<s>[INST] Different ways of solving inequation - different solutions? After 'solving' the following exercise, I entered the solution in a website that gave me the graph and I noticed that It wasn't the same as the solution it gave me for the 'original' inequation.
The inequation is:
$5-\frac{4-5x}{2}<3\left(2x-1\right)$
And I solved it like:
1)
 $\frac{10-4+5x}{2}*\frac{1}{2x-1}<3$
2)
$\frac{6+5x}{4x-2}-\frac{3(4x-2)}{4x-2}<0$
3)
$\frac{12-7x}{4x-2}<0$
Then I solved it and I got: 
Solution= $]-\infty ,\frac{1}{2}[ U [\frac{12}{7},+\infty[$
But, it's wrong, it should be: $[\frac{12}{7},+\infty[$
I then did the exercise in a different way:
1) $\frac{6+5x}{2}-3(2x-1)<0$
2)$\frac{6+5x}{2}-\frac{2*3(2x-1)}{2}<0$
3)$\frac{12-7x}{2}<0$
And the result is now the correct one: $[\frac{12}{7},+\infty[$
Can someone explain me why there are these differences between the first way of solving and the second one?
 [/INST] With inequalities you need to be careful because when you multiply for negative value inequality reverses. For this reason the first attempt gives  a wrong result and the second was correct.
Note that you can also use first method but you need to distiguish the cases, notably


*

*for $2x-1>0$


$$5-\frac{4-5x}{2}<3\left(2x-1\right) \implies \frac{10-4+5x}{2}\cdot\frac{1}{2x-1}<3$$


*

*for $2x-1<0$


$$5-\frac{4-5x}{2}<3\left(2x-1\right) \implies \frac{10-4+5x}{2}\cdot\frac{1}{2x-1}>3$$


*

*for $2x-1=0$


$$5-\frac{4-5x}{2}<3\left(2x-1\right) \implies 5-\frac{4-5\frac12}{2}<0\implies 5+\frac34<0$$
</s>