1. **State the problem:** Solve the equation \( \frac{x + 7}{x^2 + 3x - 10} - \frac{7}{x^2 + 10x + 25} = \frac{x - 7}{x^2 + 3x - 10} \). 2. **Factor the denominators:** \[ x^2 + 3x - 10 = (x + 5)(x - 2) \] \[ x^2 + 10x + 25 = (x + 5)^2 \] 3. **Rewrite the equation with factored denominators:** \[ \frac{x + 7}{(x + 5)(x - 2)} - \frac{7}{(x + 5)^2} = \frac{x - 7}{(x + 5)(x - 2)} \] 4. **Bring all terms to one side:** \[ \frac{x + 7}{(x + 5)(x - 2)} - \frac{x - 7}{(x + 5)(x - 2)} = \frac{7}{(x + 5)^2} \] 5. **Combine the left side fractions since they have the same denominator:** \[ \frac{(x + 7) - (x - 7)}{(x + 5)(x - 2)} = \frac{7}{(x + 5)^2} \] 6. **Simplify numerator:** \[ (x + 7) - (x - 7) = x + 7 - x + 7 = 14 \] 7. **So the equation becomes:** \[ \frac{14}{(x + 5)(x - 2)} = \frac{7}{(x + 5)^2} \] 8. **Cross multiply:** \[ 14 (x + 5)^2 = 7 (x + 5)(x - 2) \] 9. **Divide both sides by 7:** \[ \cancel{7} \times 2 (x + 5)^2 = \cancel{7} (x + 5)(x - 2) \] \[ 2 (x + 5)^2 = (x + 5)(x - 2) \] 10. **Divide both sides by \(x + 5\) (not zero):** \[ \cancel{(x + 5)} 2 (x + 5) = \cancel{(x + 5)} (x - 2) \] \[ 2 (x + 5) = x - 2 \] 11. **Expand left side:** \[ 2x + 10 = x - 2 \] 12. **Bring all terms to one side:** \[ 2x + 10 - x + 2 = 0 \] \[ x + 12 = 0 \] 13. **Solve for \(x\):** \[ x = -12 \] 14. **Check for restrictions:** Denominators cannot be zero: \[ x \neq -5, x \neq 2 \] 15. **Final answer:** \[ \boxed{x = -12} \]