Solving the Inequality (x-1) / (x+2) < 1: Step-by-Step Guide

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To solve the inequality (x-1) / (x+2) < 1, the user attempted various methods, including squaring both sides and factorizing. They simplified the expression to ((2x+1)/(x+2))((-3/(x+2)) < 0, leading to two cases to consider. The first case involves ((2x+1)/(x+2)) < 0 and ((-3/(x+2)) > 0, while the second case is ((2x+1)/(x+2)) > 0 and ((-3/(x+2)) < 0. The user seeks guidance on how to proceed with these cases without squaring the inequality again.
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I've tried squaring both sides, then moving the RHS to the LHS, then factorizing according to a2-b2=(a+b)(a-b).

I simplified, and got
((2x+1)/(x+2))((-3/(x+2)) < 0

Then there are 2 cases:
((2x+1)/(x+2)) < 0 and ((-3/(x+2)) > 0

or

((2x+1)/(x+2)) > 0 and ((-3/(x+2)) < 0

I'm not really sure how to go from here...
Please reply ASAP!

 
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No need to square.

Don't forget: |x| &lt; y[/tex], means \pm x &amp;lt; y[/tex]
 
thanks for telling me! now i see how much easier it is :)
 

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