document.write( "Question 142584: Mike invested $6000 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest. How much did he invest at each rate? \n" ); document.write( "
Algebra.Com's Answer #103788 by vleith(2983)\"\" \"About 
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Total invested = 6000.
\n" ); document.write( "Total earned = 624.
\n" ); document.write( "Let x be the part invested at 0.09
\n" ); document.write( "Let y be the part invested at 0.11\r
\n" ); document.write( "\n" ); document.write( "\"6000+=+x+%2B+y\"
\n" ); document.write( "\"6000-x+=+y\"\r
\n" ); document.write( "\n" ); document.write( "\"0.09x+%2B+0.11y+=+624\"
\n" ); document.write( "\"0.09x+%2B+0.11%286000-x%29+=+624\"
\n" ); document.write( "\"0.09x+%2B+660+-+0.11x+=+624\"
\n" ); document.write( "\"-0.02x+=+-36\"
\n" ); document.write( "\"x+=+36%2F%280.02%29\"
\n" ); document.write( "\"x+=+1800\"\r
\n" ); document.write( "\n" ); document.write( "\"y+=+4200\"
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