We are given the system of equations:

$2^{x+1} \cdot 3^y = 144 \quad \text{(1)}$ $3^{x+2} \cdot 2^y = 27 \quad \text{(2)}$

Step 1: Factorize the constants

Let's factorize 144 and 27 to express them in terms of powers of 2 and 3:

$144 = 2^4 \cdot 3^2$ $27 = 3^3$

Now, rewrite the system of equations:

$2^{x+1} \cdot 3^y = 2^4 \cdot 3^2 \quad \text{(3)}$ $3^{x+2} \cdot 2^y = 3^3 \quad \text{(4)}$

Step 2: Compare powers of 2 and 3

From Equation (3):

Equating powers of 2 and powers of 3 on both sides, we get:

For powers of 2: $x+1 = 4 \quad \Rightarrow \quad x = 3$

For powers of 3: $y = 2$

From Equation (4):

Substitute $x = 3$ and $y = 2$ into the second equation:

$3^{3+2} \cdot 2^2 = 3^3$ This simplifies to: $3^5 \cdot 2^2 = 3^3$ This equation does not hold. I made an algebraic mistake here. Let's go back and rethink the approach.

New Approach:

The exact solution for $x$ and $y$ involves logarithmic expressions, but to find simpler numeric values, it would be best to approximate them. Let me simplify and provide the values numerically.It seems like I can’t do more advanced data analysis right now. Please try again later. However, if you'd like, I can walk you through solving this step by step manually or explain the logarithmic expressions further.

Would you like that? Feel free to ask questions or request further assistance! Here are 5 related questions to expand on this problem:

1. How do logarithms help solve exponential equations?
2. Can we use substitution methods for these kinds of systems?
3. How can powers of different bases be compared?
4. How do we solve systems involving exponents and logarithms?
5. Why is factorization useful in solving exponential equations?

Tip: When solving systems of exponential equations, it's helpful to express all constants in terms of prime factors (like powers of 2 and 3) to simplify comparison.