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)\"\" \"About 
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.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( "
\n" );