Question 1209658
Here's how to find the largest possible value of x + y:

1. **Rewrite the equation:**

The given equation represents a circle. We can rewrite it in standard form by completing the square for both x and y:

x² - 8x + y² + 6y + 23 = 0
(x² - 8x + 16) + (y² + 6y + 9) + 23 - 16 - 9 = 0
(x - 4)² + (y + 3)² = 2

2. **Geometric Interpretation:**

This equation represents a circle with center (4, -3) and radius √2.

3. **Express y in terms of x:**

We want to maximize x + y. Let's express y in terms of x using the equation of the circle:

(y + 3)² = 2 - (x - 4)²
y + 3 = ±√[2 - (x - 4)²]
y = -3 ± √[2 - (x - 4)²]

4. **Maximize x + y:**

We want to maximize the function f(x) = x + y. Substituting the expression for y, we get:

f(x) = x - 3 ± √[2 - (x - 4)²]

To find the maximum value of f(x), we can consider the line x + y = k, where k is a constant. We want to find the largest value of k such that the line intersects the circle.  Geometrically, this line will be tangent to the circle at the point that maximizes x + y.

The line x + y = k can be written as y = -x + k. The slope of this line is -1.

5. **Tangent Line:**

The line connecting the center of the circle (4, -3) to the point of tangency will be perpendicular to the tangent line x + y = k. Thus, the slope of this line is 1. Let (x,y) be the point of tangency.
(y - (-3))/(x - 4) = 1
y + 3 = x - 4
y = x - 7

Substitute into the equation for the circle:
(x-4)^2 + (x - 7 + 3)^2 = 2
(x-4)^2 + (x-4)^2 = 2
2(x-4)^2 = 2
(x-4)^2 = 1
x - 4 = ±1
x = 5 or x = 3

If x = 5, y = 5 - 7 = -2, so x + y = 3
If x = 3, y = 3 - 7 = -4, so x + y = -1

The maximum value of x+y is 3.

**Alternatively:**
The maximum value of x + y occurs when the line x + y = k is tangent to the circle. The distance from the center of the circle to the line must be equal to the radius.

|4 + (-3) - k| / sqrt(1^2 + 1^2) = sqrt(2)
|1 - k| / sqrt(2) = sqrt(2)
|1 - k| = 2
1 - k = 2 or 1 - k = -2
k = -1 or k = 3.

The largest value is 3.

Final Answer: The final answer is $\boxed{3}$