Solve `7lt11-4xlt2x+18,x inZ`

Answer

To solve the inequality \( 7 < 11 - 4x < 2x + 18 \) where \( x \) belongs to the set of integers \( \mathbb{Z} \), we can break it down into two separate inequalities:

1. \( 7 < 11 - 4x \)
2. \( 11 - 4x < 2x + 18 \)

### Step 1: Solve the first inequality \( 7 < 11 - 4x \)

1. Start by isolating \( x \):
\[ 7 < 11 - 4x \]

2. Subtract 11 from both sides:
\[ 7 - 11 < -4x \]
\[ -4 < -4x \]

3. Divide both sides by -4 (remember to reverse the inequality sign):
\[ 1 > x \quad \text{or} \quad x < 1 \]

### Step 2: Solve the second inequality \( 11 - 4x < 2x + 18 \)

1. Start by isolating \( x \):
\[ 11 - 4x < 2x + 18 \]

2. Add \( 4x \) to both sides:
\[ 11 < 6x + 18 \]

3. Subtract 18 from both sides:
\[ 11 - 18 < 6x \]
\[ -7 < 6x \]

4. Divide both sides by 6:
\[ -\frac{7}{6} < x \quad \text{or} \quad x > -\frac{7}{6} \]

### Step 3: Combine the results

From the two inequalities, we have:
1. \( x < 1 \)
2. \( x > -\frac{7}{6} \)

### Step 4: Identify the integer solutions

Since \( x \) must be an integer, we need to find integers that satisfy both conditions:
- The integers greater than \(-\frac{7}{6}\) are \(-1, 0, 1, 2, \ldots\)
- The integers less than \( 1 \) are \(-1, 0\)

### Final Solution

The integers that satisfy both inequalities are:
\[ x \in \{-1, 0\} \]