Solve the nonlinear system of equations.
-$$\left\{ \begin{array} { r }
x ^ { 2 } + y ^ { 2 } = 40 \
2 x + 6 y = 0
\end{array} \right.$$
A) $( 6 , - 2 )$
B) $( 6 , - 2 ) , ( - 6 , - 2 )$
C) $( 6 , - 2 ) , ( - 6,2 )$
D) $( 6,2 ) , ( - 6 , - 2 )$

Question

Solve the nonlinear system of equations.
-$$\left\{ \begin{array} { r } 
x ^ { 2 } + y ^ { 2 } = 40 \
2 x + 6 y = 0
\end{array} \right.$$
A) $( 6 , - 2 )$
B) $( 6 , - 2 ) , ( - 6 , - 2 )$
C) $( 6 , - 2 ) , ( - 6,2 )$
D) $( 6,2 ) , ( - 6 , - 2 )$

Quizplus · Accepted Answer

From the second equation, $y = -\frac{1}{3}x$. Substituting this into the first equation gives $x^2 + \left(-\frac{1}{3}x\right)^2 = 40$, leading to $x^2 + \frac{1}{9}x^2 = 40$, which simplifies to $\frac{10}{9}x^2 = 40$, and further to $x^2 = 36$, giving $x = 6$ or $x = -6$. Substituting these values back into $y = -\frac{1}{3}x$ gives the pairs $(6, -2)$ and $(-6, 2)$.

Solve the Nonlinear System of Equations {x2+y2=402x+6y=0\left\{ \begin{array} { r } x ^ { 2 } + y ^ { 2 } = 40 \\2 x + 6 y = 0\end{array} \right.{x2+y2=402x+6y=0​

Solve the Nonlinear System of Equations $\left\{ \begin{array} { r } x ^ { 2 } + y ^ { 2 } = 40 \\2 x + 6 y = 0\end{array} \right.$