document.write( "Question 331859: roberto invested some money at 7%, and then invested $3000 more than twice this ammount at 9%. his total annual income for two investments was $3200. how much was invested at 9% \n" ); document.write( "
Algebra.Com's Answer #237922 by mananth(16946)\"\" \"About 
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amount invested be x at 7%
\n" ); document.write( "2x+3000 @ 9%
\n" ); document.write( "total interest for 1 year =3200\r
\n" ); document.write( "\n" ); document.write( "0.07x+0.09*(2x+3000)=3200
\n" ); document.write( "0.07x+0.18x+270=3200
\n" ); document.write( "0.25x=3200-270
\n" ); document.write( "0.25x=2930
\n" ); document.write( "x=2930/0.25
\n" ); document.write( "$11,720 at 7%
\n" ); document.write( "..
\n" ); document.write( "2x+3000
\n" ); document.write( "2*11720+300
\n" ); document.write( "23,440+300
\n" ); document.write( "=$23,740 @ 9%\r
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