If `x^(2)-6x-27=0 and y^(2)-6y-40=0`, what is the maximum value of `x+y`?

Answer

To solve the problem, we need to find the maximum value of \( x + y \) given the equations \( x^2 - 6x - 27 = 0 \) and \( y^2 - 6y - 40 = 0 \).

### Step 1: Solve the equation for \( x \)
The first equation is:
\[ x^2 - 6x - 27 = 0 \]
We can factor this equation. To do this, we need to find two numbers that multiply to \(-27\) and add to \(-6\). The numbers are \(-9\) and \(3\).

Thus, we can rewrite the equation as:
\[ (x - 9)(x + 3) = 0 \]

Setting each factor to zero gives us:
1. \( x - 9 = 0 \) → \( x = 9 \)
2. \( x + 3 = 0 \) → \( x = -3 \)

So, the possible values for \( x \) are \( 9 \) and \( -3 \).

### Step 2: Solve the equation for \( y \)
The second equation is:
\[ y^2 - 6y - 40 = 0 \]
We can factor this equation as well. We need two numbers that multiply to \(-40\) and add to \(-6\). The numbers are \(-10\) and \(4\).

Thus, we can rewrite the equation as:
\[ (y - 10)(y + 4) = 0 \]

Setting each factor to zero gives us:
1. \( y - 10 = 0 \) → \( y = 10 \)
2. \( y + 4 = 0 \) → \( y = -4 \)

So, the possible values for \( y \) are \( 10 \) and \( -4 \).

### Step 3: Find the maximum value of \( x + y \)
Now, we need to find the maximum value of \( x + y \). We have the following combinations of \( x \) and \( y \):
1. \( x = 9 \) and \( y = 10 \) → \( x + y = 9 + 10 = 19 \)
2. \( x = 9 \) and \( y = -4 \) → \( x + y = 9 - 4 = 5 \)
3. \( x = -3 \) and \( y = 10 \) → \( x + y = -3 + 10 = 7 \)
4. \( x = -3 \) and \( y = -4 \) → \( x + y = -3 - 4 = -7 \)

The maximum value of \( x + y \) from these combinations is:
\[ \text{Maximum value} = 19 \]

### Final Answer
Thus, the maximum value of \( x + y \) is \( 19 \).

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