document.write( "Question 1169265: 4x-7/5x+1≥2
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Algebra.Com's Answer #793925 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!
assuming you have\r
\n" ); document.write( "\n" ); document.write( "\"%284x-7%29%2F%285x%2B1%29%3E=2\" .....\r
\n" ); document.write( "\n" ); document.write( "first denominator cannot be zero, so if \"5x%2B1=0\" =>\"5x=-1\"=>\"x=-1%2F5\"
\n" ); document.write( "so,\"+x%3C%3E-1%2F5+\"\r
\n" ); document.write( "\n" ); document.write( "\"%284x-7%29%3E=2%285x%2B1%29\"\r
\n" ); document.write( "\n" ); document.write( "\"4x-7%3E=10x%2B2\"\r
\n" ); document.write( "\n" ); document.write( "\"-7-2%3E=10x-4x\"\r
\n" ); document.write( "\n" ); document.write( "\"-9%3E=6x\"\r
\n" ); document.write( "\n" ); document.write( "\"-9%2F6%3E=x\"\r
\n" ); document.write( "\n" ); document.write( "\"x%3C=-3%2F2\"\r
\n" ); document.write( "\n" ); document.write( " \"-3%2F2%3C=+x+%3C-1%2F5+\" ( \"x\" is less than \"-1%2F5\" because \"x%3C%3E-1%2F5\")\r
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