1. **State the problem:** Solve the equation $$\frac{1}{x + 6} + \frac{x}{x - 6} = \frac{2}{x^2 - 36}$$ for $x$. 2. **Recall the formula and rules:** The denominator $x^2 - 36$ can be factored as a difference of squares: $$x^2 - 36 = (x + 6)(x - 6)$$ This will help us find a common denominator. 3. **Rewrite the equation with a common denominator:** The common denominator is $(x + 6)(x - 6)$. Rewrite each term: $$\frac{1}{x + 6} = \frac{x - 6}{(x + 6)(x - 6)}$$ $$\frac{x}{x - 6} = \frac{x(x + 6)}{(x - 6)(x + 6)}$$ The right side is already over $(x + 6)(x - 6)$. 4. **Combine the left side:** $$\frac{x - 6}{(x + 6)(x - 6)} + \frac{x(x + 6)}{(x + 6)(x - 6)} = \frac{(x - 6) + x(x + 6)}{(x + 6)(x - 6)}$$ 5. **Simplify the numerator:** $$ (x - 6) + x(x + 6) = x - 6 + x^2 + 6x = x^2 + 7x - 6 $$ 6. **Rewrite the equation:** $$ \frac{x^2 + 7x - 6}{(x + 6)(x - 6)} = \frac{2}{(x + 6)(x - 6)} $$ 7. **Multiply both sides by the common denominator to clear fractions:** $$ \cancel{(x + 6)(x - 6)} \cdot \frac{x^2 + 7x - 6}{\cancel{(x + 6)(x - 6)}} = \cancel{(x + 6)(x - 6)} \cdot \frac{2}{\cancel{(x + 6)(x - 6)}} $$ which simplifies to $$ x^2 + 7x - 6 = 2 $$ 8. **Bring all terms to one side:** $$ x^2 + 7x - 6 - 2 = 0 $$ $$ x^2 + 7x - 8 = 0 $$ 9. **Factor the quadratic:** We look for two numbers that multiply to $-8$ and add to $7$. These are $8$ and $-1$. $$ (x + 8)(x - 1) = 0 $$ 10. **Solve for $x$:** $$ x + 8 = 0 \Rightarrow x = -8 $$ $$ x - 1 = 0 \Rightarrow x = 1 $$ 11. **Check for restrictions:** The original denominators cannot be zero: $$ x + 6 \neq 0 \Rightarrow x \neq -6 $$ $$ x - 6 \neq 0 \Rightarrow x \neq 6 $$ Neither $-8$ nor $1$ are restricted values. **Final answer:** $$ \boxed{x = -8 \text{ or } x = 1} $$