The problem asks us to find the truth set (solution set) of the following inequalities and illustrate the answer on a number line: 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 us to find the truth set (solution set) of the following inequalities and illustrate the answer on a number line:

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 by 2:

$2 > x$

or $x < 2$

Solution set: $x < 2$

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

Subtract $4x$ from both sides:

$4x+1 \le -3$

Subtract 1 from both sides:

$4x \le -4$

Divide by 4:

$x \le -1$

Solution set: $x \le -1$

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

Expand:

$4x+8 < 5x-1$

Subtract $4x$ from both sides:

$8 < x-1$

Add 1 to both sides:

$9 < x$

or $x > 9$

Solution set: $x > 9$

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

Expand:

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

Subtract $2x$ from both sides:

$-4 \ge 3x+5$

Subtract 5 from both sides:

$-9 \ge 3x$

Divide by 3:

$-3 \ge x$

or $x \le -3$

Solution set: $x \le -3$

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

Add $x$ to both sides:

$6x+1 > 10$

Subtract 1 from both sides:

$6x > 9$

Divide by 6:

$x > \frac{9}{6} = \frac{3}{2} = 1.5$

Solution set: $x > \frac{3}{2}$

3. Final Answer

a. $x < 2$

b. $x \le -1$

c. $x > 9$

d. $x \le -3$

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