The problem is to solve the following inequalities: a) $4x + 8 \le 2x - 12$ b) $3x - 2 \ge x - 14$ c) $2x - 7 < -3x + 3$ d) $5x - 4 > 2x - 3$

1. Problem Description

The problem is to solve the following inequalities:

a) $4x + 8 \le 2x - 12$

b) $3x - 2 \ge x - 14$

c) $2x - 7 < -3x + 3$

d) $5x - 4 > 2x - 3$

2. Solution Steps

a) $4x + 8 \le 2x - 12$

Subtract $2x$ from both sides:

$4x - 2x + 8 \le 2x - 2x - 12$

$2x + 8 \le -12$

Subtract $8$ from both sides:

$2x + 8 - 8 \le -12 - 8$

$2x \le -20$

Divide both sides by $2$:

$\frac{2x}{2} \le \frac{-20}{2}$

$x \le -10$

b) $3x - 2 \ge x - 14$

Subtract $x$ from both sides:

$3x - x - 2 \ge x - x - 14$

$2x - 2 \ge -14$

Add $2$ to both sides:

$2x - 2 + 2 \ge -14 + 2$

$2x \ge -12$

Divide both sides by $2$:

$\frac{2x}{2} \ge \frac{-12}{2}$

$x \ge -6$

c) $2x - 7 < -3x + 3$

Add $3x$ to both sides:

$2x + 3x - 7 < -3x + 3x + 3$

$5x - 7 < 3$

Add $7$ to both sides:

$5x - 7 + 7 < 3 + 7$

$5x < 10$

Divide both sides by $5$:

$\frac{5x}{5} < \frac{10}{5}$

$x < 2$

d) $5x - 4 > 2x - 3$

Subtract $2x$ from both sides:

$5x - 2x - 4 > 2x - 2x - 3$

$3x - 4 > -3$

Add $4$ to both sides:

$3x - 4 + 4 > -3 + 4$

$3x > 1$

Divide both sides by $3$:

$\frac{3x}{3} > \frac{1}{3}$

$x > \frac{1}{3}$

3. Final Answer

a) $x \le -10$

b) $x \ge -6$

c) $x < 2$

d) $x > \frac{1}{3}$