Solve the system of equations using elimination.
-$$\left\{ \begin{array} { l }
x ^ { 2 } + y ^ { 2 } = 64 \
\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 9 } = 1
\end{array} \right.$$
A) $x = 0 , y = - 8 ; x = 0 , y = 8$
B) $x = - 8 , y = 0 ; x = 8 , y = 0$
C) $x = 0 , y = - 3 ; x = 0 , y = 3$
D) inconsistent

Question

Solve the system of equations using elimination.
-$$\left\{ \begin{array} { l } 
x ^ { 2 } + y ^ { 2 } = 64 \
\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 9 } = 1
\end{array} \right.$$
A) $x = 0 , y = - 8 ; x = 0 , y = 8$
B) $x = - 8 , y = 0 ; x = 8 , y = 0$
C) $x = 0 , y = - 3 ; x = 0 , y = 3$
D) inconsistent

Quizplus · Accepted Answer

The first equation represents a circle with radius 8 centered at the origin, and the second equation represents an ellipse. By multiplying the second equation by 64*9 to eliminate the denominators, we get $9x^2 + 64y^2 = 576$. Substituting $x^2 + y^2 = 64$ from the first equation into this modified second equation doesn't directly solve the system but indicates that the solutions must satisfy both the circle and the ellipse equations. The correct solutions are where these two shapes intersect, which, given the symmetry and the equations, are at points where either $x$ or $y$ is zero but not both simultaneously. Solving for these intersections, we find that the solutions are $x = \pm 8, y = 0$, which corresponds to choice B.

Solve the System of Equations Using Elimination {x2+y2=64x264+y29=1\left\{ \begin{array} { l } x ^ { 2 } + y ^ { 2 } = 64 \\\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 9 } = 1\end{array} \right.{x2+y2=6464x2​+9y2​=1​

Solve the System of Equations Using Elimination $\left\{ \begin{array} { l } x ^ { 2 } + y ^ { 2 } = 64 \\\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 9 } = 1\end{array} \right.$