The problem is to solve the inequality $4 + \frac{3}{4}(x+2) \le \frac{3}{8}x + 1$.

1. Problem Description

The problem is to solve the inequality $4 + \frac{3}{4}(x+2) \le \frac{3}{8}x + 1$.

2. Solution Steps

First, distribute the $\frac{3}{4}$ in the left side of the inequality:

$4 + \frac{3}{4}x + \frac{3}{4}(2) \le \frac{3}{8}x + 1$

$4 + \frac{3}{4}x + \frac{6}{4} \le \frac{3}{8}x + 1$

$4 + \frac{3}{4}x + \frac{3}{2} \le \frac{3}{8}x + 1$

Convert all terms to have a common denominator, which is 8:

$\frac{32}{8} + \frac{6}{8}x + \frac{12}{8} \le \frac{3}{8}x + \frac{8}{8}$

Combine the constants on the left side:

$\frac{44}{8} + \frac{6}{8}x \le \frac{3}{8}x + \frac{8}{8}$

Multiply both sides by 8 to eliminate the fractions:

$44 + 6x \le 3x + 8$

Subtract $3x$ from both sides:

$44 + 6x - 3x \le 3x - 3x + 8$

$44 + 3x \le 8$

Subtract 44 from both sides:

$44 - 44 + 3x \le 8 - 44$

$3x \le -36$

Divide both sides by 3:

$\frac{3x}{3} \le \frac{-36}{3}$

$x \le -12$

3. Final Answer

$x \le -12$