https://file.aiquickdraw.com/mathbot/file/172843824549848wijhwo.jpg

The problem provided in the image is an inequality that needs to be solved and expressed in interval notation:

\[
16 + 8x \geq 3(13x + 12)
\]

### Step-by-Step Solution:
1. **Expand the Right-Hand Side:**
\[
16 + 8x \geq 3 \times 13x + 3 \times 12
\]
\[
16 + 8x \geq 39x + 36
\]

2. **Move All \(x\)-terms to the Left and Constants to the Right:**
Subtract \( 8x \) from both sides:
\[
16 \geq 39x - 8x + 36
\]
\[
16 \geq 31x + 36
\]

Next, subtract 36 from both sides:
\[
16 - 36 \geq 31x
\]
\[
-20 \geq 31x
\]

3. **Solve for \(x\):**
Divide both sides by 31:
\[
\frac{-20}{31} \geq x \quad \Rightarrow \quad x \leq \frac{-20}{31}
\]

### Interval Notation:
The solution in interval notation is:
\[
(-\infty, \frac{-20}{31}]
\]

Let me know if you need further details or have any questions.

### Related Questions:
1. How do you solve inequalities involving distribution and combining like terms?
2. What changes in the solution if the inequality sign is reversed?
3. How do you express the solution on a number line?
4. What are the rules for multiplying or dividing inequalities by negative numbers?
5. How would you solve a compound inequality like \( 2 < 3x - 5 \leq 10 \)?

**Tip:** Always remember to reverse the inequality sign when multiplying or dividing both sides by a negative number.

Solve the inequality and write your answer in interval notation: 16 + 8x ≥ 3(13x + 12)

Learn how to solve the inequality 16 + 8x ≥ 3(13x + 12) with a detailed step-by-step solution, covering key algebraic concepts such as distribution and inequality solving rules. Suitable for students in Grades 9-11.

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