document.write( "Question 183535: Mike invested $6000.00 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624.00 in interest. How much did he
\n" );
document.write( "invest at each rate? \n" );
document.write( "
Algebra.Com's Answer #137773 by checkley77(12844) ![]() You can put this solution on YOUR website! .11x+.09(6000-x)=624 \n" ); document.write( ".11x+540-.09x=624 \n" ); document.write( ".02x=624-540 \n" ); document.write( ".02x=84 \n" ); document.write( "x=84/.02 \n" ); document.write( "x=4200 invested @ 11%. \n" ); document.write( "6000-4200=1800 invested @ 9%. \n" ); document.write( "Proof: \n" ); document.write( ".11*4200+.09*1800=624 \n" ); document.write( "462+162=624 \n" ); document.write( "624=624 \n" ); document.write( " \n" ); document.write( " |