Let's work through the questions one by one from your uploaded image.

1. Solve and graph the solution set on a number line:

$\frac{4x - 3}{6} \geq \frac{2x - 1}{12} - 2$

Step 1: Simplify both sides

Multiply both sides by 12 to eliminate denominators: $12 \cdot \frac{4x - 3}{6} \geq 12 \cdot \left( \frac{2x - 1}{12} - 2 \right)$ This simplifies to: $2(4x - 3) \geq 2x - 1 - 24$ $8x - 6 \geq 2x - 25$

Step 2: Combine like terms

Move all terms involving $x$ to one side and constant terms to the other: $8x - 2x \geq -25 + 6$ $6x \geq -19$

Step 3: Solve for $x$

$x \geq -\frac{19}{6}$

Graph:

• Draw a number line with an open circle at $-\frac{19}{6}$, shading everything to the right, as the inequality is $x \geq -\frac{19}{6}$.

Solution:

$x \geq -\frac{19}{6}$ In interval notation: $\left[ -\frac{19}{6}, \infty \right)$

2. Solve and give solution in interval notation. Graph solution on a number line.

(a) $-9 \leq 3x + 3 < 21$

Step 1: Solve for $x$

First, subtract 3 from all parts of the inequality: $-9 - 3 \leq 3x < 21 - 3$ $-12 \leq 3x < 18$

Now, divide by 3: $-4 \leq x < 6$

Solution:

In interval notation: $[-4, 6)$ Graph: Closed circle at -4, open circle at 6, shading in between.

(b) $5x - 3 \leq x + 1 \ \text{or} \ -4x \leq -8$

For $5x - 3 \leq x + 1$:

$5x - x \leq 1 + 3$ $4x \leq 4 \quad \Rightarrow \quad x \leq 1$

For $-4x \leq -8$:

Divide by -4, and reverse the inequality: $x \geq 2$

Solution:

Since this is an "or" inequality, the solution is: $(-\infty, 1] \cup [2, \infty)$ Graph: Shaded from $-\infty$ to 1 (closed circle), and from 2 to $\infty$ (closed circle).

3. Solve: $|2x - 1| = |-3x + 4|$

Step 1: Consider two cases for absolute value

Case 1: $2x - 1 = -3x + 4$

$2x + 3x = 4 + 1 \quad \Rightarrow \quad 5x = 5 \quad \Rightarrow \quad x = 1$

Case 2: $2x - 1 = 3x - 4$

$2x - 3x = -4 + 1 \quad \Rightarrow \quad -x = -3 \quad \Rightarrow \quad x = 3$

Solution:

$x = 1 \quad \text{or} \quad x = 3$

4. Solve and graph the solution set on a number line. Give solutions in interval notation.

$| -3x + 7 | < 11$

Step 1: Break the absolute value into two inequalities

$-11 < -3x + 7 < 11$

Step 2: Solve the inequality

Subtract 7 from all parts: $-11 - 7 < -3x < 11 - 7$ $-18 < -3x < 4$

Now divide by -3 (and reverse the inequalities): $\frac{-18}{-3} > x > \frac{4}{-3}$ $6 > x > -\frac{4}{3}$

Solution:

$\left( -\frac{4}{3}, 6 \right)$ Graph: Open circles at $-\frac{4}{3}$ and 6, shading in between.

Would you like further details or clarifications on any of these problems? Here are some related questions:

1. How do you graph compound inequalities on a number line?
2. What are the rules for solving absolute value equations and inequalities?
3. How can interval notation be used to represent solution sets of inequalities?
4. How do you solve inequalities involving multiple steps (like in question 1)?
5. What happens when you divide by a negative number in an inequality?

Tip: Always remember to reverse the inequality when multiplying or dividing by a negative number!