document.write( "Question 186653: please solve this equation thank you\r
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Algebra.Com's Answer #139932 by Edwin McCravy(20063)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Get 0 on the right\r\n" );
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document.write( "\"1%2F%28x%2B1%29-1%2F%28x-1%29%3C0\"\r\n" );
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document.write( "Get LCD of \"%28x%2B1%29%28x-1%29\"\r\n" );
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document.write( "\"%28%28x-1%29-%28x%2B1%29%29%2F%28%28x%2B1%29%28x-1%29%29%3C0\"\r\n" );
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document.write( "\"%28x-1-x-1%29%2F%28%28x%2B1%29%28x-1%29%29%3C0\"\r\n" );
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document.write( "\"-2%2F%28x%2B1%29%28x-1%29%3C0\"\r\n" );
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document.write( "Critical values are gotten by setting\r\n" );
document.write( "numerators and denominators = 0 and solving.\r\n" );
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document.write( "We can't set the numerator = 0 since it's -2\r\n" );
document.write( "We set the denominator = 0\r\n" );
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document.write( "\"%28x%2B1%29%28x-1%29=0\"\r\n" );
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document.write( "\"x=-1\" and \"x=1\"\r\n" );
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document.write( "Now we make a number line and mark those critical values\r\n" );
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document.write( " -----------o---------o-----------\r\n" );
document.write( " -3   -2   -1    0    1    2    3\r\n" );
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document.write( "Now we choose any value to the left of -1, say -2\r\n" );
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document.write( "Substitute it into\r\n" );
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document.write( "\"-2%2F%28x%2B1%29%28x-1%29%3C0\"\r\n" );
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document.write( "\"-2%2F%28-2%2B1%29%28-2-1%29%3C0\"\r\n" );
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document.write( "\"-2%2F%28%28-1%29%28-3%29%29%3C0\"\r\n" );
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document.write( "\"-2%2F3%3C0\"\r\n" );
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document.write( "That is true so we shade the part of\r\n" );
document.write( "the number line to the left of -1:\r\n" );
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document.write( "<===========o---------o-----------\r\n" );
document.write( " -3   -2   -1    0    1    2    3\r\n" );
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document.write( "Next we choose any value between -1 and 1, \r\n" );
document.write( "say, 0 and substitute it into\r\n" );
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document.write( "\"-2%2F%28x%2B1%29%28x-1%29%3C0\"\r\n" );
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document.write( "\"-2%2F%280%2B1%29%280-1%29%3C0\"\r\n" );
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document.write( "\"-2%2F%28%281%29%28-1%29%29%3C0\"\r\n" );
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document.write( "\"-2%2F%28-1%29%3C0\"\r\n" );
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document.write( "\"2+%3C+0\"\r\n" );
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document.write( "That is false so we do not shade the part of\r\n" );
document.write( "the number line between -1 and 1, so we still\r\n" );
document.write( "have:\r\n" );
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document.write( "<===========o---------o-----------\r\n" );
document.write( " -3   -2   -1    0    1    2    3\r\n" );
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document.write( "Next we choose any value to the right of 1, say 2\r\n" );
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document.write( "Substitute it into\r\n" );
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document.write( "\"-2%2F%282%2B1%29%282-1%29%3C0\"\r\n" );
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document.write( "\"-2%2F%282%2B1%29%282-1%29%3C0\"\r\n" );
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document.write( "\"-2%2F%28%283%29%281%29%29%3C0\"\r\n" );
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document.write( "\"-2%2F3%3C0\"\r\n" );
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document.write( "That is true so we shade the part of\r\n" );
document.write( "the number line to the right of 1:\r\n" );
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document.write( "<===========o---------o==========>\r\n" );
document.write( " -3   -2   -1    0    1    2    3\r\n" );
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document.write( "The critical values do not satisfy the inequality\r\n" );
document.write( "because they cause 0's in the denominator, so\r\n" );
document.write( "we leave them open.\r\n" );
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document.write( "Now we translate this graph into interval notation:\r\n" );
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document.write( "   \r\n" );
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document.write( "Edwin
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