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

We start by distributing $\frac{3}{4}$ to $(x+2)$.

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

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

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

Combine the constant terms on the left side.

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

$\frac{11}{2} + \frac{3}{4}x \le \frac{3}{8}x + 1$

Subtract $\frac{3}{8}x$ from both sides.

$\frac{11}{2} + \frac{3}{4}x - \frac{3}{8}x \le 1$

$\frac{11}{2} + \frac{6}{8}x - \frac{3}{8}x \le 1$

$\frac{11}{2} + \frac{3}{8}x \le 1$

Subtract $\frac{11}{2}$ from both sides.

$\frac{3}{8}x \le 1 - \frac{11}{2}$

$\frac{3}{8}x \le \frac{2}{2} - \frac{11}{2}$

$\frac{3}{8}x \le -\frac{9}{2}$

Multiply both sides by $\frac{8}{3}$.

$x \le -\frac{9}{2} \cdot \frac{8}{3}$

$x \le -\frac{72}{6}$

$x \le -12$

3. Final Answer

$x \le -12$