document.write( "Question 224152: paul invested $6000 in two accounts. one pay 5% simple interest and other pays 3%. if the interestearned on the 3% investment is $60 more than the interest earned on the 5% investment, how much is investedat each rate? \n" ); document.write( "
Algebra.Com's Answer #167463 by checkley77(12844)\"\" \"About 
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.03x=.05(6,000-x)+60
\n" ); document.write( ".03x=300-.05x+60
\n" ); document.write( ".03x+.05x=360
\n" ); document.write( ".08x=360
\n" ); document.write( "x=360/.08
\n" ); document.write( "x=4,500 invested @ 3%.
\n" ); document.write( "6,000-4,500=1,500 invested @ 5%.
\n" ); document.write( "Proof:
\n" ); document.write( ".03*4,500=.05*1,500+60
\n" ); document.write( "135=75+60
\n" ); document.write( "135=135
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