@dharr thank you, that looks very promissing and I understand what you are doing. I could replicate the procedure with another symbolic number but the second number I tryd brought this nasty error in the line: 
new_indexed := select(x -> evala(Normal(x - sys2)) = 0, all)[];
Error, (in evala/Normal/preproc0) index should be a positive integer less than 2
It just came out of nowhere and I dont even understand on what problem the error is refering since I dont see any diffrence between the structure of those two symbolix expressions.  Do you know why?

Thank you, Regards, Felix

Download RootOf_with_Error.mw

@dharr As I was already expecting, there are cases where index 1 isnt true. Please have a look in the attacked file. This huge sybolic expression which I got from SemiAlgebraic should become shorter and more direct. I really dont know why it got so huge in the first case (also note, that the results is a fraction). However if you just evaluate the new expression ("new") you get a wrong solution. It turns out (as you can see in the second programm in this document) that index 3 would be necesarry. So ok, I could go for all values and than search for the right one but it takes a really long time with symbolic expressions and it is just seems unnecesarry. So it would just be easier if the expression "new" alway would have an index coding. So I can live with the fact, that maybe in 95% of the cases it gives back index 1 but it would make sure that I dont make any errors by always expecting index 1 to be right. Do you see my problem? 

Maybe you have another idea. Thank you. 

Download index_1_isnt_always_the_right_index.mw

Regards Felix

@dharr that seems to work out. But the new RootOf expression lost its precise describtion, which of the roots to take. so _Z^5 may hav 5 real values which make the polynom 0. normaly you get a further information telling you something like index=real_3 or 1.234 ... 1.235. So how do I get the right root?

Thank you

Awesome, thank you @dharr !

@acer Yes youre right. But what about the first part? Do you understand my problem with that solution? 

@Carl Love I used the SemiAlgebraic command to solve systems of polynomial equations and inequalities. Now I take those solutions and go on working with them. 

Is there a chance to tell SemiAlgebraic, that I want the solutions in a diffrent form then RootOf? Im not sure if its always possible but often you see that there would be a diffrent way to tell the solution. For exmaple you can desrcibe RootOf(Z^2-y,index=2) and 1<y<2 as x=t+1, y=t^2 with 1<t<2^0,5 . Sometimes, but very rarely I saw those solutions. I dont like that you need an extra parameter and its not directly clear how the graph looks but its easier to handle and to be honest you dont know how the graph looks in many RootOf expressions as well. 

@acer thank you for your answer. The problem with your approach is, that the solution misses the intervalls of y. So for me the given solution om Maple looks wrong. Because if you look at the solution you might think, that you can choose any y. But I need y to be between 1 and 2. Why is it lost? Since x depedence on y (via the RootOf) it should be part of the solution. 

What I mean with the cited sentence of me is, that set, described by the function and the intervall for y is a real subset of the area. So intersecting both areas (for me that is hwat happens if you put both sets in the solve command) should give me back the subset, so the RootOf function with the intervall 1<y, y<2. I just choose this example because you can see the right solution quite easily. In general the RootOf functions are little bit more complicated. 

@rcorless That is an interesting approach. Thank you. So this process transform my system of equations and inequalities in an easier system? Because it does not give the solution immediately. So I know would like to take the new system and use solve, is that right? However I dont know how to exctract the equations and inequalities from the solutions of regularchains. How can I do that, because there are multiple lines. Do you have advice? RegularChains.mw

Download RegularChains.mw

@rcorless Yes you are right. But when that solutions fullfills the equations under the restrictions, than it should appear if I solve the system without restrictions as well, since with restrictions you get a subset of solutions. So why is the marked solution you gave not part of the solutions of Sol_traditinal and Sol_backsolve? Thats my main concern. THey should be there, right? Since PolynomialSystem (under some engignes) does not find a solution which has to be there I am loosing trust into the function. Do you see my point?

Thank you for your resonse @Scot Gould . I see your point. But why does adding 0<s makes the problem and not 0<x and 0<y? And shouldnt Maple find out at itself, that although I ask for 8 variables and it is a parametric solution that it can give me just 7 variables back?

I would be glad if you have time to look at my second approach. In the attached file you can see the comparison of three solving approaches. In the first case I solve equations and inequalities together and get a nice solution, this is the one I want to replicate. In the second and thirsd case you can see the two stepped approch. If I take the solutions found while solving 8 equations and put it together with the inequalities it does not find any solution. But as we know from sol_1 there is one. Why is that? 

In the third case I asked for allvalues first and now he finds the solution which is missing in case 2. However the way the solution is shown is weird. The conditions are really complicated and not necesarry. You could summarisze them since they are all depend from x. furthermore it is given in two rows, but the second row is useless. I ask for solutions and it tells me the area where it cant find a solution. How can I change the way Maple gives back solutions?

By the way sol_1 and sol_3 are equal, you can show that k and t in both cases have the same value although they look diffrent.

Thank you

Download Strange_Output_form_and_missing_solutions.mw

@dharr Thank you for the fast response. That is a good explination. I didnt thought of that. Does that mean, that in first place alle the solutions of the RootOf expression are possible solutions? If Maple would mean one specifc RootOf solution the programm would make that clear either with an index or with a numeric expression like 0,355...0,356 behind the RootOf expression right? 

And does this also explain, why solving the second step with the symbolic expression of RootOf does take so much more time? Because there is more the the on value you get with evalf? I just changed to the numeric expression via evalf because it took to much time and I already saw the "error" ,you just explained, inbetween.