## Precalculus (6th Edition) Blitzer

see graph; $(0,-2)$ and $(0,2)$
Step 1. Given equations $4x^2+y^2=4$ and $y^2-4x^2=4$, we can graph both equations as shown in the figure. Step 2. We can identify two intersection points $(0,-2)$ and $(0,2)$ as solutions to the system. Step 3. Check the solutions with the first equation; we have $4(0)^2+(-2)^2=4$ and $4(0)^2+(2)^2=4$ Step 4. Check the solutions with the second equation; we have $(-2)^2-4(0)^2=4$ and $(2)^2-4(0)^2=4$