The problem asks to find the truth set of the following inequalities and illustrate the answer on a number line. The inequalities are: a. $x + 4 > 3x$ b. $8x + 1 \le 4x - 3$ c. $4(x+2) < 5x - 1$ d. $2(x-2) \ge 5(x+1)$ e. $5x + 1 > 10 - x$

1. Problem Description

The problem asks to find the truth set of the following inequalities and illustrate the answer on a number line. The inequalities are:

a. $x + 4 > 3x$

b. $8x + 1 \le 4x - 3$

c. $4(x+2) < 5x - 1$

d. $2(x-2) \ge 5(x+1)$

e. $5x + 1 > 10 - x$

2. Solution Steps

a. $x + 4 > 3x$

Subtract $x$ from both sides:

$4 > 2x$

Divide both sides by 2:

$2 > x$

So, $x < 2$

b. $8x + 1 \le 4x - 3$

Subtract $4x$ from both sides:

$4x + 1 \le -3$

Subtract 1 from both sides:

$4x \le -4$

Divide both sides by 4:

$x \le -1$

c. $4(x+2) < 5x - 1$

Distribute the 4:

$4x + 8 < 5x - 1$

Subtract $4x$ from both sides:

$8 < x - 1$

Add 1 to both sides:

$9 < x$

So, $x > 9$

d. $2(x-2) \ge 5(x+1)$

Distribute the 2 and the 5:

$2x - 4 \ge 5x + 5$

Subtract $2x$ from both sides:

$-4 \ge 3x + 5$

Subtract 5 from both sides:

$-9 \ge 3x$

Divide both sides by 3:

$-3 \ge x$

So, $x \le -3$

e. $5x + 1 > 10 - x$

Add $x$ to both sides:

$6x + 1 > 10$

Subtract 1 from both sides:

$6x > 9$

Divide both sides by 6:

$x > \frac{9}{6}$

Simplify the fraction:

$x > \frac{3}{2}$

$x > 1.5$

3. Final Answer

a. $x < 2$

b. $x \le -1$

c. $x > 9$

d. $x \le -3$

e. $x > \frac{3}{2}$ (or $x > 1.5$)