Question: Simplify each rational expression.
1. $\frac{3a+1}{6}$ · $\frac{36}{3a+1}$
2. $\frac{15x+3}{x(x-4y)+4y^2}$ · $\frac{3x-6y}{5x+1}$
3. $\frac{6}{5(d-1)}$ ÷ $\frac{d+1}{10(d-1)}$
4. $\frac{6x^2+23x+20}{8x^2+20x}$ ÷ $\frac{3x^2-23x-36}{x^2-18x+81}$
5. $\frac{11}{7ab}$ + $\frac{3}{7ab}$
6. $\frac{2b-1}{2b-5}$ - $\frac{3b-2}{3b-4}$
PRACTICE
Simplify each rational expression.
1. $\frac{x^2-25}{26}$ · $\frac{13}{x-5}$
2. $\frac{2k^2-9k+4}{3k^2-(10k+8)}$ · $\frac{3k^2+11k+6}{-2k^2-5k+3}$
make the number 4 simpler but in an understandable way
1. **State the problem:** Simplify the expression $$\frac{6x^2+23x+20}{8x^2+20x} \div \frac{3x^2-23x-36}{x^2-18x+81}$$
2. **Rewrite division as multiplication:**
$$\frac{6x^2+23x+20}{8x^2+20x} \times \frac{x^2-18x+81}{3x^2-23x-36}$$
3. **Factor all polynomials:**
- Numerator 1: $6x^2+23x+20$
- Factors as $(2x+5)(3x+4)$
- Denominator 1: $8x^2+20x$
- Factor out $4x$: $4x(2x+5)$
- Numerator 2: $x^2-18x+81$
- Recognize perfect square: $(x-9)^2$
- Denominator 2: $3x^2-23x-36$
- Factor as $(3x+4)(x-9)$
4. **Substitute factored forms:**
$$\frac{(2x+5)(3x+4)}{4x(2x+5)} \times \frac{(x-9)^2}{(3x+4)(x-9)}$$
5. **Cancel common factors:**
- Cancel $(2x+5)$ from numerator and denominator
- Cancel $(3x+4)$ from numerator and denominator
- Cancel one $(x-9)$ from numerator and denominator
Intermediate step with cancellations:
$$\frac{\cancel{(2x+5)}\cancel{(3x+4)}}{4x\cancel{(2x+5)}} \times \frac{(x-9)\cancel{(x-9)}}{\cancel{(3x+4)}\cancel{(x-9)}} = \frac{1}{4x} \times (x-9)$$
6. **Multiply remaining terms:**
$$\frac{1}{4x} \times (x-9) = \frac{x-9}{4x}$$
7. **Final simplified expression:**
$$\boxed{\frac{x-9}{4x}}$$
This is the simplest form of the original expression.