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)![]() ![]() You can put this solution on YOUR website! 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 \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |