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)\"\" \"About 
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( "
\n" );