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?

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 $2x-1=0$


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