Rewrite the rational expressions $\frac{3x}{x^2-3x-10}$ and $\frac{5}{x^2+3x+2}$ as equivalent expressions with the lowest common denominator $(x-5)(x+2)(x+1)$.

Step 1 — Factor each denominator. The denominators factor into $(x-5)(x+2)$ and $(x+1)(x+2)$, respectively.

Step 2 — Multiply the numerator and denominator of each expression by the missing LCD factor. For the first expression, the missing factor from the LCD is $(x+1)$: $\frac{3x}{(x-5)(x+2)} \cdot \frac{(x+1)}{(x+1)} = \frac{3x(x+1)}{(x-5)(x+2)(x+1)}$

For the second expression, the missing factor from the LCD is $(x-5)$: $\frac{5}{(x+1)(x+2)} \cdot \frac{(x-5)}{(x-5)} = \frac{5(x-5)}{(x-5)(x+2)(x+1)}$

Step 3 — Simplify the numerators. Distribute the terms in the numerators while keeping the denominators factored. The resulting equivalent rational expressions are: $\frac{3x^2+3x}{(x-5)(x+2)(x+1)}$ and $\frac{5x-25}{(x-5)(x+2)(x+1)}$