document.write( "Question 64818: Roberto invested some money at 7%, and then investes $5000 more than twice this amount at 11%. His total annual income from the two investments was $3450. How much was invested at 11%? \n" ); document.write( "
Algebra.Com's Answer #45329 by checkley71(8403)![]() ![]() ![]() You can put this solution on YOUR website! .07x+.11(2x+5000)=3450 \n" ); document.write( ".07x+.22x+550=3450 \n" ); document.write( ".29x=3450-550 \n" ); document.write( ".29x=2900 \n" ); document.write( "x=2900/.29 \n" ); document.write( "x=10000 amount invested @ 7% \n" ); document.write( "2*10000+5000=20000+5000=25000 amount invested @ 11% \n" ); document.write( "proof \n" ); document.write( ".07*10000+.11*25000=3450 \n" ); document.write( "700+2750=3450 \n" ); document.write( "3450=3450 \n" ); document.write( " |