1. **State the problem:** Find the least common denominator (LCD) for each pair of rational expressions given. 2. **Recall the rule:** The LCD of rational expressions is the least common multiple (LCM) of their denominators. Factor each denominator completely to find the LCD. --- **1a. Expressions:** $\frac{y}{12x^5}$ and $\frac{5}{9x^2}$ - Factor denominators: - $12x^5 = 2^2 \times 3 \times x^5$ - $9x^2 = 3^2 \times x^2$ - LCD must include the highest powers of each prime factor: - For 2: $2^2$ - For 3: $3^2$ - For $x$: $x^5$ - So, $$\text{LCD} = 2^2 \times 3^2 \times x^5 = 4 \times 9 \times x^5 = 36x^5$$ --- **1b. Expressions:** $\frac{p}{10mn}$ and $\frac{7}{15}$ - Factor denominators: - $10mn = 2 \times 5 \times m \times n$ - $15 = 3 \times 5$ - LCD must include: - 2, 3, 5, $m$, $n$ - So, $$\text{LCD} = 2 \times 3 \times 5 \times m \times n = 30mn$$ --- **2a. Expressions:** $\frac{y}{3y+6}$ and $\frac{5}{4y+8}$ - Factor denominators: - $3y+6 = 3(y+2)$ - $4y+8 = 4(y+2)$ - LCD must include: - 3, 4, and $(y+2)$ - So, $$\text{LCD} = 3 \times 4 \times (y+2) = 12(y+2)$$ --- **2b. Expressions:** $\frac{9}{x^2 - 2x + 1}$ and $\frac{17}{2x^2 - 2x}$ - Factor denominators: - $x^2 - 2x + 1 = (x-1)^2$ - $2x^2 - 2x = 2x(x-1)$ - LCD must include: - 2, $x$, and $(x-1)^2$ - So, $$\text{LCD} = 2 \times x \times (x-1)^2 = 2x(x-1)^2$$ --- **Final answers:** 1a. LCD = $36x^5$ 1b. LCD = $30mn$ 2a. LCD = $12(y+2)$ 2b. LCD = $2x(x-1)^2$