1. **State the problem:** Simplify the expression $$\frac{2y + 1}{2y^2 + y - 1} + \frac{3}{2 - 3y + 2y^2}$$. 2. **Rewrite the denominators in a standard polynomial form:** - First denominator: $$2y^2 + y - 1$$ - Second denominator: $$2 - 3y + 2y^2$$ can be rewritten as $$2y^2 - 3y + 2$$. 3. **Factor the denominators:** - Factor $$2y^2 + y - 1$$: Find two numbers that multiply to $$2 \times (-1) = -2$$ and add to $$1$$. These are $$2$$ and $$-1$$. So, $$2y^2 + y - 1 = 2y^2 + 2y - y - 1 = 2y(y + 1) -1(y + 1) = (2y - 1)(y + 1)$$. - Factor $$2y^2 - 3y + 2$$: Find two numbers that multiply to $$2 \times 2 = 4$$ and add to $$-3$$. These are $$-1$$ and $$-2$$. So, $$2y^2 - 3y + 2 = 2y^2 - 2y - y + 2 = 2y(y - 1) -1(y - 2)$$ but this does not factor nicely this way. Let's try the quadratic formula or check for factorization: The discriminant is $$(-3)^2 - 4 \times 2 \times 2 = 9 - 16 = -7$$, which is negative, so it does not factor over the reals. 4. **Rewrite the expression with factored denominators where possible:** $$\frac{2y + 1}{(2y - 1)(y + 1)} + \frac{3}{2y^2 - 3y + 2}$$. 5. **Find a common denominator:** Since the second denominator does not factor nicely, the common denominator is $$ (2y - 1)(y + 1)(2y^2 - 3y + 2) $$. 6. **Rewrite each fraction with the common denominator:** - First fraction numerator multiplied by $$2y^2 - 3y + 2$$: $$ (2y + 1)(2y^2 - 3y + 2) $$ - Second fraction numerator multiplied by $$ (2y - 1)(y + 1) $$: $$ 3(2y - 1)(y + 1) $$ 7. **Expand the numerators:** - Expand $$ (2y + 1)(2y^2 - 3y + 2) $$: $$2y \times 2y^2 = 4y^3$$ $$2y \times (-3y) = -6y^2$$ $$2y \times 2 = 4y$$ $$1 \times 2y^2 = 2y^2$$ $$1 \times (-3y) = -3y$$ $$1 \times 2 = 2$$ Sum: $$4y^3 - 6y^2 + 4y + 2y^2 - 3y + 2 = 4y^3 - 4y^2 + y + 2$$ - Expand $$3(2y - 1)(y + 1)$$: First expand $$(2y - 1)(y + 1)$$: $$2y \times y = 2y^2$$ $$2y \times 1 = 2y$$ $$-1 \times y = -y$$ $$-1 \times 1 = -1$$ Sum: $$2y^2 + 2y - y - 1 = 2y^2 + y - 1$$ Multiply by 3: $$3(2y^2 + y - 1) = 6y^2 + 3y - 3$$ 8. **Add the numerators:** $$4y^3 - 4y^2 + y + 2 + 6y^2 + 3y - 3 = 4y^3 + 2y^2 + 4y - 1$$ 9. **Final expression:** $$\frac{4y^3 + 2y^2 + 4y - 1}{(2y - 1)(y + 1)(2y^2 - 3y + 2)}$$ 10. **Summary:** The sum of the two rational expressions simplifies to $$\frac{4y^3 + 2y^2 + 4y - 1}{(2y - 1)(y + 1)(2y^2 - 3y + 2)}$$. This is the simplified form since the denominator cannot be factored further over the reals.