1. **Stating the problem:** Solve the equation $$s = \frac{s(z-1)}{\frac{1}{3}(z+1)}$$ for $s$ or simplify it. 2. **Rewrite the denominator:** The denominator is $$\frac{1}{3}(z+1) = \frac{z+1}{3}$$. 3. **Rewrite the equation:** $$s = \frac{s(z-1)}{\frac{z+1}{3}} = s(z-1) \times \frac{3}{z+1}$$ 4. **Simplify the right side:** $$s = s \times \frac{3(z-1)}{z+1}$$ 5. **Isolate terms:** Subtract the right side from the left side: $$s - s \times \frac{3(z-1)}{z+1} = 0$$ 6. **Factor out $s$:** $$s \left(1 - \frac{3(z-1)}{z+1}\right) = 0$$ 7. **Simplify inside the parentheses:** $$1 - \frac{3(z-1)}{z+1} = \frac{z+1}{z+1} - \frac{3(z-1)}{z+1} = \frac{z+1 - 3(z-1)}{z+1}$$ 8. **Expand numerator:** $$z + 1 - 3z + 3 = -2z + 4$$ 9. **Rewrite:** $$s \times \frac{-2z + 4}{z+1} = 0$$ 10. **Solve for $s$ or $z$:** For the product to be zero, either $$s = 0$$ or $$\frac{-2z + 4}{z+1} = 0$$ 11. **Solve the fraction equal to zero:** $$-2z + 4 = 0 \implies 2z = 4 \implies z = 2$$ **Final answer:** $$s = 0 \quad \text{or} \quad z = 2$$