Solve the nonlinear system of equations for real solutions.
-$\left\{ \begin{array} { l }
y = x ^ { 2 } + 2 \
y = - x ^ { 2 } + 4
\end{array} \right.$

A) $( 1,3 ) , ( - 1,3 )$
B) $( 1,3 ) , ( 1 , - 3 ) , ( - 1,3 ) , ( - 1 , - 3 )$
C) $\varnothing$
D) $( 3,1 ) , ( 3 , - 1 )$

Question

Solve the nonlinear system of equations for real solutions.
-$\left\{ \begin{array} { l } 
y = x ^ { 2 } + 2 \
y = - x ^ { 2 } + 4
\end{array} \right.$

A) $( 1,3 ) , ( - 1,3 )$
B) $( 1,3 ) , ( 1 , - 3 ) , ( - 1,3 ) , ( - 1 , - 3 )$
C) $\varnothing$
D) $( 3,1 ) , ( 3 , - 1 )$

Quizplus · Accepted Answer

To find the real solutions, set the two equations equal to each other: $x^2 + 2 = -x^2 + 4$. Solving for $x$, we get $2x^2 = 2$, so $x^2 = 1$, which gives $x = \pm 1$. Substituting $x = 1$ into the first equation, we get $y = 1^2 + 2 = 3$, and similarly for $x = -1$, $y = 3$. Thus, the solutions are $(1, 3)$ and $(-1, 3)$.

Solve the Nonlinear System of Equations for Real Solutions {y=x2+2y=−x2+4\left\{ \begin{array} { l } y = x ^ { 2 } + 2 \\y = - x ^ { 2 } + 4\end{array} \right.{y=x2+2y=−x2+4​

Solve the Nonlinear System of Equations for Real Solutions $\left\{ \begin{array} { l } y = x ^ { 2 } + 2 \\y = - x ^ { 2 } + 4\end{array} \right.$