Why are some solutions excluded if we simply multiply the denominator in an fraction inequality? I have the next inequality for which I have to find the solutions : $$\frac{2x-5}{3x-1}\geq 1, \text{ where } x\in\mathbb{R}$$ I know I have to subtract $1$ and then I have to analyse the sign for the grade 1 function for both numerator and denominator and then see for which intervals they have different signs... Why the approach of multiplying on both sides with $3x-1$ is incorrect?

Because multiplying by a negative quantity reverse the inequality, so you have to separe the two case: if the denominator is $>0$ or $<0$ ( and obviously exclude the case that it is $=0$).