The problem asks to find the truth set for each of the given 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 for each of the given 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, $x < 2$

Number line: Draw a number line and place an open circle at

2. Shade the line to the left of

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: $x \le -1$

Number line: Draw a number line and place a closed circle at -

1. Shade the line to the left of -

1.

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

Distribute 4: $4x + 8 < 5x - 1$

Subtract $4x$ from both sides: $8 < x - 1$

Add 1 to both sides: $9 < x$

So, $x > 9$

Number line: Draw a number line and place an open circle at

9. Shade the line to the right of

9.

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

Distribute 2 and 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, $x \le -3$

Number line: Draw a number line and place a closed circle at -

3. Shade the line to the left of -

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

So, $x > \frac{3}{2} = 1.5$

Number line: Draw a number line and place an open circle at 1.

5. Shade the line to the right of 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 $x > 1.5$

(Number lines as described above)