Let's solve both inequalities step-by-step.

### Inequality 1: \( 45 > 4x + 9 \)
1. Subtract 9 from both sides:
   \[
   45 - 9 > 4x
   \]
   \[
   36 > 4x
   \]
2. Divide both sides by 4:
   \[
   \frac{36}{4} > x
   \]
   \[
   9 > x \quad \text{or} \quad x < 9
   \]
   
### Inequality 2: \( 53 \leq 4x + 9 \)
1. Subtract 9 from both sides:
   \[
   53 - 9 \leq 4x
   \]
   \[
   44 \leq 4x
   \]
2. Divide both sides by 4:
   \[
   \frac{44}{4} \leq x
   \]
   \[
   11 \leq x \quad \text{or} \quad x \geq 11
   \]

### Final Solution:
From the first inequality, \( x < 9 \), and from the second inequality, \( x \geq 11 \). There is **no value of \( x \)** that can satisfy both conditions at the same time. Therefore, **no solution** exists that satisfies both inequalities simultaneously.

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Do you want further details or have any questions about this? 

Here are five related questions to extend your understanding:
1. How do we solve compound inequalities with "and" or "or" conditions?
2. What happens when there is no solution for a system of inequalities?
3. How do we graph inequalities on a number line?
4. What are the steps to solve quadratic inequalities?
5. Can we solve inequalities involving absolute values?

**Tip:** When solving inequalities, always reverse the inequality sign when multiplying or dividing by a negative number.

Discover how to solve compound linear inequalities such as 45 > 4x + 9 and 53 ≤ 4x + 9 with detailed steps. Learn the key concepts of inequality solving, including properties of inequalities and operations. Suitable for students in Grades 7-9.

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