simplifying rational inequality results
The command
solve(abs((2*x-2)/(x-5)) <= 2/3, x)
yields
#0: solve_rat_ineq(ineq=2*abs(x-1)/abs(x-5)-2/3 <= 0)
[[x == -1, -6 != 0, -6 != 0], [x == -1, -6 != 0, -6 != 0, -6 != 0], [x
== -1, -6 != 0, -6 != 0], [x == -1, -6 != 0, -6 != 0, -6 != 0], [x == 2,
-3 != 0, -3 != 0], [x == 2, -3 != 0, -3 != 0, -3 != 0], [x == 2, -3 !=
0, -3 != 0], [x == 2, -3 != 0, -3 != 0, -3 != 0], [x == 1], [1 < x, x
< 2], [-1 < x, x < 1]]
Is there a way to simplify that output to get something like
[[-1 <= x, x <= 2]]
?
Not directly within Sage, I think, though as you'll note the union of all those results is indeed the answer you are looking for. Even `solve(abs((x-1)/x)<=1,x)` gives several answers, which union to the correct one. That said, it would be nice to "sanitize" the ones above so that it looks more like `[[x == -1],[-1<x,x\<1],[x =="1],[1<x,x<2]]`" (where="" `<`="" means="" <.<="" p="">
For other readers - note the same question was asked [here on the Maxima list](http://www.math.utexas.edu/pipermail/maxima/2013/033339.html), with inconclusive results for now (other than what does/doesn't work in Maxima, and which Maxima function creates this).
@kcrisman, since my question was just a boolean one ("is there a way ..."), I would accept your "not directly within Sage" as an Askbot answer, if you care.
Well, I guess I lied, since Python is presumably Turing-complete! But it's not an easy one-liner, though probably someone could write a tricky one-liner.