The problem asks us to solve two inequalities for $x$ and fill in the boxes with the solution. (1) $2(x+3) \ge 3(x+1)$ (2) $\frac{x+3}{10} - \frac{5x-2}{8} \le -\frac{x}{4}$

1. Problem Description

The problem asks us to solve two inequalities for $x$ and fill in the boxes with the solution.

(1) $2(x+3) \ge 3(x+1)$

(2) $\frac{x+3}{10} - \frac{5x-2}{8} \le -\frac{x}{4}$

2. Solution Steps

(1) $2(x+3) \ge 3(x+1)$

Expand both sides:

$2x + 6 \ge 3x + 3$

Subtract $2x$ from both sides:

$6 \ge x + 3$

Subtract 3 from both sides:

$3 \ge x$

$x \le 3$

Thus, box 28 contains

3.

(2) $\frac{x+3}{10} - \frac{5x-2}{8} \le -\frac{x}{4}$

To eliminate the fractions, multiply both sides by the least common multiple of 10, 8, and 4, which is 40:

$40 \cdot \left( \frac{x+3}{10} - \frac{5x-2}{8} \right) \le 40 \cdot \left( -\frac{x}{4} \right)$

$4(x+3) - 5(5x-2) \le -10x$

$4x + 12 - 25x + 10 \le -10x$

$-21x + 22 \le -10x$

Add $21x$ to both sides:

$22 \le 11x$

Divide both sides by 11:

$2 \le x$

$x \ge 2$

Thus, box 29 contains

2.

3. Final Answer

Box 28 contains

3. Box 29 contains 2.