Solve the system by the substitution method.
-$x^{2}+y^{2}=4$
$\mathrm{x}^{2}-\mathrm{y}^{2}=4$
A) $(2,0),(-2,0)$
B) $(2,0),(2,2)$
C) $(2,2),(-2,2)$
D) No solution

Question

Quizplus · Accepted Answer

Solving the system by substitution, we can add the two equations together to eliminate $y^2$, getting $2x^2 = 8$, which simplifies to $x^2 = 4$. Solving for $x$, we get $x = 2$ or $x = -2$. Substituting these values back into one of the original equations, we find that $y = 0$ for both cases, leading to the solutions $(2,0)$ and $(-2,0)$.

Solve the System by the Substitution Method x2+y2=4x^{2}+y^{2}=4x2+y2=4 x2−y2=4\mathrm{x}^{2}-\mathrm{y}^{2}=4x2−y2=4 A) (2,0),(−2,0)(2,0),(-2,0)(2,0),(−2,0) B) (2,0),(2,2)(2,0),(2,2)(2,0),(2,2)

Solve the System by the Substitution Method $x^{2}+y^{2}=4$ $\mathrm{x}^{2}-\mathrm{y}^{2}=4$ A) $(2,0),(-2,0)$ B) $(2,0),(2,2)$