Simplify $ \dfrac{x^2+13x+36}{x+4} $ to $ x+9$.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+4}$.

$$ \begin{aligned} \frac{x^2+13x+36}{x+4} & =\frac{ \left( x+9 \right) \cdot \color{blue}{ \left( x+4 \right) }}{ 1 \cdot \color{blue}{ \left( x+4 \right) }} = \\[1ex] &= \frac{x+9}{1} =x+9 \end{aligned} $$

Multiply $ \dfrac{x}{x^2+4x-45} $ by $ x+9 $ to get $ \dfrac{ x }{ x-5 } $.

Step 1: Write $ x+9 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Factor numerators and denominators.

Step 3: Cancel common factors.

Step 4: Multiply numerators and denominators.

Step 5: Simplify numerator and denominator.

$$ \begin{aligned} \frac{x}{x^2+4x-45} \cdot x+9 & \xlongequal{\text{Step 1}} \frac{x}{x^2+4x-45} \cdot \frac{x+9}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ x }{ \left( x-5 \right) \cdot \color{red}{ \left( x+9 \right) } } \cdot \frac{ 1 \cdot \color{red}{ \left( x+9 \right) } }{ 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x }{ x-5 } \cdot \frac{ 1 }{ 1 } \xlongequal{\text{Step 4}} \frac{ x \cdot 1 }{ \left( x-5 \right) \cdot 1 } \xlongequal{\text{Step 5}} \frac{ x }{ x-5 } \end{aligned} $$