1. **State the problem:** Solve the rational equation $$\frac{3}{x - 7} = \frac{x + 1}{3}$$ and check for extraneous solutions. 2. **Identify the domain restrictions:** The denominator $x - 7$ cannot be zero, so $x \neq 7$. 3. **Use the cross-multiplication method:** Multiply both sides by the product of denominators to eliminate fractions: $$3 \cdot 3 = (x + 1)(x - 7)$$ 4. **Write the equation:** $$9 = (x + 1)(x - 7)$$ 5. **Expand the right side:** $$9 = x^2 - 7x + x - 7 = x^2 - 6x - 7$$ 6. **Bring all terms to one side:** $$0 = x^2 - 6x - 7 - 9 = x^2 - 6x - 16$$ 7. **Solve the quadratic equation:** Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=-6$, $c=-16$: $$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 + 64}}{2} = \frac{6 \pm \sqrt{100}}{2}$$ 8. **Calculate the roots:** $$x = \frac{6 \pm 10}{2}$$ 9. **Find each solution:** - $$x = \frac{6 + 10}{2} = \frac{16}{2} = 8$$ - $$x = \frac{6 - 10}{2} = \frac{-4}{2} = -2$$ 10. **Check for extraneous solutions:** - $x=8$ is valid since $8 \neq 7$. - $x=-2$ is valid since $-2 \neq 7$. **Final answer:** $$\boxed{x = 8 \text{ or } x = -2}$$