Rational Expressions 8E0D6C

1. **Problem 2:** Simplify $$\frac{15x + 3}{x(x - 4y) + 4y^2} \cdot \frac{3x - 6y}{5x + 1}$$ 2. **Step 1:** Factor all expressions where possible. - Numerator of first fraction: $15x + 3 = 3(5x + 1)$ - Denominator of first fraction: $x(x - 4y) + 4y^2 = x^2 - 4xy + 4y^2 = (x - 2y)^2$ - Numerator of second fraction: $3x - 6y = 3(x - 2y)$ - Denominator of second fraction: $5x + 1$ (already factored) 3. **Step 2:** Rewrite the expression with factored forms: $$\frac{3(5x + 1)}{(x - 2y)^2} \cdot \frac{3(x - 2y)}{5x + 1}$$ 4. **Step 3:** Multiply the numerators and denominators: $$\frac{3(5x + 1) \cdot 3(x - 2y)}{(x - 2y)^2 \cdot (5x + 1)} = \frac{9(5x + 1)(x - 2y)}{(x - 2y)^2 (5x + 1)}$$ 5. **Step 4:** Cancel common factors $5x + 1$ and one $(x - 2y)$: $$\frac{9\cancel{(5x + 1)}\cancel{(x - 2y)}}{\cancel{(x - 2y)}(x - 2y)\cancel{(5x + 1)}} = \frac{9}{x - 2y}$$ 6. **Answer for problem 2:** $$\boxed{\frac{9}{x - 2y}}$$ --- 7. **Problem 4:** Simplify $$\frac{6x^2 + 23x + 20}{8x^2 + 20x} \div \frac{3x^2 - 23x - 36}{x^2 - 18x + 81}$$ 8. **Step 1:** Factor all polynomials. - Numerator of first fraction: $6x^2 + 23x + 20$ - Factors of $6 \times 20 = 120$ that sum to 23 are 8 and 15. - Rewrite: $6x^2 + 8x + 15x + 20 = 2x(3x + 4) + 5(3x + 4) = (2x + 5)(3x + 4)$ - Denominator of first fraction: $8x^2 + 20x = 4x(2x + 5)$ - Numerator of second fraction: $3x^2 - 23x - 36$ - Factors of $3 \times (-36) = -108$ that sum to -23 are -27 and 4. - Rewrite: $3x^2 - 27x + 4x - 36 = 3x(x - 9) + 4(x - 9) = (3x + 4)(x - 9)$ - Denominator of second fraction: $x^2 - 18x + 81 = (x - 9)^2$ 9. **Step 2:** Rewrite the division as multiplication by reciprocal: $$\frac{(2x + 5)(3x + 4)}{4x(2x + 5)} \times \frac{(x - 9)^2}{(3x + 4)(x - 9)}$$ 10. **Step 3:** Multiply numerators and denominators: $$\frac{(2x + 5)(3x + 4)(x - 9)^2}{4x(2x + 5)(3x + 4)(x - 9)}$$ 11. **Step 4:** Cancel common factors $(2x + 5)$, $(3x + 4)$, and one $(x - 9)$: $$\frac{\cancel{(2x + 5)}\cancel{(3x + 4)}(x - 9)\cancel{(x - 9)}}{4x\cancel{(2x + 5)}\cancel{(3x + 4)}\cancel{(x - 9)}} = \frac{x - 9}{4x}$$ 12. **Answer for problem 4:** $$\boxed{\frac{x - 9}{4x}}$$