1. **State the problem:** Simplify the expression $$\frac{8y}{2 - 3y} - \frac{4}{-3y^2 + 2y} + \frac{3y + 2}{y}$$. 2. **Rewrite the expression and factor denominators:** Note that $$-3y^2 + 2y = y(-3y + 2)$$. 3. **Rewrite the expression with factored denominators:** $$\frac{8y}{2 - 3y} - \frac{4}{y(-3y + 2)} + \frac{3y + 2}{y}$$ 4. **Notice that $$2 - 3y = -(3y - 2)$$, so rewrite the first term:** $$\frac{8y}{2 - 3y} = \frac{8y}{-(3y - 2)} = -\frac{8y}{3y - 2}$$ 5. **Rewrite the second term denominator:** $$y(-3y + 2) = y(-(3y - 2)) = -y(3y - 2)$$ So, $$- \frac{4}{y(-3y + 2)} = - \frac{4}{-y(3y - 2)} = \frac{4}{y(3y - 2)}$$ 6. **Rewrite the entire expression:** $$-\frac{8y}{3y - 2} + \frac{4}{y(3y - 2)} + \frac{3y + 2}{y}$$ 7. **Find common denominator:** The common denominator is $$y(3y - 2)$$. 8. **Rewrite each term with the common denominator:** - First term: $$-\frac{8y}{3y - 2} = -\frac{8y \cdot y}{y(3y - 2)} = -\frac{8y^2}{y(3y - 2)}$$ - Second term is already over the common denominator: $$\frac{4}{y(3y - 2)}$$ - Third term: $$\frac{3y + 2}{y} = \frac{(3y + 2)(3y - 2)}{y(3y - 2)}$$ 9. **Combine all terms:** $$\frac{-8y^2 + 4 + (3y + 2)(3y - 2)}{y(3y - 2)}$$ 10. **Expand numerator:** $$(3y + 2)(3y - 2) = 9y^2 - 4$$ So numerator becomes: $$-8y^2 + 4 + 9y^2 - 4 = (-8y^2 + 9y^2) + (4 - 4) = y^2 + 0 = y^2$$ 11. **Simplify the fraction:** $$\frac{y^2}{y(3y - 2)}$$ 12. **Cancel common factor $$y$$:** $$\frac{\cancel{y} y}{\cancel{y}(3y - 2)} = \frac{y}{3y - 2}$$ **Final answer:** $$\boxed{\frac{y}{3y - 2}}$$