document.write( "Question 992074: Fred invested some money at 3%, and then invested $5,000 more than twice this amount at 6%. His total annual income from the two investments was $2250. How much was invested at 6%? \n" ); document.write( "
Algebra.Com's Answer #611863 by macston(5194)\"\" \"About 
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\n" ); document.write( "x=amount invested at 3%; 2x+$5000= amount invested at 6%
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\n" ); document.write( "0.03x+0.06(2x+$5000)=$2250
\n" ); document.write( "0.03x+0.12x+$300=$2250
\n" ); document.write( "0.15x=$1950
\n" ); document.write( "x=$13000
\n" ); document.write( ".
\n" ); document.write( "Amount at 6%=2x+$5000=2($13000)+$5000=$26000+$5000=$31000
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\n" ); document.write( "CHECK:
\n" ); document.write( "0.03x+0.06(2x+$5000)=$2250
\n" ); document.write( "0.03($13000)+0.06($31000)=$2250
\n" ); document.write( "$390+$1860=$2250
\n" ); document.write( "$2250=$2250
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