Solve the system by the elimination method or a combination of the elimination and substitution methods.
-$x^{2}+3 x y+y^{2}=4$
$x^{2}-3 x y-y^{2}=4$
A) $\{(2,0),(-2,0)\}$
B) $\{(2,0),(-2,0),(2,-6),(-2,6)\}$
C) $\{(0,2),(0,-2),(2,-6),(-2,6)\}$
D) $\{(2,-6),(-2,6)\}$

Question

Quizplus · Accepted Answer

Adding the two equations eliminates $y$, resulting in $2x^2 = 8$, which simplifies to $x^2 = 4$, giving $x = 2$ or $x = -2$. Substituting $x = 2$ and $x = -2$ back into either original equation allows solving for $y$, yielding solutions $(2, 0)$, $(-2, 0)$, $(2, -6)$, and $(-2, 6)$.

Solve the System by the Elimination Method or a Combination x2+3xy+y2=4x^{2}+3 x y+y^{2}=4x2+3xy+y2=4

Solve the System by the Elimination Method or a Combination $x^{2}+3 x y+y^{2}=4$