Let p(x), q(x) denote the following open statements. p(x): x <= 3 and q(x)= x + 1 is odd

We have the two open statements:

1. \( p(x): x \leq 3 \)
2. \( q(x): x + 1 \text{ is odd} \)

### Interpretation:
1. \( p(x) \) states that \( x \) is less than or equal to 3.
2. \( q(x) \) states that \( x + 1 \) is odd, which implies \( x \) is even because adding 1 to an even number yields an odd number.

### Combining the Statements
We can examine \( p(x) \land q(x) \), which requires both conditions to be true simultaneously:
1. \( x \leq 3 \) (from \( p(x) \)).
2. \( x \) is even (from \( q(x) \)).

### Solution:
The integers \( x \) that satisfy both conditions are:
- \( x = 2 \): \( 2 \leq 3 \) and \( 2 + 1 = 3 \), which is odd.
- \( x = 0 \): \( 0 \leq 3 \) and \( 0 + 1 = 1 \), which is odd.
- \( x = -2 \): \( -2 \leq 3 \) and \( -2 + 1 = -1 \), which is odd.

Thus, the integers satisfying \( p(x) \land q(x) \) are \( \{ -2, 0, 2 \} \).

Let me know if you want details or have questions!

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**Related Questions:**
1. What happens when \( p(x) \) is changed to \( x \geq 3 \)?
2. What is the result if \( q(x) \) is redefined as \( x - 1 \text{ is odd} \)?
3. How would the result change if non-integer \( x \) were allowed?
4. What are the solutions to \( p(x) \lor q(x) \)?
5. Can you prove \( x \) must always be even for \( q(x) \) to hold?

**Tip:** Always check whether constraints apply to integers, rationals, or real numbers when solving logical statements involving \( x \).

Let p(x), q(x) denote the following open statements. p(x): x <= 3 and q(x): x + 1 is odd

Explore a detailed solution to the logical problem involving the open statements p(x): x <= 3 and q(x): x + 1 is odd. Learn how to combine conditions to find integers satisfying both, while understanding properties of even and odd numbers. Ideal for Grades 8-10 students.

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