To solve the equation
\[ \frac{6}{x + 5} + \frac{3}{x - 5} = \frac{30}{x^2 - 25}, \]
we first note that the denominator \(x^2 - 25\) can be factored:
\[ x^2 - 25 = (x + 5)(x - 5). \]
Thus, we can rewrite the equation as follows:
\[ \frac{6}{x + 5} + \frac{3}{x - 5} = \frac{30}{(x + 5)(x - 5)}. \]
To eliminate the fractions, we will multiply through by \((x + 5)(x - 5)\):
\[ 6(x - 5) + 3(x + 5) = 30. \]
Now we distribute on the left-hand side:
\[ 6x - 30 + 3x + 15 = 30. \]
Combining the like terms gives us:
\[ 9x - 15 = 30. \]
Next, we will add 15 to both sides:
\[ 9x = 45. \]
Now, divide both sides by 9:
\[ x = 5. \]
Now we need to check whether \(x = 5\) is a valid solution in the original equation. We substitute \(x = 5\) back into the original equation:
The terms \(\frac{6}{x + 5}\) and \(\frac{3}{x - 5}\) would become:
\[ \frac{6}{5 + 5} + \frac{3}{5 - 5} \rightarrow \frac{6}{10} + \frac{3}{0}. \]
The second term \(\frac{3}{0}\) is undefined, implying that \(x = 5\) is not a valid solution.
Since \(x = 5\) leads to an undefined expression, let's check if there are any other potential values. We see that the original equation involves fractions that have denominators either \(x + 5\) or \(x - 5\), which cannot be zero. This means:
- \(x + 5 \neq 0 \implies x \neq -5\)
- \(x - 5 \neq 0 \implies x \neq 5\)
Thus, the only solution we calculated, \(x = 5\), is not in the domain of the original equation.
Therefore, the final conclusion is:
\[ \text{No solution.} \]