document.write( "Question 581657: How do you solve the inequality:\r
\n" ); document.write( "\n" ); document.write( "[(x+1)/(x-1)] + [(1-x)/(1+x)] is less than -1 \n" ); document.write( "
Algebra.Com's Answer #371786 by Edwin McCravy(20054) $\"\"$ $\"About$
You can put this solution on YOUR website!
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document.write( " +  < -1\r\n" );
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document.write( "Get 0 on the right by adding 1 to both sides:\r\n" );
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document.write( " +  + 1 < 0\r\n" );
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document.write( "Th LCD is (x-1)(1+x).  \r\n" );
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document.write( "· + · + 1· < 0\r\n" );
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document.write( " +  +  < 0\r\n" );
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document.write( "Multiply out the tops but not the bottoms:\r\n" );
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document.write( " +  +  < 0\r\n" );
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document.write( " +  +  < 0\r\n" );
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document.write( "Write the sum of the numerators over the LCD:\r\n" );
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document.write( " < 0\r\n" );
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document.write( "Take away the parentheses on top:\r\n" );
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document.write( " < 0\r\n" );
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document.write( " < 0\r\n" );
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document.write( "We find the critical numbers by setting the numerator and\r\n" );
document.write( "the denominator = 0 and solving:\r\n" );
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document.write( "Numerator = 0\r\n" );
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document.write( "x² + 4x - 1 = 0\r\n" );
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document.write( "That does not factor so we must use the quadratic formula:\r\n" );
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document.write( "Fcator out 2 on the top\r\n" );
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document.write( " is approximately 0.24 and is approximately -4.24  \r\n" );
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document.write( "That's two of the critical numbers.\r\n" );
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document.write( "Set denominator = 0\r\n" );
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document.write( "(x - 1)(1 + x) = 0\r\n" );
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document.write( "x - 1 = 0;  1 + x = 0\r\n" );
document.write( "    x = 1;      x = -1\r\n" );
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document.write( "So the four critical numbers are\r\n" );
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document.write( " which is approximately 0.24 \r\n" );
document.write( " which is approximately -4.24\r\n" );
document.write( "1\r\n" );
document.write( "-1\r\n" );
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document.write( "We put those in order smallest to largest:\r\n" );
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document.write( " which is approximately -4.24\r\n" );
document.write( "-1\r\n" );
document.write( " which is approximately 0.24\r\n" );
document.write( "1\r\n" );
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document.write( "The possible solution intervals are the intervals between and beyond\r\n" );
document.write( "the critical numbers.  None of the critical numbers are solutions, since\r\n" );
document.write( "the inequality is < and not < so all the possible intervals are \r\n" );
document.write( "open. They are:\r\n" );
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document.write( "We pick a test value in each interval:\r\n" );
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document.write( ", pick test value -5\r\n" );
document.write( ", pick test value -2\r\n" );
document.write( ", pick test value 0 \r\n" );
document.write( ", pick test value .5\r\n" );
document.write( " 2\r\n" );
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document.write( "We substitute each test value into\r\n" );
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document.write( " < 0\r\n" );
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document.write( "Substituting test value -5\r\n" );
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document.write( " < 0\r\n" );
document.write( " < 0\r\n" );
document.write( " < 0\r\n" );
document.write( " < 0\r\n" );
document.write( "That is false so \r\n" );
document.write( "is not part of the solution\r\n" );
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document.write( "Substituting test value -2\r\n" );
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document.write( " < 0\r\n" );
document.write( " < 0\r\n" );
document.write( " < 0\r\n" );
document.write( " < 0\r\n" );
document.write( "That is true so  \r\n" );
document.write( "is part of the solution\r\n" );
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document.write( "Substituting test value 0\r\n" );
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document.write( " < 0\r\n" );
document.write( " < 0\r\n" );
document.write( " < 0\r\n" );
document.write( "1 < 0\r\n" );
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document.write( "That is false so  \r\n" );
document.write( "is not part of the solution\r\n" );
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document.write( "Substituting test value .5\r\n" );
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document.write( " < 0\r\n" );
document.write( " < 0\r\n" );
document.write( " < 0\r\n" );
document.write( " < 0\r\n" );
document.write( " < 0\r\n" );
document.write( "That is true so ,  \r\n" );
document.write( "is part of the solution\r\n" );
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document.write( "Substituting test value 2\r\n" );
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document.write( " < 0\r\n" );
document.write( " < 0\r\n" );
document.write( " < 0\r\n" );
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document.write( "That is false so  \r\n" );
document.write( "is not part of the solution.\r\n" );
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document.write( "So the solution is\r\n" );
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document.write( " ⋃ \r\n" );
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document.write( "Edwin
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