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I want to find the relationship between several parameters. But I don't know how to define an Intermediate Variable. For example: I want to know the relationship between x and {y,z} which will satisfy the constraints with the variables {q1,q2}. How should I define {q1,q2} to make them just Intermediate Variables and make sure that they will not show up at the final results.

Reduce[x q1 + y q2 + 3 z > 0 && q1 + q2 == x - y && z q1 + y q2 == 0, x]

I have figured out the question above, but I don't know why I can not get a result for the following code:

Reduce[1/(
6 m) (6 q1 (r - w) m + q1^3 (t (-1 + m) + 4 (-r + s) m) + 
  q2 (6 (-r + w) m + q2^2 (t + 4 r m - 4 s m))) > 
 0 && -((c q1^2 (-1 + m) + 2 (-r + w) m + 
     q1^2 (r + t + 2 r m - 3 s m - t m))/(2 m)) == 0 && (
   c q2^2 + 2 (r - w) m - q2^2 (r + t + 3 r m - 3 s m))/(2 m) == 0 && 
  r > w > s > 0 && t >= 0 && c > t + m r + (1 - m) s && 0 < m < 1 && 
  0 < q2 < q1 < m/(1 + m), c, {q1, q2}]
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    $\begingroup$ To be more specific, Reduce[x q1 + y q2 + 3 z > 0 && q1 + q2 == x - y && z q1 + y q2 == 0, x, {q1, q2}] $\endgroup$
    – xzczd
    Commented Mar 23, 2020 at 10:28
  • $\begingroup$ Yes! Thank you very much! But it seems that I cannot find the solution with too many complex constraints. Do you know how to handle this? $\endgroup$
    – cclinoom
    Commented Mar 23, 2020 at 12:19
  • $\begingroup$ Please show us a specific example. $\endgroup$
    – xzczd
    Commented Mar 23, 2020 at 12:32
  • $\begingroup$ Depending on what exactly is wanted, another contender is Resolve[Exists[{q1, q2}, x q1 + y q2 + 3 z > 0 && q1 + q2 == x - y && z q1 + y q2 == 0], Reals]. $\endgroup$ Commented Mar 23, 2020 at 13:20
  • $\begingroup$ The new system is probably too complicated. As a start, just try Solve[-c q1^2 (-1 + m) + 2 (-r + w) m + q1^2 (r + t + 2 r m - 3 s m - t m) == 0 && c q2^2 + 2 (r - w) m - q2^2 (r + t + 3 r m - 3 s m) == 0, {q1, q2}, Reals]. $\endgroup$
    – xzczd
    Commented Mar 24, 2020 at 10:17

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