document.write( "Question 930503: Solve this equation:
\n" ); document.write( "|x+1| + |1-x| = 2
\n" ); document.write( "Thanks.
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Algebra.Com's Answer #565060 by josgarithmetic(39625)\"\" \"About 
You can put this solution on YOUR website!
Analyze the equation logically. The absolute value inputs can be greater-than-or-equal to zero, or less than zero.\r
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\n" ); document.write( "\n" ); document.write( "If both nonzero, then \"x%2B1%2B1-x=2\"
\n" ); document.write( "\"2=2\"; and you can find what must be x for both inputs to be nonzero.
\n" ); document.write( "\"x%2B1%3E=0\"
\n" ); document.write( "\"x%3E=-1\"
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\n" ); document.write( "\"1-x%3E=0\"
\n" ); document.write( "\"1%3E=x\"
\n" ); document.write( "\"x%3C=1\"
\n" ); document.write( "Meaning:
\n" ); document.write( "\"highlight%28-1%3C=x%3C=1%29\"\r
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\n" ); document.write( "\n" ); document.write( "IF both are negative, then \"-x-1%2B%28-1%29%2Bx=2\"
\n" ); document.write( "\"-2=2\" FALSE statement. Obviously both expressions in the absolute value inputs must not be simultaneously negative.\r
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\n" ); document.write( "\n" ); document.write( "If x+1>=0 and 1-x<0, then \"x%2B1%2B%28-1%29%2Bx=2\"
\n" ); document.write( "\"2x=2\"
\n" ); document.write( "\"x=1\", which is only a piece of the solution found in \"-1%3C=x%3C=1\".\r
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\n" ); document.write( "\n" ); document.write( "If x+1<0 and 1-x>=0, then \"-x-1%2B1-x=2\"
\n" ); document.write( "\"-2x=2\"
\n" ); document.write( "\"x=-1\", again this one is consistent with the first found inequality solution but is only one piece of it.\r
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\n" ); document.write( "\n" ); document.write( "FINAL FINISHED ANSWER RESULT: \"highlight%28-1%3C=x%3C=1%29\"
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