Simplify $ \dfrac{x^2+4x+3}{x^2+5x+6} $ to $ \dfrac{x+1}{x+2} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+3}$.

$$ \begin{aligned} \frac{x^2+4x+3}{x^2+5x+6} & =\frac{ \left( x+1 \right) \cdot \color{blue}{ \left( x+3 \right) }}{ \left( x+2 \right) \cdot \color{blue}{ \left( x+3 \right) }} = \\[1ex] &= \frac{x+1}{x+2} \end{aligned} $$

Multiply $ \dfrac{x+1}{x+2} $ by $ \dfrac{x^2-3x-10}{x^2+x} $ to get $ \dfrac{ x-5 }{ x } $.

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} \frac{x+1}{x+2} \cdot \frac{x^2-3x-10}{x^2+x} & \xlongequal{\text{Step 1}} \frac{ 1 \cdot \color{blue}{ \left( x+1 \right) } }{ 1 \cdot \color{red}{ \left( x+2 \right) } } \cdot \frac{ \left( x-5 \right) \cdot \color{red}{ \left( x+2 \right) } }{ x \cdot \color{blue}{ \left( x+1 \right) } } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 1 }{ 1 } \cdot \frac{ x-5 }{ x } \xlongequal{\text{Step 3}} \frac{ 1 \cdot \left( x-5 \right) }{ 1 \cdot x } \xlongequal{\text{Step 4}} \frac{ x-5 }{ x } \end{aligned} $$