1. **State the problem:** Simplify the expression $$\frac{x-6}{5x} \cdot \frac{x^2-49}{x^2+x-42}$$. 2. **Factor all polynomials:** - Factor $x^2-49$ as a difference of squares: $$x^2-49 = (x-7)(x+7)$$. - Factor $x^2+x-42$ by finding two numbers that multiply to $-42$ and add to $1$: these are $7$ and $-6$, so $$x^2+x-42 = (x+7)(x-6)$$. 3. **Rewrite the expression with factored forms:** $$\frac{x-6}{5x} \cdot \frac{(x-7)(x+7)}{(x+7)(x-6)}$$ 4. **Cancel common factors:** - Cancel $(x-6)$ from numerator and denominator. - Cancel $(x+7)$ from numerator and denominator. Intermediate step showing cancellation: $$\frac{\cancel{x-6}}{5x} \cdot \frac{(x-7)\cancel{(x+7)}}{\cancel{(x+7)}\cancel{(x-6)}} = \frac{1}{5x} \cdot (x-7)$$ 5. **Multiply remaining terms:** $$\frac{1}{5x} \cdot (x-7) = \frac{x-7}{5x}$$ 6. **Final answer:** $$\boxed{\frac{x-7}{5x}}$$