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i try to do sth. like this

A=Solve[{x>=5, y>=3, x<13,y<13} ,{x,y}]

which gives me a convex set and then

Solve[{x==y, y==3}, {x,y}]

with (x,y) element of A.

I know that i could just use one solve for all, but for faster computation it would be better if i could calculate A first and then use it for the next step.

How could i do that? Thank you very much

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    $\begingroup$ Your Solve syntax doesn't seem correct to me. $\endgroup$ – Sjoerd C. de Vries Mar 4 '16 at 16:19
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I don't think there is much benefit to doing this in multiple steps, but if you want to do it that way I would suggest solving these equations with Reduce and combinations of And:

equation1 = And[x >= 5, y >= 3, x < 13, y < 13];
equation2 = And[x==y,y==3];
Reduce[ And[ equation1, equation2 ], {x,y} ]

which gives False because the second equation leads to x and y both being 3:

Reduce[And[x == y, y == 3], {x, y}] (* gives x == 3 && y == 3 *)

And that falls outside of the region defined by the first equation:

RegionPlot[ And[x >= 5, y >= 3, x < 13, y < 13], {x, 0, 15}, {y, 0, 15}]

enter image description here

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  • $\begingroup$ Thx Arnoud, i just tried to create a fast example code so that my problem is clear. If i get you r code right than still all the calculation is done in the "Reduce" line, right. So that if i have a do slop over 15k equations the convex room needs to be calculated everytime? $\endgroup$ – user34047 Mar 4 '16 at 20:17
  • $\begingroup$ I think so. It would be extremely useful if you edit your original question so it includes correct Wolfram Language code in the first place (people tend to vote to close questions otherwise) and also a full problem description (you did not mention the 15K equations before, so my answer did not take this into account. The more effort you put into asking a clear question, the more likely you are going to get a great answer! $\endgroup$ – Arnoud Buzing Mar 4 '16 at 21:59

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