Add $ \dfrac{1}{x^2-5x} $ and $ \dfrac{3}{x^2+2x-35} $ to get $ \dfrac{ \color{purple}{ 4x+7 } }{ x^3+2x^2-35x }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ x+7 }$ and the second by $\color{blue}{ x }$.

$$ \begin{aligned} \frac{1}{x^2-5x} + \frac{3}{x^2+2x-35} & = \frac{ 1 \cdot \color{blue}{ \left( x+7 \right) }}{ \left( x^2-5x \right) \cdot \color{blue}{ \left( x+7 \right) }} + \frac{ 3 \cdot \color{blue}{ x }}{ \left( x^2+2x-35 \right) \cdot \color{blue}{ x }} = \\[1ex] &=\frac{ \color{purple}{ x+7 } }{ x^3+7x^2-5x^2-35x } + \frac{ \color{purple}{ 3x } }{ x^3+7x^2-5x^2-35x } = \\[1ex] &=\frac{ \color{purple}{ 4x+7 } }{ x^3+2x^2-35x } \end{aligned} $$