SOLUTION: please solve this equation thank you {{{1/(x+1)<1/(x-1)}}}

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: please solve this equation thank you {{{1/(x+1)<1/(x-1)}}}      Log On


   



Question 186653: please solve this equation thank you

1%2F%28x%2B1%29%3C1%2F%28x-1%29

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!


1%2F%28x%2B1%29%3C1%2F%28x-1%29

Get 0 on the right

1%2F%28x%2B1%29-1%2F%28x-1%29%3C0

Get LCD of %28x%2B1%29%28x-1%29

%28%28x-1%29-%28x%2B1%29%29%2F%28%28x%2B1%29%28x-1%29%29%3C0

%28x-1-x-1%29%2F%28%28x%2B1%29%28x-1%29%29%3C0

-2%2F%28x%2B1%29%28x-1%29%3C0

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

%28x%2B1%29%28x-1%29=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%2F%28x%2B1%29%28x-1%29%3C0

-2%2F%28-2%2B1%29%28-2-1%29%3C0

-2%2F%28%28-1%29%28-3%29%29%3C0

-2%2F3%3C0

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%2F%28x%2B1%29%28x-1%29%3C0

-2%2F%280%2B1%29%280-1%29%3C0

-2%2F%28%281%29%28-1%29%29%3C0

-2%2F%28-1%29%3C0

2+%3C+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%2F%282%2B1%29%282-1%29%3C0

-2%2F%282%2B1%29%282-1%29%3C0

-2%2F%28%283%29%281%29%29%3C0

-2%2F3%3C0

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:

   

Edwin