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Could you guys solve them and show me the solutions and all working so that I learn? This is another practice question and I have not been taught (I haven't been taught how to do it with 3 sides). I find that when I read the solution to a type of problem I understand how to do it so then I can do other problems like it. Thanks in advance.

$$-1<\frac{2x+3}{x-1}<1$$

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For $x-1\ne0,(x-1)^2>0$

$$\dfrac{2x+3}{x-1}<1\iff\dfrac{2x+3}{x-1}-1<0\iff\dfrac{x+4}{x-1}<0\iff\dfrac{x+4}{x-1}\cdot(x-1)^2<0$$

$$\iff(x+4)(x-1)<0$$

Now for $(x-a)(x-b)<0$ where $a<b; a<x<b$

Please follow the same method for $$-1<\dfrac{2x+3}{x-1}\iff0<\dfrac{2x+3}{x-1}+1=\dfrac{3x+2}{x-1}$$

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  • $\begingroup$ what happened to -1< part? $\endgroup$
    – jb3navides
    Commented Aug 23, 2015 at 4:31
  • $\begingroup$ @jb3navides, Can't you follow the same method? $\endgroup$
    – lab bhattacharjee
    Commented Aug 23, 2015 at 4:31
  • $\begingroup$ Oh so we do it piece by piece thanks. $\endgroup$
    – jb3navides
    Commented Aug 23, 2015 at 4:37

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