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

1. Problem Description

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

2. Solution Steps

First, we distribute the $\frac{3}{4}$ in the term $\frac{3}{4}(x+2)$:

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

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

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

Now, we can combine the constants on the left side:

$4 + \frac{3}{2} = \frac{8}{2} + \frac{3}{2} = \frac{11}{2}$

So, the inequality becomes:

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

Next, we subtract $\frac{3}{8}x$ from both sides:

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

To combine the x terms, we need a common denominator, which is 8:

$\frac{3}{4}x = \frac{6}{8}x$

So, $\frac{6}{8}x - \frac{3}{8}x = \frac{3}{8}x$

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

Now, subtract $\frac{11}{2}$ from both sides:

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

$1 - \frac{11}{2} = \frac{2}{2} - \frac{11}{2} = -\frac{9}{2}$

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

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

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

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

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

$x \le -12$

3. Final Answer

$x \le -12$