1. **State the problem:** Find the excluded values of the expression $$\frac{6x^2y^3}{2x^2y^2} \times \frac{10x^3y^4}{18y^2}$$. 2. **Identify excluded values:** Excluded values occur where any denominator is zero because division by zero is undefined. 3. **Analyze denominators:** The denominators are $2x^2y^2$ and $18y^2$. 4. **Set denominators not equal to zero:** $$2x^2y^2 \neq 0 \implies x^2 \neq 0 \text{ and } y^2 \neq 0 \implies x \neq 0, y \neq 0$$ $$18y^2 \neq 0 \implies y^2 \neq 0 \implies y \neq 0$$ 5. **Combine conditions:** The excluded values are where $x=0$ or $y=0$. 6. **Simplify the expression:** $$\frac{6x^2y^3}{2x^2y^2} \times \frac{10x^3y^4}{18y^2} = \left(\frac{6}{2} \times \frac{x^2}{x^2} \times \frac{y^3}{y^2}\right) \times \left(\frac{10}{18} \times x^3 \times \frac{y^4}{y^2}\right)$$ 7. **Cancel common factors:** $$\frac{6}{2} = 3, \quad \frac{x^2}{x^2} = \cancel{1}, \quad \frac{y^3}{y^2} = y$$ $$\frac{10}{18} = \frac{5}{9}, \quad y^{4-2} = y^2$$ 8. **Rewrite:** $$3 \times y \times \frac{5}{9} \times x^3 \times y^2 = \frac{15}{9} x^3 y^{3}$$ 9. **Simplify fraction:** $$\frac{15}{9} = \frac{\cancel{15}}{\cancel{9}} = \frac{5}{3}$$ 10. **Final simplified expression:** $$\frac{5}{3} x^3 y^3$$ **Answer:** The excluded values are $x=0$ and $y=0$. The simplified expression is $$\frac{5}{3} x^3 y^3$$.