document.write( "Question 104549: A total of $8000 is invested, part at 6% and the remainder at 9%. How much is invested at each rate if the annual interest is $540? \n" ); document.write( "
Algebra.Com's Answer #76065 by checkley75(3666)![]() ![]() ![]() You can put this solution on YOUR website! .09X+.06(8000-X)=540 \n" ); document.write( ".09X+480-.06X=540 \n" ); document.write( ".03X=540-480 \n" ); document.write( ".03X=60 \n" ); document.write( "X=60/.03 \n" ); document.write( "X=2000 INVESTED @ 9% \n" ); document.write( "8000-2000=6000 INVESTED @ 6%. \n" ); document.write( "PROOF \n" ); document.write( ".09*2000+.06*6000=540 \n" ); document.write( "180+360=540 \n" ); document.write( "540=540 \n" ); document.write( " |