document.write( "Question 160601: Mike invested $6000 for one year. he invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest. How much did he invest at each rate? \n" ); document.write( "
Algebra.Com's Answer #118449 by checkley77(12844)\"\" \"About 
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.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 @8%.
\n" ); document.write( "Proof:
\n" ); document.write( ".11*4200+.09*1800=624
\n" ); document.write( "462+162=624
\n" ); document.write( "624=624
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