \(3x=-15\) dus \(x=-5\)

\(\begin{rcases}6x+4y=-42 \\ x=-5\end{rcases}\begin{matrix}-30+4y=-42 \\ 4y=-12 \\ y=-3\end{matrix}\)

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

Optellen geeft \(11x=55\) dus \(x=5\)

\(\begin{rcases}8x+5y=30 \\ x=5\end{rcases}\begin{matrix}40+5y=30 \\ 5y=-10 \\ y=-2\end{matrix}\)

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

Aftrekken geeft \(-4a=-8\) dus \(a=2\)

\(\begin{rcases}4a-6b=-16 \\ a=2\end{rcases}\begin{matrix}8-6b=-16 \\ -6b=-24 \\ b=4\end{matrix}\)

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

\(\begin{cases}3a+3b=-3 \\ 5a+4b=-3\end{cases}\) \(\begin{vmatrix}4 \\ 3\end{vmatrix}\) geeft \(\begin{cases}12a+12b=-12 \\ 15a+12b=-9\end{cases}\)

Aftrekken geeft \(-3a=-3\) dus \(a=1\)

\(\begin{rcases}3a+3b=-3 \\ a=1\end{rcases}\begin{matrix}3+3b=-3 \\ 3b=-6 \\ b=-2\end{matrix}\)

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

\(\begin{cases}4p-3q=11 \\ -5p+5q=-10\end{cases}\) \(\begin{vmatrix}5 \\ 3\end{vmatrix}\) geeft \(\begin{cases}20p-15q=55 \\ -15p+15q=-30\end{cases}\)

Optellen geeft \(5p=25\) dus \(p=5\)

\(\begin{rcases}4p-3q=11 \\ p=5\end{rcases}\begin{matrix}20-3q=11 \\ -3q=-9 \\ q=3\end{matrix}\)

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

\(\begin{cases}3a-2b=2 \\ -4a-5b=-41\end{cases}\) \(\begin{vmatrix}5 \\ 2\end{vmatrix}\) geeft \(\begin{cases}15a-10b=10 \\ -8a-10b=-82\end{cases}\)

Aftrekken geeft \(23a=92\) dus \(a=4\)

\(\begin{rcases}3a-2b=2 \\ a=4\end{rcases}\begin{matrix}12-2b=2 \\ -2b=-10 \\ b=5\end{matrix}\)

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

Gelijk stellen geeft \(6a-23=8a-29\)

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

\(\begin{rcases}b=6a-23 \\ a=3\end{rcases}\begin{matrix}b=6⋅3-23 \\ b=-5\end{matrix}\)

De oplossing is \((a, b)=(3, -5)\text{.}\)

Gelijk stellen geeft \(8y-21=6y-17\)

\(2y=4\) dus \(y=2\)

\(\begin{rcases}x=8y-21 \\ y=2\end{rcases}\begin{matrix}x=8⋅2-21 \\ x=-5\end{matrix}\)

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

Substitutie geeft \(2p+4(9p+21)=-30\)

Haakjes wegwerken geeft \(2p+36p+84=-30\)
ofwel \(38p=-114\) dus \(p=-3\text{.}\)

\(\begin{rcases}q=9p+21 \\ p=-3\end{rcases}\begin{matrix}q=9⋅-3+21 \\ q=-6\end{matrix}\)

De oplossing is \((p, q)=(-3, -6)\text{.}\)

Substitutie geeft \(2(9y-22)+8y=34\)

Haakjes wegwerken geeft \(18y-44+8y=34\)
ofwel \(26y=78\) dus \(y=3\text{.}\)

\(\begin{rcases}x=9y-22 \\ y=3\end{rcases}\begin{matrix}x=9⋅3-22 \\ x=5\end{matrix}\)

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

Substitutie geeft \(y=2(9y+22)-10\)

Haakjes wegwerken geeft \(y=18y+44-10\)
ofwel \(-17y=34\) dus \(y=-2\text{.}\)

\(\begin{rcases}x=9y+22 \\ y=-2\end{rcases}\begin{matrix}x=9⋅-2+22 \\ x=4\end{matrix}\)

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

Substitutie geeft \(x=3(9x-40)+16\)

Haakjes wegwerken geeft \(x=27x-120+16\)
ofwel \(-26x=-104\) dus \(x=4\text{.}\)

\(\begin{rcases}y=9x-40 \\ x=4\end{rcases}\begin{matrix}y=9⋅4-40 \\ y=-4\end{matrix}\)

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