document.write( "Question 212493: Joe invested a portion of $6000 at 4.5% interest and the balance at 6% interest. How much did he invest at 6% if his total income from both investments is $279? \n" ); document.write( "
Algebra.Com's Answer #160482 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( ".015X=9
\n" ); document.write( "X=9/.015
\n" ); document.write( "X=$600 WAS INVESTED @ 6%.
\n" ); document.write( "6,000-600=5,400 INVESTED @ 4.5%
\n" ); document.write( "PROOF:
\n" ); document.write( ".06*600+.045*5,400=279
\n" ); document.write( "36+243=279
\n" ); document.write( "279=279
\n" ); document.write( "
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