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)![]() ![]() ![]() You can put this solution on YOUR website! .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( " |