document.write( "Question 113012: Roberto invested some money at 7%, and then invested $2000 more than twice this amount at 11%. His total annual income from the two investments was $3990. How much was invested at 11%? \n" ); document.write( "
Algebra.Com's Answer #82319 by SHUgrad05(58)![]() ![]() ![]() You can put this solution on YOUR website! .07x+.11y=3990 \n" ); document.write( "Let x=amount invested at 7% \n" ); document.write( "Let y=amount invested at 11%; y=2x+2000\r \n" ); document.write( "\n" ); document.write( ".07x+(2x+2000).11=3990 \n" ); document.write( ".07x+.22x+220=3990 \n" ); document.write( ".07x+.22x=3770 \n" ); document.write( ".29x=3770 \n" ); document.write( "x=$13,000-> plug back into original equation\r \n" ); document.write( "\n" ); document.write( ".07(13,000)+.11y=3990 \n" ); document.write( "910+.11y=3990 \n" ); document.write( ".11y=3080 \n" ); document.write( "y=$28,000\r \n" ); document.write( "\n" ); document.write( "$28,000 invested at 11%\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |