\[ \frac{6}{x-4}=\frac{-3}{x+5}+\frac{27}{(x-4)(x+5)} \] Select the correct choice below and fill in any answer boxes in your choice. A. The solution set is \( \{\square \). (Simplify your answer.) B. There is no solution.

To solve the equation \[ \frac{6}{x-4}=\frac{-3}{x+5}+\frac{27}{(x-4)(x+5)}, \] first, we can combine the right side over a common denominator: \[ \frac{-3}{x+5}+\frac{27}{(x-4)(x+5)} = \frac{-3(x-4) + 27}{(x-4)(x+5)} = \frac{-3x + 12 + 27}{(x-4)(x+5)} = \frac{-3x + 39}{(x-4)(x+5)}. \] Now, we have \[ \frac{6}{x-4} = \frac{-3x + 39}{(x-4)(x+5)}. \] Next, multiply both sides by \((x-4)(x+5)\) to eliminate the denominators: \[ 6(x+5) = -3x + 39. \] Now, distribute on the left side: \[ 6x + 30 = -3x + 39. \] Adding \(3x\) to both sides gives: \[ 9x + 30 = 39. \] Subtracting \(30\) from both sides results in: \[ 9x = 9. \] Dividing by \(9\) yields: \[ x = 1. \] Lastly, we should check that \(x = 1\) doesn't make any original denominators zero. Since neither \(x-4\) nor \(x+5\) equals zero at \(x = 1\), we confirm the solution. A. The solution set is \( \{1\} \).