Solve the system by the substitution or elimination method. $$\left( \begin{array} { r l }
2 x^{2}-3 y^{2}=-1 \
2 x^{2}+3 y^{2}=5
\end{array} \right)$$
A) (1, 1), (1, $-$ 1)
B) (1, 1), ( $-$ 1, $-$ 1)
C) ( $-$ 1, 1), (1, $-$ 1)
D) (1, 1), ( $-$ 1, 1), (1, $-$ 1), ( $-$ 1, $-$ 1)

Question

Solve the system by the substitution or elimination method. $$\left( \begin{array} { r l } 
2 x^{2}-3 y^{2}=-1 \
2 x^{2}+3 y^{2}=5
\end{array} \right)$$
A) (1, 1), (1, $-$ 1)
B) (1, 1), ( $-$ 1, $-$ 1)
C) ( $-$ 1, 1), (1, $-$ 1)
D) (1, 1), ( $-$ 1, 1), (1, $-$ 1), ( $-$ 1, $-$ 1)

Quizplus · Accepted Answer

Adding the two equations gives $2x^2 - 3y^2 + 2x^2 + 3y^2 = -1 + 5$, which simplifies to $4x^2 = 4$, or $x^2 = 1$. This means $x = \pm 1$. Substituting $x^2 = 1$ into either original equation, for example, the first one, gives $2(1) - 3y^2 = -1$, leading to $2 - 3y^2 = -1$, which simplifies to $3y^2 = 3$, or $y^2 = 1$, meaning $y = \pm 1$. Thus, the solutions are (1, 1), (-1, 1), (1, -1), and (-1, -1).

Solve the System by the Substitution or Elimination Method (2x2−3y2=−12x2+3y2=5)\left( \begin{array} { r l } 2 x^{2}-3 y^{2}=-1 \\2 x^{2}+3 y^{2}=5\end{array} \right)(2x2−3y2=−12x2+3y2=5​)

Solve the System by the Substitution or Elimination Method $\left( \begin{array} { r l } 2 x^{2}-3 y^{2}=-1 \\2 x^{2}+3 y^{2}=5\end{array} \right)$