You have the following system of equations:

$x^2 + y^2 = 16 \quad \text{(1)}$ $x^2 - y^2 = 16 \quad \text{(2)}$

To solve for $x$ and $y$, we can add and subtract these two equations.

Step 1: Add the equations

$(x^2 + y^2) + (x^2 - y^2) = 16 + 16$ $2x^2 = 32 \quad \Rightarrow \quad x^2 = 16 \quad \Rightarrow \quad x = \pm 4$

Step 2: Subtract the equations

$(x^2 + y^2) - (x^2 - y^2) = 16 - 16$ $2y^2 = 0 \quad \Rightarrow \quad y^2 = 0 \quad \Rightarrow \quad y = 0$

Thus, the solutions are: $x = 4 \quad \text{or} \quad x = -4, \quad y = 0$

Final Answer:

The solutions are $(x, y) = (4, 0)$ and $(x, y) = (-4, 0)$.

Would you like more details on the solution, or do you have any other questions?

1. How do we manipulate systems of equations with exponents?
2. What is the geometric interpretation of this system?
3. How would this approach change if we had a different equation type (e.g., linear)?
4. Can you solve this type of system using substitution instead of elimination?
5. What happens if you add or subtract in a different order?

Tip: When solving systems of equations with squares, always look for opportunities to add or subtract to simplify the expressions.