Question: Solve the following equation for $x$.
$$\frac{2}{x+3} + \frac{2}{x-4} = \frac{2}{x^2 - x - 12}$$
Answer in simplified fraction form.
1. **State the problem:** Solve the equation $$\frac{2}{x+3} + \frac{2}{x-4} = \frac{2}{x^2 - x - 12}$$ for $x$.
2. **Factor the denominator on the right side:** Note that $$x^2 - x - 12 = (x+3)(x-4)$$.
3. **Rewrite the equation using the factorization:**
$$\frac{2}{x+3} + \frac{2}{x-4} = \frac{2}{(x+3)(x-4)}$$
4. **Find a common denominator for the left side:** The common denominator is $(x+3)(x-4)$.
5. **Rewrite each fraction on the left with the common denominator:**
$$\frac{2(x-4)}{(x+3)(x-4)} + \frac{2(x+3)}{(x-4)(x+3)} = \frac{2}{(x+3)(x-4)}$$
6. **Combine the left side:**
$$\frac{2(x-4) + 2(x+3)}{(x+3)(x-4)} = \frac{2}{(x+3)(x-4)}$$
7. **Simplify the numerator on the left:**
$$2(x-4) + 2(x+3) = 2x - 8 + 2x + 6 = 4x - 2$$
8. **Rewrite the equation:**
$$\frac{4x - 2}{(x+3)(x-4)} = \frac{2}{(x+3)(x-4)}$$
9. **Multiply both sides by $(x+3)(x-4)$ to clear denominators:**
$$\cancel{(x+3)(x-4)} \cdot \frac{4x - 2}{\cancel{(x+3)(x-4)}} = \cancel{(x+3)(x-4)} \cdot \frac{2}{\cancel{(x+3)(x-4)}}$$
This simplifies to:
$$4x - 2 = 2$$
10. **Solve for $x$:**
$$4x - 2 = 2$$
$$4x = 2 + 2$$
$$4x = 4$$
$$x = \frac{4}{4} = 1$$
11. **Check for restrictions:** The denominators cannot be zero, so $x \neq -3$ and $x \neq 4$. Since $x=1$ does not violate these, it is valid.
**Final answer:**
$$x = 1$$