Question 581657
<pre>
{{{(x+1)/(x-1)}}} + {{{(1-x)/(1+x)}}} < -1

Get 0 on the right by adding 1 to both sides:

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

Th LCD is (x-1)(1+x).  

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

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

Multiply out the tops but not the bottoms:

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

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

Write the sum of the numerators over the LCD:

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

Take away the parentheses on top:

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

{{{(x^2+ 4x-1)/((x-1)(1+x))}}} < 0

We find the critical numbers by setting the numerator and
the denominator = 0 and solving:

Numerator = 0

x² + 4x - 1 = 0

That does not factor so we must use the quadratic formula:

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}

{{{x = (-4 +- sqrt( 4^2-4*1*(-1) ))/(2*1) }}}

{{{x = (-4 +- sqrt(16+4))/2 }}}

{{{x = (-4 +- sqrt(20))/2 }}}

{{{x = (-4 +- sqrt(4*5))/2 }}}

{{{x = (-4 +- 2sqrt(5))/2 }}}

Fcator out 2 on the top

{{{x = (2(-2 +- sqrt(5)))/2 }}}

{{{x = (cross(2)(-2 +- sqrt(5)))/cross(2) }}}

{{{x = -2 +- sqrt(5) }}}

{{{-2 + sqrt(5) }}} is approximately 0.24 and{{{-2 - sqrt(5) }}} is approximately -4.24  

That's two of the critical numbers.

Set denominator = 0

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

x - 1 = 0;  1 + x = 0
    x = 1;      x = -1

So the four critical numbers are

{{{-2 + sqrt(5) }}} which is approximately 0.24 
{{{-2 - sqrt(5) }}} which is approximately -4.24
1
-1

We put those in order smallest to largest:

{{{-2 - sqrt(5) }}} which is approximately -4.24
-1
{{{-2 + sqrt(5) }}} which is approximately 0.24
1

The possible solution intervals are the intervals between and beyond
the critical numbers.  None of the critical numbers are solutions, since
the inequality is < and not <u><</u> so all the possible intervals are 
open. They are:


{{{(matrix(1,3,      -infinity,   ",", -2-sqrt(5)))}}}
{{{(matrix(1,3,    -2-sqrt(5), ",", -1 ))}}}
{{{(matrix(1,3,      -1,   ",", -2+sqrt(5)))}}}
{{{(matrix(1,3,    -2+sqrt(5), ",", 1 ))}}}
{{{(matrix(1,3,    1, ",", infinity ))}}}



We pick a test value in each interval:

{{{(matrix(1,3,      -infinity,   ",", -2-sqrt(5)))}}}, pick test value -5
{{{(matrix(1,3,    -2-sqrt(5), ",", -1 ))}}}, pick test value -2
{{{(matrix(1,3,      -1,   ",", -2+sqrt(5)))}}}, pick test value 0 
{{{(matrix(1,3,    -2+sqrt(5), ",", 1 ))}}}, pick test value .5
{{{(matrix(1,3,    -2+sqrt(5), ",", infinity ))}}} 2

We substitute each test value into

{{{(x^2+ 4x-1)/((x-1)(1+x))}}} < 0

Substituting test value -5

{{{((-5)^2+ 4(-5)-1)/(((-5)-1)(1+(-5)))}}} < 0
{{{(25-20-1)/((-5-1)(1-5))}}} < 0
{{{4/((-6)(-4)) < 0
{{{4/24}}} < 0
{{{1/6}}} < 0
That is false so {{{(matrix(1,3,      -infinity,   ",", -2-sqrt(5)))}}}
is not part of the solution

Substituting test value -2

{{{((-2)^2+ 4(-2)-1)/(((-2)-1)(1+(-2)))}}} < 0
{{{(4-8-1)/((-2-1)(1-2))}}} < 0
{{{(-5)/((-3)(-1)) < 0
{{{(-5)/3}}} < 0
{{{-5/3}}} < 0
That is true so {{{(matrix(1,3,    -2-sqrt(5), ",", -1 ))}}} 
is part of the solution

Substituting test value 0

{{{((0)^2+ 4(0)-1)/(((0)-1)(1+(0)))}}} < 0
{{{(0+0-1)/((0-1)(1+0))}}} < 0
{{{(-1)/(-1)}}} < 0
1 < 0

That is false so {{{(matrix(1,3,      -1,   ",", -2+sqrt(5)))}}} 
is not part of the solution

Substituting test value .5

{{{((.5)^2+ 4(.5)-1)/(((.5)-1)(1+(.5)))}}} < 0
{{{(.25+2-1)/((.5-1)(1+.5))}}} < 0
{{{1.25/((-.5)(1.5)) < 0
{{{1.25/(-.75)}}} < 0
{{{125/(-75)}}} < 0
{{{-5/3}}} < 0
That is true so {{{(matrix(1,3,    -2+sqrt(5), ",", 1 ))}}},  
is part of the solution

Substituting test value 2

{{{((2)^2+ 4(2)-1)/(((2)-1)(1+(2)))}}} < 0
{{{(4+8-1)/((2-1)(1+2))}}} < 0
{{{11/((1)(3)) < 0
{{{11/3}}} < 0

That is false so {{{(matrix(1,3,    -2+sqrt(5), ",", infinity ))}}} 
is not part of the solution.

So the solution is

{{{(matrix(1,3,    -2-sqrt(5), ",", -1 ))}}} &#8899; {{{(matrix(1,3,    -2+sqrt(5), ",", 1 ))}}}

Edwin</pre>