document.write( "Question 1115838: Solve 2/x (lesser to equal than sign) x/2\r
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Algebra.Com's Answer #730679 by josgarithmetic(39794)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\"2%2Fx\" is less than or equal to \"x%2F2\".
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\n" ); document.write( "\n" ); document.write( "\"2%2Fx%3C=x%2F2\"\r
\n" ); document.write( "\n" ); document.write( "-\r
\n" ); document.write( "\n" ); document.write( "\"2%2Fx-x%2F2%3C=0\"\r
\n" ); document.write( "\n" ); document.write( "\"%282%2Fx%29%282%2F2%29-%28x%2F2%29%28x%2Fx%29%3C=0\"\r
\n" ); document.write( "\n" ); document.write( "\"4%2F%282x%29-x%5E2%2F%282x%29%3C=0\"\r
\n" ); document.write( "\n" ); document.write( "\"%284-x%5E2%29%2F%282x%29%3C=0\"\r
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\n" ); document.write( "\n" ); document.write( "\"highlight_green%28%28%282-x%29%282%2Bx%29%29%2F%282x%29%3C=0%29\"\r
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\n" ); document.write( "\n" ); document.write( "The critical values of x can be found from that inequality.\r
\n" ); document.write( "\n" ); document.write( "These critical x values are -2, 0, 2.\r
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\n" ); document.write( "\n" ); document.write( "check the intervals on x in \"-infinity%3Cx%3C=-2\", \"-2%3C=x%3C0\",\"0%3Cx%3C=2\", \"2%3C=x%3Cinfinity\".
\n" ); document.write( "See in which intervals the ORIGINAL inequality is true and in which it is false.\r
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\n" ); document.write( "\n" ); document.write( "TRUE FOR \"-2%3C=x%3C0\" and for \"2%3C=x%3Cinfinity\".
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