Solve the system.
-$$\begin{array} { l }
x ^ { 2 } + y ^ { 2 } = 10 \
3 y ^ { 2 } - x ^ { 2 } = 14
\end{array}$$
A) $\{ ( 2 , \sqrt { 6 } ) , ( 2 , - \sqrt { 6 } ) , ( - 2 , \sqrt { 6 } ) , ( - 2 , - \sqrt { 6 } ) \}$
B) $\{ ( 2 , \sqrt { 6 } ) \}$
C){ }
D) $\{ ( 2 , \sqrt { 6 } ) , ( - 2 , - \sqrt { 6 } ) \}$

Question

Solve the system.
-$$\begin{array} { l } 
x ^ { 2 } + y ^ { 2 } = 10 \
3 y ^ { 2 } - x ^ { 2 } = 14
\end{array}$$
A) $\{ ( 2 , \sqrt { 6 } ) , ( 2 , - \sqrt { 6 } ) , ( - 2 , \sqrt { 6 } ) , ( - 2 , - \sqrt { 6 } ) \}$
B) $\{ ( 2 , \sqrt { 6 } ) \}$
C){  }
D) $\{ ( 2 , \sqrt { 6 } ) , ( - 2 , - \sqrt { 6 } ) \}$

Quizplus · Accepted Answer

Adding the two equations gives $4y^2 = 24$, leading to $y^2 = 6$. Substituting $y^2$ back into one of the equations gives four possible solutions for $x$ and $y$, making option A correct.

Solve the System x2+y2=103y2−x2=14\begin{array} { l } x ^ { 2 } + y ^ { 2 } = 10 \\3 y ^ { 2 } - x ^ { 2 } = 14\end{array}x2+y2=103y2−x2=14​

Solve the System $\begin{array} { l } x ^ { 2 } + y ^ { 2 } = 10 \\3 y ^ { 2 } - x ^ { 2 } = 14\end{array}$