document.write( "Question 196351: Joe invested a portion of $6000 at 4.5% interest and the balance at 6% interest. How much did he invest at each rate if his total income from both investments is $279?
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Algebra.Com's Answer #147200 by checkley77(12844)\"\" \"About 
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.06x+.045(6,000-x)=279
\n" ); document.write( ".06x+270-.045x=279
\n" ); document.write( ".015x=279-270
\n" ); document.write( "x=9/.015
\n" ); document.write( "x=600
\n" ); document.write( "6,000-600=5,400
\n" ); document.write( ".06*600+.045*5,400=279
\n" ); document.write( "36+243=279
\n" ); document.write( "279=279
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