1. **State the problem:** Simplify the expression $$\frac{9y^2-4}{4y-12}x \cdot \frac{18-6y}{9y^2+12y+4}$$. 2. **Factor all polynomials:** - Numerator of first fraction: $$9y^2-4 = (3y-2)(3y+2)$$ (difference of squares). - Denominator of first fraction: $$4y-12 = 4(y-3)$$. - Numerator of second fraction: $$18-6y = 6(3-y)$$. - Denominator of second fraction: $$9y^2+12y+4 = (3y+2)^2$$ (perfect square trinomial). 3. **Rewrite the expression with factored forms:** $$\frac{(3y-2)(3y+2)}{4(y-3)} x \cdot \frac{6(3-y)}{(3y+2)^2}$$ 4. **Simplify the factor $(3-y)$:** Note that $$3-y = -(y-3)$$. 5. **Substitute and combine:** $$\frac{(3y-2)(3y+2)}{4(y-3)} x \cdot \frac{6 \cdot -(y-3)}{(3y+2)^2} = \frac{(3y-2)(3y+2)}{4(y-3)} x \cdot \frac{-6(y-3)}{(3y+2)^2}$$ 6. **Multiply the fractions:** $$\frac{(3y-2)(3y+2)}{4(y-3)} \cdot \frac{-6(y-3)}{(3y+2)^2} x = \frac{(3y-2)(3y+2) \cdot -6(y-3)}{4(y-3)(3y+2)^2} x$$ 7. **Cancel common factors:** - Cancel $(y-3)$ in numerator and denominator: $$\frac{(3y-2)(3y+2) \cdot -6 \cancel{(y-3)}}{4 \cancel{(y-3)} (3y+2)^2} x$$ - Cancel one $(3y+2)$ factor: $$\frac{(3y-2) \cancel{(3y+2)} \cdot -6}{4 (3y+2) \cancel{(3y+2)}} x = \frac{(3y-2)(-6)}{4(3y+2)} x$$ 8. **Simplify constants:** $$\frac{-6}{4} = -\frac{3}{2}$$ 9. **Final simplified expression:** $$-\frac{3}{2} \cdot \frac{3y-2}{3y+2} x = -\frac{3(3y-2)}{2(3y+2)} x$$ **Answer:** $$-\frac{3(3y-2)}{2(3y+2)} x$$