To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $2\left(-2x+1\right)$ and $1-2x$ is $2\left(-2x+1\right)$. Multiply $\frac{5x}{1-2x}$ times $\frac{2}{2}$.
Since $\frac{6}{2\left(-2x+1\right)}$ and $\frac{2\times 5x}{2\left(-2x+1\right)}$ have the same denominator, subtract them by subtracting their numerators.
For the quotient to be $≤0$, one of the values $3-5x$ and $1-2x$ has to be $≥0$, the other has to be $≤0$, and $1-2x$ cannot be zero. Consider the case when $3-5x\geq 0$ and $1-2x$ is negative.
$$3-5x\geq 0$$ $$1-2x<0$$
The solution satisfying both inequalities is $x\in \left(\frac{1}{2},\frac{3}{5}\right]$.
$$x\in (\frac{1}{2},\frac{3}{5}]$$
Consider the case when $3-5x\leq 0$ and $1-2x$ is positive.
$$3-5x\leq 0$$ $$1-2x>0$$
This is false for any $x$.
$$x\in \emptyset$$
The final solution is the union of the obtained solutions.
