We are given the following system of equations involving $x$ and $y$:

$3x + 2y = 8$

$(3x + 6)^2 + (2y - 2)^2 = 144$

We need to find $m + n$, where $m$ is the sum of all possible values of $x$ and $n$ is the sum of all possible values of $y$.

Step 1: Solve the first equation for one variable.

We can solve the first equation for $y$:

$3x + 2y = 8 \quad \Rightarrow \quad 2y = 8 - 3x \quad \Rightarrow \quad y = \frac{8 - 3x}{2}$

Step 2: Substitute into the second equation.

Now substitute $y = \frac{8 - 3x}{2}$ into the second equation:

$(3x + 6)^2 + (2y - 2)^2 = 144$

First, let's express $2y - 2$ in terms of $x$:

$2y = 8 - 3x \quad \Rightarrow \quad 2y - 2 = 8 - 3x - 2 = 6 - 3x$

Now the second equation becomes:

$(3x + 6)^2 + (6 - 3x)^2 = 144$

Step 3: Simplify the equation.

Expanding both squares:

$(3x + 6)^2 = 9x^2 + 36x + 36$ $(6 - 3x)^2 = 9x^2 - 36x + 36$

Now substitute these into the equation:

$9x^2 + 36x + 36 + 9x^2 - 36x + 36 = 144$

Simplifying:

$18x^2 + 72 = 144 \quad \Rightarrow \quad 18x^2 = 72 \quad \Rightarrow \quad x^2 = 4$

Thus:

$x = 2 \quad \text{or} \quad x = -2$

Step 4: Find corresponding values of $y$.

For $x = 2$:

$y = \frac{8 - 3(2)}{2} = \frac{8 - 6}{2} = 1$

For $x = -2$:

$y = \frac{8 - 3(-2)}{2} = \frac{8 + 6}{2} = 7$

Step 5: Calculate $m + n$.

The possible values of $x$ are $2$ and $-2$, so $m = 2 + (-2) = 0$.

The possible values of $y$ are $1$ and $7$, so $n = 1 + 7 = 8$.

Thus, $m + n = 0 + 8 = 8$.

Final Answer:

$m + n = 8$

Would you like further clarification or details on any step?

Here are 5 related questions you might explore:

1. How would the problem change if the second equation were different?
2. Can this system of equations be solved graphically? If so, how?
3. What happens if we vary the coefficients of $x$ and $y$ in the equations?
4. Are there other methods to solve this type of system, such as matrix methods?
5. What if the equations involved three variables instead of two?

Tip: Always check if quadratic equations lead to multiple solutions, as the structure of the problem might depend on more than one set of values for $x$ and $y$.