1. **State the problem:** Solve the rational equation $$\frac{4}{3x+3} = \frac{12}{x^2-1}$$. 2. **Identify the denominators and restrictions:** The denominators are $3x+3$ and $x^2-1$. - Note that $3x+3 = 3(x+1)$, so $x \neq -1$ to avoid division by zero. - Also, $x^2-1 = (x-1)(x+1)$, so $x \neq 1$ and $x \neq -1$. 3. **Rewrite the equation with factored denominators:** $$\frac{4}{3(x+1)} = \frac{12}{(x-1)(x+1)}$$ 4. **Cross-multiply to eliminate denominators:** $$4 \cdot (x-1)(x+1) = 12 \cdot 3(x+1)$$ 5. **Simplify both sides:** Left side: $$4(x^2 - 1) = 4x^2 - 4$$ Right side: $$36(x+1) = 36x + 36$$ 6. **Set up the equation:** $$4x^2 - 4 = 36x + 36$$ 7. **Bring all terms to one side:** $$4x^2 - 4 - 36x - 36 = 0$$ Simplify: $$4x^2 - 36x - 40 = 0$$ 8. **Divide the entire equation by 4 to simplify:** $$\cancel{4}x^2 - \cancel{36}x - \cancel{40} = 0 \Rightarrow x^2 - 9x - 10 = 0$$ 9. **Factor the quadratic:** $$x^2 - 9x - 10 = (x - 10)(x + 1) = 0$$ 10. **Solve for $x$:** $$x - 10 = 0 \Rightarrow x = 10$$ $$x + 1 = 0 \Rightarrow x = -1$$ 11. **Check for restrictions:** Recall $x \neq -1$ because it makes denominators zero. 12. **Final solution:** $$\boxed{x = 10}$$