document.write( "Question 177072This question is from textbook
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document.write( ": 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" );
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Algebra.Com's Answer #132142 by checkley75(3666)![]() ![]() ![]() 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=4,200 invested @ 11% \n" ); document.write( "6,000-4,200=1,800 invested @ 9% \n" ); document.write( "Proof: \n" ); document.write( ".11*4,200+.09*1,800=624 \n" ); document.write( "462+162=624 \n" ); document.write( "624=624 \n" ); document.write( " |