Stelsels vergelijkingen

a

Aftrekken geeft \(-3x=-9\) dus \(x=3\)

1p

\(\begin{rcases}5x+7y=-6 \\ x=3\end{rcases}\begin{matrix}15+7y=-6 \\ 7y=-21 \\ y=-3\end{matrix}\)

1p

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

1p

a

Optellen geeft \(14x=-14\) dus \(x=-1\)

1p

\(\begin{rcases}5x+3y=-17 \\ x=-1\end{rcases}\begin{matrix}-5+3y=-17 \\ 3y=-12 \\ y=-4\end{matrix}\)

1p

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

1p

a

Aftrekken geeft \(3x=-9\) dus \(x=-3\)

1p

\(\begin{rcases}7x-9y=24 \\ x=-3\end{rcases}\begin{matrix}-21-9y=24 \\ -9y=45 \\ y=-5\end{matrix}\)

1p

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

1p

a

\(\begin{cases}3x+2y=4 \\ 5x+5y=0\end{cases}\) \(\begin{vmatrix}5 \\ 2\end{vmatrix}\) geeft \(\begin{cases}15x+10y=20 \\ 10x+10y=0\end{cases}\)

1p

Aftrekken geeft \(5x=20\) dus \(x=4\)

1p

\(\begin{rcases}3x+2y=4 \\ x=4\end{rcases}\begin{matrix}12+2y=4 \\ 2y=-8 \\ y=-4\end{matrix}\)

1p

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

1p

a

\(\begin{cases}4x-2y=0 \\ -5x+3y=-1\end{cases}\) \(\begin{vmatrix}3 \\ 2\end{vmatrix}\) geeft \(\begin{cases}12x-6y=0 \\ -10x+6y=-2\end{cases}\)

1p

Optellen geeft \(2x=-2\) dus \(x=-1\)

1p

\(\begin{rcases}4x-2y=0 \\ x=-1\end{rcases}\begin{matrix}-4-2y=0 \\ -2y=4 \\ y=-2\end{matrix}\)

1p

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

1p

a

\(\begin{cases}5x-2y=-13 \\ -6x-5y=-14\end{cases}\) \(\begin{vmatrix}5 \\ 2\end{vmatrix}\) geeft \(\begin{cases}25x-10y=-65 \\ -12x-10y=-28\end{cases}\)

1p

Aftrekken geeft \(37x=-37\) dus \(x=-1\)

1p

\(\begin{rcases}5x-2y=-13 \\ x=-1\end{rcases}\begin{matrix}-5-2y=-13 \\ -2y=-8 \\ y=4\end{matrix}\)

1p

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

1p

a

Gelijk stellen geeft \(7x-9=4x-6\)

1p

\(3x=3\) dus \(x=1\)

1p

\(\begin{rcases}y=7x-9 \\ x=1\end{rcases}\begin{matrix}y=7⋅1-9 \\ y=-2\end{matrix}\)

1p

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

1p

a

Gelijk stellen geeft \(6y-8=3y-5\)

1p

\(3y=3\) dus \(y=1\)

1p

\(\begin{rcases}x=6y-8 \\ y=1\end{rcases}\begin{matrix}x=6⋅1-8 \\ x=-2\end{matrix}\)

1p

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

1p

a

Substitutie geeft \(6x+4(9x+17)=-16\)

1p

Haakjes wegwerken geeft \(6x+36x+68=-16\)
ofwel \(42x=-84\) dus \(x=-2\text{.}\)

1p

\(\begin{rcases}y=9x+17 \\ x=-2\end{rcases}\begin{matrix}y=9⋅-2+17 \\ y=-1\end{matrix}\)

1p

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

1p

a

Substitutie geeft \(7(9y-56)+6y=22\)

1p

Haakjes wegwerken geeft \(63y-392+6y=22\)
ofwel \(69y=414\) dus \(y=6\text{.}\)

1p

\(\begin{rcases}x=9y-56 \\ y=6\end{rcases}\begin{matrix}x=9⋅6-56 \\ x=-2\end{matrix}\)

1p

De oplossing is \((x, y)=(-2, 6)\text{.}\)

1p

a

Substitutie geeft \(y=2(8y+27)-9\)

1p

Haakjes wegwerken geeft \(y=16y+54-9\)
ofwel \(-15y=45\) dus \(y=-3\text{.}\)

1p

\(\begin{rcases}x=8y+27 \\ y=-3\end{rcases}\begin{matrix}x=8⋅-3+27 \\ x=3\end{matrix}\)

1p

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

1p

a

Substitutie geeft \(x=5(3x-18)+34\)

1p

Haakjes wegwerken geeft \(x=15x-90+34\)
ofwel \(-14x=-56\) dus \(x=4\text{.}\)

1p

\(\begin{rcases}y=3x-18 \\ x=4\end{rcases}\begin{matrix}y=3⋅4-18 \\ y=-6\end{matrix}\)

1p

De oplossing is \((x, y)=(4, -6)\text{.}\)

1p