1. **State the problem:** Simplify and solve the equation $$\frac{3x}{x^2 + 7x + 10} - \frac{x - 6}{x + 5} = - \frac{x^2 - 7x - 12}{(x + 2)(x + 5)}$$ 2. **Factor denominators and numerators where possible:** - Factor the quadratic in the first denominator: $$x^2 + 7x + 10 = (x + 5)(x + 2)$$ - Factor the numerator on the right side: $$x^2 - 7x - 12 = (x - 4)(x + 3)$$ 3. **Rewrite the equation with factored forms:** $$\frac{3x}{(x + 5)(x + 2)} - \frac{x - 6}{x + 5} = - \frac{(x - 4)(x + 3)}{(x + 2)(x + 5)}$$ 4. **Find a common denominator for the left side:** The common denominator is $(x + 5)(x + 2)$. Rewrite the second fraction on the left: $$\frac{x - 6}{x + 5} = \frac{(x - 6)(x + 2)}{(x + 5)(x + 2)}$$ 5. **Rewrite the left side as a single fraction:** $$\frac{3x - (x - 6)(x + 2)}{(x + 5)(x + 2)} = - \frac{(x - 4)(x + 3)}{(x + 5)(x + 2)}$$ 6. **Multiply both sides by the common denominator $(x + 5)(x + 2)$ to eliminate denominators:** $$\cancel{(x + 5)(x + 2)} \cdot \frac{3x - (x - 6)(x + 2)}{\cancel{(x + 5)(x + 2)}} = - (x - 4)(x + 3)$$ Simplifies to: $$3x - (x - 6)(x + 2) = - (x - 4)(x + 3)$$ 7. **Expand the products:** - Expand $(x - 6)(x + 2)$: $$x^2 + 2x - 6x - 12 = x^2 - 4x - 12$$ - Expand $(x - 4)(x + 3)$: $$x^2 + 3x - 4x - 12 = x^2 - x - 12$$ 8. **Substitute expansions back into the equation:** $$3x - (x^2 - 4x - 12) = - (x^2 - x - 12)$$ 9. **Distribute the minus signs:** $$3x - x^2 + 4x + 12 = - x^2 + x + 12$$ 10. **Combine like terms on the left:** $$- x^2 + 7x + 12 = - x^2 + x + 12$$ 11. **Add $x^2$ to both sides to cancel $-x^2$ terms:** $$\cancel{- x^2} + 7x + 12 + x^2 = \cancel{- x^2} + x + 12 + x^2$$ Simplifies to: $$7x + 12 = x + 12$$ 12. **Subtract 12 from both sides:** $$7x + \cancel{12} - \cancel{12} = x + \cancel{12} - \cancel{12}$$ Simplifies to: $$7x = x$$ 13. **Subtract $x$ from both sides:** $$7x - x = x - x$$ Simplifies to: $$6x = 0$$ 14. **Divide both sides by 6:** $$\frac{6x}{\cancel{6}} = \frac{0}{\cancel{6}}$$ Simplifies to: $$x = 0$$ 15. **Check for restrictions:** The denominators cannot be zero: - $x + 5 \neq 0 \Rightarrow x \neq -5$ - $x + 2 \neq 0 \Rightarrow x \neq -2$ Since $x=0$ does not violate these, it is a valid solution. **Final answer:** $$\boxed{x = 0}$$