a) x 2  y 2  25; x  {0; 1; 2; 3; 4; 5

Let's solve the equation \( x^2 + y^2 = 25 \) for \( x \in \{0, 1, 2, 3, 4, 5\} \).

1. For \( x = 0 \):
   \[
   0^2 + y^2 = 25 \implies y^2 = 25 \implies y = \pm 5
   \]
   Solutions: \( (0, 5) \) and \( (0, -5) \)

2. For \( x = 1 \):
   \[
   1^2 + y^2 = 25 \implies 1 + y^2 = 25 \implies y^2 = 24 \implies y = \pm \sqrt{24} = \pm 2\sqrt{6}
   \]
   Solutions: \( (1, 2\sqrt{6}) \) and \( (1, -2\sqrt{6}) \)

3. For \( x = 2 \):
   \[
   2^2 + y^2 = 25 \implies 4 + y^2 = 25 \implies y^2 = 21 \implies y = \pm \sqrt{21}
   \]
   Solutions: \( (2, \sqrt{21}) \) and \( (2, -\sqrt{21}) \)

4. For \( x = 3 \):
   \[
   3^2 + y^2 = 25 \implies 9 + y^2 = 25 \implies y^2 = 16 \implies y = \pm 4
   \]
   Solutions: \( (3, 4) \) and \( (3, -4) \)

5. For \( x = 4 \):
   \[
   4^2 + y^2 = 25 \implies 16 + y^2 = 25 \implies y^2 = 9 \implies y = \pm 3
   \]
   Solutions: \( (4, 3) \) and \( (4, -3) \)

6. For \( x = 5 \):
   \[
   5^2 + y^2 = 25 \implies 25 + y^2 = 25 \implies y^2 = 0 \implies y = 0
   \]
   Solution: \( (5, 0) \)

So, the complete set of solutions for \( x \in \{0, 1, 2, 3, 4, 5\} \) are:
\[
(0, 5), (0, -5), (1, 2\sqrt{6}), (1, -2\sqrt{6}), (2, \sqrt{21}), (2, -\sqrt{21}), (3, 4), (3, -4), (4, 3), (4, -3), (5, 0)
\]

Would you like any further details or have any questions?

Here are five more questions you might want to ask:
1. Can you explain the process of solving quadratic equations?
2. How do you solve systems of linear equations?
3. What are the properties of trigonometric functions?
4. Can you provide an example of solving a calculus problem?
5. How do you find the roots of a polynomial?

**Tip:** When solving equations, always check if the solutions satisfy the original equation to verify their correctness.

Learn how to solve the equation x^2 + y^2 = 25 for different values of x, including 0, 1, 2, 3, 4, and 5. Detailed step-by-step solutions are provided for each case. This problem involves applying the Pythagorean Theorem and is suitable for Grades 7-9.

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