Subtract $ \dfrac{7}{x} $ from $ \dfrac{x}{x+5} $ to get $ \dfrac{ \color{purple}{ x^2-7x-35 } }{ x^2+5x }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ x }$ and the second by $\color{blue}{ x+5 }$.

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