The problem requires solving the inequality $|x+1| \ge 2|x+2|$. Additionally, there is an expression $12\frac{11}{18}+3\frac{4}{15}x$ mentioned, but there is no relation between the inequality and the expression. So, we solve the inequality $|x+1| \ge 2|x+2|$.

1. Problem Description

The problem requires solving the inequality $|x+1| \ge 2|x+2|$. Additionally, there is an expression $12\frac{11}{18}+3\frac{4}{15}x$ mentioned, but there is no relation between the inequality and the expression. So, we solve the inequality $|x+1| \ge 2|x+2|$.

2. Solution Steps

We consider the inequality $|x+1| \ge 2|x+2|$. To solve this inequality, we consider different cases based on the values of $x$.

Case 1: $x \ge -1$. Then $x+1 \ge 0$ and $x+2 > 0$, so $|x+1| = x+1$ and $|x+2| = x+2$.

The inequality becomes $x+1 \ge 2(x+2)$, which simplifies to $x+1 \ge 2x+4$.

Subtracting $x$ from both sides gives $1 \ge x+4$, so $x \le -3$.

Since we assumed $x \ge -1$, and now $x \le -3$, there is no solution in this case because $-1 \le x \le -3$ is not possible.

Case 2: $-2 \le x < -1$. Then $x+1 < 0$, so $|x+1| = -(x+1) = -x-1$. Also, $x+2 \ge 0$, so $|x+2| = x+2$.

The inequality becomes $-x-1 \ge 2(x+2)$, which simplifies to $-x-1 \ge 2x+4$.

Adding $x$ to both sides gives $-1 \ge 3x+4$.

Subtracting 4 from both sides gives $-5 \ge 3x$, so $x \le -\frac{5}{3} = -1\frac{2}{3}$.

Since we assumed $-2 \le x < -1$ and we have $x \le -\frac{5}{3}$, the solution in this case is $-2 \le x \le -\frac{5}{3}$.

Case 3: $x < -2$. Then $x+1 < 0$, so $|x+1| = -(x+1) = -x-1$. Also, $x+2 < 0$, so $|x+2| = -(x+2) = -x-2$.

The inequality becomes $-x-1 \ge 2(-x-2)$, which simplifies to $-x-1 \ge -2x-4$.

Adding $2x$ to both sides gives $x-1 \ge -4$.

Adding 1 to both sides gives $x \ge -3$.

Since we assumed $x < -2$ and we have $x \ge -3$, the solution in this case is $-3 \le x < -2$.

Combining the solutions from the cases:

$-2 \le x \le -\frac{5}{3}$ and $-3 \le x < -2$.

Thus, the solution to the inequality is $-3 \le x \le -\frac{5}{3}$.

3. Final Answer

The solution to the inequality $|x+1| \ge 2|x+2|$ is $-3 \le x \le -\frac{5}{3}$.