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RegionPlot[ImplicitRegion[(2 x - 1)/(x - 1) < 3/2, {x, y}], 
 PlotRange -> {{-9, 9}, {-9, 9}}]

enter image description here

Just checking whether anyone else can replicate this. Can anybody confirm such behavior and, maybe, confirm this as a bug?

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  • $\begingroup$ Same result with Linux 11.0.1 $\endgroup$ – grbl Feb 8 '17 at 8:34
  • $\begingroup$ It does not work on 10.4.1 too. The following inequalities also do not work: 1/x < 1 and 1/x <-1. $\endgroup$ – Anjan Kumar Feb 8 '17 at 8:50
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    $\begingroup$ Yes, I think that's a bug and it would be good if you could report it. $\endgroup$ – user21 Feb 8 '17 at 9:50
  • $\begingroup$ The bug is still present in 11.1.0 $\endgroup$ – user58955 Mar 23 '17 at 17:50
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Another thing to do is just cut ImplicitRegion out of the loop altogether. RegionPlot is naturally designed to take a predicate of inequalities. 

RegionPlot[(2 x - 1)/(x - 1) < 3/2, {x, -9, 9}, {y, -9, 9}]

Mathematica graphics

There have been other examples of RegionPlot having trouble with ImplicitRegion, and usually the answer is just to give the inequalities to RegionPlot directly.

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  • $\begingroup$ much simpler (and natural) +1. Hope WA is going well $\endgroup$ – ubpdqn Feb 9 '17 at 0:39
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This can worked around by using Reduce,e.g:

p = Plot[(2 x - 1)/(x - 1), {x, -5, 5}, 
   GridLines -> {{-1, 1}, {3/2, {2, Red}}}, PlotRange -> {-5, 5}, 
   Frame -> True];
rp = RegionPlot[
   ImplicitRegion[Reduce[(2 x - 1)/(x - 1) < 3/2, x], {x, y}], 
   PlotRange -> {{-9, 9}, {-9, 9}}, PlotStyle -> {Pink, Opacity[0.2]}];
Show[p, rp]

enter image description here

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