Let's solve the system step by step using the elimination method.

Given system of equations:

1. $x^2 + y^2 = 5$
2. $x^2 - y^2 = 3$

Step 1: Eliminate one variable

Add the two equations: $(x^2 + y^2) + (x^2 - y^2) = 5 + 3$ $2x^2 = 8 \quad \implies \quad x^2 = 4$

Now subtract the second equation from the first: $(x^2 + y^2) - (x^2 - y^2) = 5 - 3$ $2y^2 = 2 \quad \implies \quad y^2 = 1$

Step 2: Solve for $x$ and $y$

From $x^2 = 4$, we have: $x = \pm 2$

From $y^2 = 1$, we have: $y = \pm 1$

Step 3: Determine all combinations

Now we combine the signs of $x$ and $y$ to find the four solutions:

1. $x = -2, y = -1$
2. $x = -2, y = 1$
3. $x = 2, y = -1$
4. $x = 2, y = 1$

Step 4: Assign solutions to the quadrants

Based on the question:

• $x < 0, y < 0$: $x = -2, y = -1$
• $x < 0, y > 0$: $x = -2, y = 1$
• $x > 0, y < 0$: $x = 2, y = -1$
• $x > 0, y > 0$: $x = 2, y = 1$

Final Answer:

1. $x = -2, y = -1$
2. $x = -2, y = 1$
3. $x = 2, y = -1$
4. $x = 2, y = 1$

Would you like me to explain any step in more detail or discuss a related concept?

Further questions:

1. Why does eliminating variables simplify the solution process?
2. How would the solution change if the equations had different constants?
3. What does the geometric interpretation of this system look like?
4. How can we confirm our solutions using substitution?
5. Can this method be generalized to three variables?

Tip:

Always double-check your solutions by substituting them back into the original equations to ensure accuracy!