Given `P = {x: 5 lt 2x - 1 le 11, x in R } ` and `Q = {x : -1 le 3 +4x lt 23, x in I} `
where R = {real number } and I = { integers}.

Answer

To solve the given sets \( P \) and \( Q \), we will break down the inequalities step by step.

### Step 1: Solve for Set \( P \)

Given:
\[ P = \{ x : 5 < 2x - 1 \leq 11, x \in \mathbb{R} \} \]

We have two inequalities to solve:

1. **First Inequality**:
\[ 5 < 2x - 1 \]
Adding 1 to both sides:
\[ 5 + 1 < 2x \]
\[ 6 < 2x \]
Dividing both sides by 2:
\[ 3 < x \]
or
\[ x > 3 \]

2. **Second Inequality**:
\[ 2x - 1 \leq 11 \]
Adding 1 to both sides:
\[ 2x \leq 11 + 1 \]
\[ 2x \leq 12 \]
Dividing both sides by 2:
\[ x \leq 6 \]

Combining both results:
\[ 3 < x \leq 6 \]

Thus, the set \( P \) can be expressed as:
\[ P = (3, 6] \]

### Step 2: Solve for Set \( Q \)

Given:
\[ Q = \{ x : -1 \leq 3 + 4x < 23, x \in \mathbb{I} \} \]

We will solve the two parts of the inequality:

1. **First Inequality**:
\[ -1 \leq 3 + 4x \]
Subtracting 3 from both sides:
\[ -1 - 3 \leq 4x \]
\[ -4 \leq 4x \]
Dividing both sides by 4:
\[ -1 \leq x \]
or
\[ x \geq -1 \]

2. **Second Inequality**:
\[ 3 + 4x < 23 \]
Subtracting 3 from both sides:
\[ 4x < 23 - 3 \]
\[ 4x < 20 \]
Dividing both sides by 4:
\[ x < 5 \]

Combining both results:
\[ -1 \leq x < 5 \]

Since \( x \) belongs to integers \( \mathbb{I} \), the integer solutions are:
\[ x = -1, 0, 1, 2, 3, 4 \]

Thus, the set \( Q \) can be expressed as:
\[ Q = \{-1, 0, 1, 2, 3, 4\} \]

### Final Results

- The solution set for \( P \) is:
\[ P = (3, 6] \]

- The solution set for \( Q \) is:
\[ Q = \{-1, 0, 1, 2, 3, 4\} \]