document.write( "Question 986580: a woman invested a sum of money at 6% and 3000 dollars more than 9%. If the total interest earned in one year is 4710 dollars, how much was invested at each rate? \n" ); document.write( "
Algebra.Com's Answer #607376 by macston(5194)\"\" \"About 
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\n" ); document.write( "x=amount invested at 6%
\n" ); document.write( ".
\n" ); document.write( ".06x+.09(x+$3000)=$4710
\n" ); document.write( "0.06x+0.09x+$270=$4710
\n" ); document.write( "0.15x=$4440
\n" ); document.write( "x=$29600
\n" ); document.write( "ANSWER 1:$29600 was invested at 6%.
\n" ); document.write( "Amount at 9%=x+$3000
\n" ); document.write( "Amount At 9%=$29600+$3000=$32600
\n" ); document.write( "ANSWER 2: $32600 was invested at 9%.
\n" ); document.write( ".
\n" ); document.write( "CHECK:
\n" ); document.write( ".06($29600)+0.09($32600)=$4710
\n" ); document.write( "$1776+$2934=$4710
\n" ); document.write( "$4710=$4710
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