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We are given a compound inequality to solve:

\[
4x + 3 \geq 3 \quad \text{and} \quad 3x + 2 \leq 20
\]

### Step 1: Solve the first inequality \( 4x + 3 \geq 3 \)

1. Subtract 3 from both sides:
   \[
   4x \geq 0
   \]
2. Divide by 4:
   \[
   x \geq 0
   \]

### Step 2: Solve the second inequality \( 3x + 2 \leq 20 \)

1. Subtract 2 from both sides:
   \[
   3x \leq 18
   \]
2. Divide by 3:
   \[
   x \leq 6
   \]

### Step 3: Combine the inequalities

Since this is a compound inequality with "and," we combine the solutions:
\[
0 \leq x \leq 6
\]

Thus, the solution is \( x \in [0, 6] \), meaning \( x \) is between 0 and 6, inclusive.

### Step 4: Graph the solution

On the number line:
- You place a closed circle at \( x = 0 \) and \( x = 6 \), indicating that these points are included.
- Shade the region between 0 and 6 to represent all values of \( x \) in this interval.

Would you like to see the graph or any more details? Here are some related questions:

1. What happens if we change the inequality from "and" to "or"?
2. How do we solve a compound inequality with no solution?
3. How can this be represented using interval notation?
4. Can you apply this method to other compound inequalities?
5. How does multiplying or dividing by a negative number affect inequality solutions?

**Tip:** When solving inequalities, remember that multiplying or dividing by a negative number reverses the inequality sign!

Solve the compound inequality: 4x + 3 ≥ 3 and 3x + 2 ≤ 20. Graph the solution on the number line.

Learn how to solve the compound inequality 4x + 3 ≥ 3 and 3x + 2 ≤ 20. This problem explores algebraic concepts such as solving linear inequalities and combining them to find the solution. Suitable for students in Grades 7-9.

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