Do the multiplications in $2x-1-\left(x+5\right)$.
$$\frac{2x-1-x-5}{x+5}<\frac{7}{2x+10}$$
Combine like terms in $2x-1-x-5$.
$$\frac{x-6}{x+5}<\frac{7}{2x+10}$$
Subtract $\frac{7}{2x+10}$ from both sides.
$$\frac{x-6}{x+5}-\frac{7}{2x+10}<0$$
Factor $2x+10$.
$$\frac{x-6}{x+5}-\frac{7}{2\left(x+5\right)}<0$$
To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $x+5$ and $2\left(x+5\right)$ is $2\left(x+5\right)$. Multiply $\frac{x-6}{x+5}$ times $\frac{2}{2}$.
Since $\frac{2\left(x-6\right)}{2\left(x+5\right)}$ and $\frac{7}{2\left(x+5\right)}$ have the same denominator, subtract them by subtracting their numerators.
Use the distributive property to multiply $2$ by $x+5$.
$$\frac{2x-19}{2x+10}<0$$
For the quotient to be negative, $2x-19$ and $2x+10$ have to be of the opposite signs. Consider the case when $2x-19$ is positive and $2x+10$ is negative.
$$2x-19>0$$ $$2x+10<0$$
This is false for any $x$.
$$x\in \emptyset$$
Consider the case when $2x+10$ is positive and $2x-19$ is negative.
$$2x+10>0$$ $$2x-19<0$$
The solution satisfying both inequalities is $x\in \left(-5,\frac{19}{2}\right)$.
$$x\in \left(-5,\frac{19}{2}\right)$$
The final solution is the union of the obtained solutions.
