Solve the nonlinear system of equations for real solutions.
-$\left\{ \begin{array} { r }
x ^ { 2 } + y ^ { 2 } = 65 \
4 x + 7 y = 0
\end{array} \right.$
A) $( 7 , - 4 )$
B) $( 7 , - 4 ) , ( - 7 , - 4 )$
C) $( 7 , - 4 ) , ( - 7,4 )$
D) $( 7,4 ) , ( - 7 , - 4 )$

Question

Solve the nonlinear system of equations for real solutions.
-$\left\{ \begin{array} { r } 
x ^ { 2 } + y ^ { 2 } = 65 \
4 x + 7 y = 0
\end{array} \right.$
A) $( 7 , - 4 )$
B) $( 7 , - 4 ) , ( - 7 , - 4 )$
C) $( 7 , - 4 ) , ( - 7,4 )$
D) $( 7,4 ) , ( - 7 , - 4 )$

Quizplus · Accepted Answer

To solve the system, we can express $y$ in terms of $x$ from the second equation: $y = -\frac{4}{7}x$. Substituting this into the first equation gives $x^2 + \left(-\frac{4}{7}x\right)^2 = 65$, which simplifies to $x^2 + \frac{16}{49}x^2 = 65$, or $49x^2 + 16x^2 = 3185$, leading to $65x^2 = 3185$, and thus $x^2 = 49$, giving $x = \pm 7$. Substituting $x = 7$ and $x = -7$ back into $y = -\frac{4}{7}x$ yields $y = -4$ and $y = 4$, respectively, resulting in the solutions $(7, -4)$ and $(-7, 4)$.

Solve the Nonlinear System of Equations for Real Solutions {x2+y2=654x+7y=0\left\{ \begin{array} { r } x ^ { 2 } + y ^ { 2 } = 65 \\4 x + 7 y = 0\end{array} \right.{x2+y2=654x+7y=0​

Solve the Nonlinear System of Equations for Real Solutions $\left\{ \begin{array} { r } x ^ { 2 } + y ^ { 2 } = 65 \\4 x + 7 y = 0\end{array} \right.$