To solve this system of non-linear equations:

$x^2 + y^2 = 25$ $x^2 + 4y^2 = 52$

We can use substitution or elimination to find the solution. The easiest approach would be to subtract the first equation from the second to eliminate $x^2$.

Step 1: Subtract the two equations

$(x^2 + 4y^2) - (x^2 + y^2) = 52 - 25$ $x^2 + 4y^2 - x^2 - y^2 = 27$ $3y^2 = 27$ $y^2 = 9$ $y = \pm 3$

Step 2: Substitute $y = 3$ and $y = -3$ into the first equation to find $x$

Case 1: $y = 3$

Substitute $y = 3$ into $x^2 + y^2 = 25$: $x^2 + 3^2 = 25$ $x^2 + 9 = 25$ $x^2 = 16$ $x = \pm 4$

Thus, the solutions for $y = 3$ are $x = 4$ and $x = -4$, giving the points $(4, 3)$ and $(-4, 3)$.

Case 2: $y = -3$

Substitute $y = -3$ into $x^2 + y^2 = 25$: $x^2 + (-3)^2 = 25$ $x^2 + 9 = 25$ $x^2 = 16$ $x = \pm 4$

Thus, the solutions for $y = -3$ are $x = 4$ and $x = -4$, giving the points $(4, -3)$ and $(-4, -3)$.

Final Solution

The solution set for the system is: $(4, 3), (-4, 3), (4, -3), (-4, -3)$

Would you like further details or have any questions?

Here are five related questions to deepen your understanding:

1. How do you solve non-linear systems with more complex terms?
2. What is the geometric interpretation of this system of equations?
3. How can substitution be used to solve more complicated systems?
4. What are the properties of the circle and ellipse equations?
5. How would the solution change if the second equation were $x^2 + 5y^2 = 60$?

Tip: Always check the nature of equations before choosing a method, as some methods may simplify the process greatly (like elimination here).