Find the solution set for the system by graphing both of the system's equations in the same rectangular coordinate
system and finding points of intersection.
-$$\begin{aligned}
16 x ^ { 2 } + y ^ { 2 } & = 16 \
y ^ { 2 } - 16 x ^ { 2 } & = 16
\end{aligned}$$

A) $\{ ( 0 , - 4 ) , ( 0,4 ) \}$
B) $\{ ( 0 , - 4 ) \}$
C) $\{ ( 0,16 ) \}$
D) $\{ ( 4,0 ) , ( 4,0 ) \}$

Question

Find the solution set for the system by graphing both of the system's equations in the same rectangular coordinate
system and finding points of intersection.
-$$\begin{aligned}
16 x ^ { 2 } + y ^ { 2 } & = 16 \
y ^ { 2 } - 16 x ^ { 2 } & = 16
\end{aligned}$$

A) $\{ ( 0 , - 4 ) , ( 0,4 ) \}$ 
B) $\{ ( 0 , - 4 ) \}$ 
C) $\{ ( 0,16 ) \}$ 
D) $\{ ( 4,0 ) , ( 4,0 ) \}$

Quizplus · Accepted Answer

We can start by graphing both equations on the same rectangular coordinate system.

The graph looks like two circles intersecting at two points.

To find the points of intersection, we can solve the system of equations:

$$\begin{aligned}
16 x ^ { 2 } + y ^ { 2 } & = 16 \
y ^ { 2 } - 16 x ^ { 2 } & = 16
\end{aligned}$$

We can use the second equation to solve for $y ^2$ as follows:

$$y ^ 2 = 16 x ^ 2 + 16$$

Substitute this expression into the first equation:

$$16 x ^ 2 + 16 x ^ 2 + 16 = 16$$

Simplify:

$$32 x ^ 2 = 0$$

Solve for $x$:

$$x = 0$$

Substitute this value into either equation to solve for $y$:

$$y ^ 2 = 16$$
$$y = \pm 4$$

Therefore, the solution set is $\{ ( 0 , - 4 ) , ( 0,4 ) \}$.

Find the Solution Set for the System by Graphing Both 16x2+y2=16y2−16x2=16\begin{aligned}16 x ^ { 2 } + y ^ { 2 } & = 16 \\y ^ { 2 } - 16 x ^ { 2 } & = 16\end{aligned}16x2+y2y2−16x2​=16=16​

Find the Solution Set for the System by Graphing Both $\begin{aligned}16 x ^ { 2 } + y ^ { 2 } & = 16 \\y ^ { 2 } - 16 x ^ { 2 } & = 16\end{aligned}$