Question 186653
<pre><font size = 4 color = "indigo"><b>

{{{1/(x+1)<1/(x-1)}}}

Get 0 on the right

{{{1/(x+1)-1/(x-1)<0}}}

Get LCD of {{{(x+1)(x-1)}}}

{{{((x-1)-(x+1))/((x+1)(x-1))<0}}}

{{{(x-1-x-1)/((x+1)(x-1))<0}}}

{{{-2/(x+1)(x-1)<0}}}

Critical values are gotten by setting
numerators and denominators = 0 and solving.

We can't set the numerator = 0 since it's -2
We set the denominator = 0

{{{(x+1)(x-1)=0}}}

{{{x=-1}}} and {{{x=1}}}

Now we make a number line and mark those critical values

 -----------o---------o-----------
 -3   -2   -1    0    1    2    3

Now we choose any value to the left of -1, say -2

Substitute it into

{{{-2/(x+1)(x-1)<0}}}

{{{-2/(-2+1)(-2-1)<0}}}

{{{-2/((-1)(-3))<0}}}

{{{-2/3<0}}}

That is true so we shade the part of
the number line to the left of -1:

<===========o---------o-----------
 -3   -2   -1    0    1    2    3

Next we choose any value between -1 and 1, 
say, 0 and substitute it into

{{{-2/(x+1)(x-1)<0}}}

{{{-2/(0+1)(0-1)<0}}}

{{{-2/((1)(-1))<0}}}

{{{-2/(-1)<0}}}

{{{2 < 0}}}

That is false so we do not shade the part of
the number line between -1 and 1, so we still
have:

<===========o---------o-----------
 -3   -2   -1    0    1    2    3

Next we choose any value to the right of 1, say 2

Substitute it into

{{{-2/(2+1)(2-1)<0}}}

{{{-2/(2+1)(2-1)<0}}}

{{{-2/((3)(1))<0}}}

{{{-2/3<0}}}

That is true so we shade the part of
the number line to the right of 1:

<===========o---------o==========>
 -3   -2   -1    0    1    2    3

The critical values do not satisfy the inequality
because they cause 0's in the denominator, so
we leave them open.

Now we translate this graph into interval notation:

{{{matrix(1,11,

"(",    -infinity, ",",  -1,  ")",  U,  "(",  1,  ",",  infinity,  ")" )}}}   

Edwin</pre>