Kamil, can you expand a tad more on that? How would you identify x-3 as a positive/negative in this case (so you could decide whether to change the inequality brackets)
You can split it into cases, like TT did, assume x-3>0 and get a solution for this domain. Then assume x-3<0 and likewise find the solutions for this domain. Then finally take the union of the two.
Actually looking at his method I think this way is more closer to what Them initially did:
Now assuming
we get this:
)
However we have assumed that x>3, and so we cannot say that all x>-2. But what we have shown is that we require x>-2, which already is certain for all x>3. Hence all x>3 solve the inequality(alternatively this can more cleverly be spotted by noticing that if x>3 then our fraction is positive and hence greater than -1).
Now solving for the other case:
)
Hence what we have shown is that if x<3, then we require x<-2. Therefore we conclude that all x<-2 solve the inequality.
Taking both cases into account the complete solution is:
 \cup (3,\infty))
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