To solve the equation
\[ \frac{x + 7}{x^2 - 4x + 3} - \frac{8}{x^2 - 7x + 6} = \frac{x - 7}{x^2 - 9x + 18}, \]
we will first factor the denominators.
- Factor \(x^2 - 4x + 3\):
\[ x^2 - 4x + 3 = (x - 3)(x - 1). \]
- Factor \(x^2 - 7x + 6\):
\[ x^2 - 7x + 6 = (x - 6)(x - 1). \]
- Factor \(x^2 - 9x + 18\):
\[ x^2 - 9x + 18 = (x - 6)(x - 3). \]
Now, substitute these factorizations back into the equation:
\[ \frac{x + 7}{(x - 3)(x - 1)} - \frac{8}{(x - 6)(x - 1)} = \frac{x - 7}{(x - 6)(x - 3)}. \]
Next, we'll find a common denominator for the left-hand side of the equation. The common denominator is \((x - 3)(x - 1)(x - 6)\):
\[ \frac{(x + 7)(x - 6)}{(x - 3)(x - 1)(x - 6)} - \frac{8(x - 3)}{(x - 6)(x - 1)(x - 3)}. \]
Now rewrite the left-hand side:
\[ \frac{(x + 7)(x - 6) - 8(x - 3)}{(x - 3)(x - 1)(x - 6)}. \]
Expand the numerators:
- Expand \((x + 7)(x - 6)\):
\[ (x + 7)(x - 6) = x^2 - 6x + 7x - 42 = x^2 + x - 42. \]
- Expand \(8(x - 3)\):
\[ 8(x - 3) = 8x - 24. \]
Now combine everything in the numerator:
\[ x^2 + x - 42 - (8x - 24) = x^2 + x - 42 - 8x + 24 = x^2 - 7x - 18. \]
Therefore, the equation becomes:
\[ \frac{x^2 - 7x - 18}{(x - 3)(x - 1)(x - 6)} = \frac{x - 7}{(x - 6)(x - 3)}. \]
Cross-multiply:
\[ (1)(x^2 - 7x - 18) = (x - 7)(x - 1). \]
Expand \( (x - 7)(x - 1) \):
\[ x^2 - x - 7x + 7 = x^2 - 8x + 7. \]
Now set the equation:
\[ x^2 - 7x - 18 = x^2 - 8x + 7. \]
Subtract \( x^2 \) from both sides:
\[ -7x - 18 = -8x + 7. \]
Rearranging gives:
\[ -7x + 8x = 7 + 18 \implies x = 25. \]
Now we need to check for \( x = 25 \) in the original equation and ensure it does not result in any zeroes in the denominators.
Check the validity of \( x = 25 \):
Denominators:
- \(x^2 - 4x + 3 = 25^2 - 4(25) + 3 = 625 - 100 + 3 = 528 \neq 0\).
- \(x^2 - 7x + 6 = 25^2 - 7(25) + 6 = 625 - 175 + 6 = 456 \neq 0\).
- \(x^2 - 9x + 18 = 25^2 - 9(25) + 18 = 625 - 225 + 18 = 418 \neq 0\).
Since none of the denominators are zero, the solution is valid:
\[ \boxed{25}. \]