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.