Stelsels vergelijkingen

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

\(\begin{rcases}6a+3b=-24 \\ a=-5\end{rcases}\begin{matrix}-30+3b=-24 \\ 3b=6 \\ b=2\end{matrix}\)

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

Optellen geeft \(8p=24\) dus \(p=3\)

\(\begin{rcases}2p+9q=-39 \\ p=3\end{rcases}\begin{matrix}6+9q=-39 \\ 9q=-45 \\ q=-5\end{matrix}\)

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

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

\(\begin{rcases}8x-7y=61 \\ x=5\end{rcases}\begin{matrix}40-7y=61 \\ -7y=21 \\ y=-3\end{matrix}\)

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

\(\begin{cases}4a+3b=-26 \\ 5a+5b=-35\end{cases}\) \(\begin{vmatrix}5 \\ 3\end{vmatrix}\) geeft \(\begin{cases}20a+15b=-130 \\ 15a+15b=-105\end{cases}\)

Aftrekken geeft \(5a=-25\) dus \(a=-5\)

\(\begin{rcases}4a+3b=-26 \\ a=-5\end{rcases}\begin{matrix}-20+3b=-26 \\ 3b=-6 \\ b=-2\end{matrix}\)

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

\(\begin{cases}5x-3y=34 \\ -6x+4y=-42\end{cases}\) \(\begin{vmatrix}4 \\ 3\end{vmatrix}\) geeft \(\begin{cases}20x-12y=136 \\ -18x+12y=-126\end{cases}\)

Optellen geeft \(2x=10\) dus \(x=5\)

\(\begin{rcases}5x-3y=34 \\ x=5\end{rcases}\begin{matrix}25-3y=34 \\ -3y=9 \\ y=-3\end{matrix}\)

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

\(\begin{cases}5x-3y=12 \\ -6x-4y=-22\end{cases}\) \(\begin{vmatrix}4 \\ 3\end{vmatrix}\) geeft \(\begin{cases}20x-12y=48 \\ -18x-12y=-66\end{cases}\)

Aftrekken geeft \(38x=114\) dus \(x=3\)

\(\begin{rcases}5x-3y=12 \\ x=3\end{rcases}\begin{matrix}15-3y=12 \\ -3y=-3 \\ y=1\end{matrix}\)

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

Gelijk stellen geeft \(8x-21=3x-11\)

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

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

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

Gelijk stellen geeft \(7b-2=2b+3\)

\(5b=5\) dus \(b=1\)

\(\begin{rcases}a=7b-2 \\ b=1\end{rcases}\begin{matrix}a=7⋅1-2 \\ a=5\end{matrix}\)

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

Substitutie geeft \(8a+9(4a-13)=59\)

Haakjes wegwerken geeft \(8a+36a-117=59\)
ofwel \(44a=176\) dus \(a=4\text{.}\)

\(\begin{rcases}b=4a-13 \\ a=4\end{rcases}\begin{matrix}b=4⋅4-13 \\ b=3\end{matrix}\)

De oplossing is \((a, b)=(4, 3)\text{.}\)

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

Haakjes wegwerken geeft \(24q-156+2q=-26\)
ofwel \(26q=130\) dus \(q=5\text{.}\)

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

De oplossing is \((p, q)=(-6, 5)\text{.}\)

Substitutie geeft \(b=4(7b-24)+15\)

Haakjes wegwerken geeft \(b=28b-96+15\)
ofwel \(-27b=-81\) dus \(b=3\text{.}\)

\(\begin{rcases}a=7b-24 \\ b=3\end{rcases}\begin{matrix}a=7⋅3-24 \\ a=-3\end{matrix}\)

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

Substitutie geeft \(p=4(7p+16)+17\)

Haakjes wegwerken geeft \(p=28p+64+17\)
ofwel \(-27p=81\) dus \(p=-3\text{.}\)

\(\begin{rcases}q=7p+16 \\ p=-3\end{rcases}\begin{matrix}q=7⋅-3+16 \\ q=-5\end{matrix}\)

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