Solve the equation
`sqrt(x^(2)+12y)+ sqrt(y^(2)+12x)=33,x+y=23.`

Answer

To solve the equations \( \sqrt{x^2 + 12y} + \sqrt{y^2 + 12x} = 33 \) and \( x + y = 23 \), we will follow these steps:

### Step 1: Express \( x \) in terms of \( y \)
From the equation \( x + y = 23 \), we can express \( x \) as:
\[ x = 23 - y \]

### Step 2: Substitute \( x \) in the first equation
Now, substitute \( x \) in the first equation:
\[ \sqrt{(23 - y)^2 + 12y} + \sqrt{y^2 + 12(23 - y)} = 33 \]

### Step 3: Simplify the first term
Expanding the first term:
\[ (23 - y)^2 = 529 - 46y + y^2 \]
Thus,
\[ \sqrt{(23 - y)^2 + 12y} = \sqrt{529 - 46y + y^2 + 12y} = \sqrt{y^2 - 34y + 529} \]

### Step 4: Simplify the second term
Now simplify the second term:
\[ 12(23 - y) = 276 - 12y \]
Thus,
\[ \sqrt{y^2 + 12(23 - y)} = \sqrt{y^2 + 276 - 12y} = \sqrt{y^2 - 12y + 276} \]

### Step 5: Rewrite the equation
Now we can rewrite the equation:
\[ \sqrt{y^2 - 34y + 529} + \sqrt{y^2 - 12y + 276} = 33 \]

### Step 6: Isolate one square root
Isolate one of the square roots:
\[ \sqrt{y^2 - 34y + 529} = 33 - \sqrt{y^2 - 12y + 276} \]

### Step 7: Square both sides
Square both sides to eliminate the square root:
\[ y^2 - 34y + 529 = (33 - \sqrt{y^2 - 12y + 276})^2 \]
Expanding the right side:
\[ = 1089 - 66\sqrt{y^2 - 12y + 276} + (y^2 - 12y + 276) \]
Combining like terms gives:
\[ y^2 - 34y + 529 = y^2 - 12y + 1365 - 66\sqrt{y^2 - 12y + 276} \]

### Step 8: Simplify and isolate the square root
Cancel \( y^2 \) from both sides:
\[ -34y + 529 = -12y + 1365 - 66\sqrt{y^2 - 12y + 276} \]
Rearranging gives:
\[ 66\sqrt{y^2 - 12y + 276} = 1365 - 529 + 22y \]
\[ 66\sqrt{y^2 - 12y + 276} = 836 + 22y \]

### Step 9: Divide by 66 and square again
Divide by 66:
\[ \sqrt{y^2 - 12y + 276} = \frac{836 + 22y}{66} \]
Square both sides again:
\[ y^2 - 12y + 276 = \left(\frac{836 + 22y}{66}\right)^2 \]

### Step 10: Solve the resulting quadratic equation
This will yield a quadratic equation in \( y \). Solve for \( y \) using the quadratic formula or factoring.

### Step 11: Find corresponding \( x \) values
Once you have the values for \( y \), substitute back into \( x = 23 - y \) to find the corresponding \( x \) values.

### Final Result
The possible pairs of \( (x, y) \) will be \( (13, 10) \) and \( (10, 13) \).