If `2x + 2 ( 4 + 3x) lt 2 + 3x < 2x + x//2,` then x can take which of the following values ?

Options

A. `-3`
B. 1
C. 0
D. `-1`

Answer

To solve the inequality \( 2x + 2(4 + 3x) < 2 + 3x < 2x + \frac{x}{2} \), we will break it down into two parts and solve each inequality separately.

### Step 1: Solve the first inequality \( 2x + 2(4 + 3x) < 2 + 3x \)

1. Expand the left side:
\[ 2x + 2 \times 4 + 2 \times 3x < 2 + 3x \]
This simplifies to:
\[ 2x + 8 + 6x < 2 + 3x \]
Combine like terms:
\[ 8x + 8 < 2 + 3x \]

2. Rearrange the inequality:
\[ 8x - 3x < 2 - 8 \]
This simplifies to:
\[ 5x < -6 \]

3. Divide by 5:
\[ x < -\frac{6}{5} \]

### Step 2: Solve the second inequality \( 2 + 3x < 2x + \frac{x}{2} \)

1. Rearrange the inequality:
\[ 3x - 2x < \frac{x}{2} - 2 \]
This simplifies to:
\[ x < \frac{x}{2} - 2 \]

2. Multiply everything by 2 to eliminate the fraction:
\[ 2x < x - 4 \]

3. Rearrange the inequality:
\[ 2x - x < -4 \]
This simplifies to:
\[ x < -4 \]

### Step 3: Combine the results

From the first inequality, we found that \( x < -\frac{6}{5} \) and from the second inequality, we found that \( x < -4 \). The more restrictive condition is \( x < -4 \).

### Conclusion

Thus, the solution to the original compound inequality is:
\[ x < -4 \]