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)![]() ![]() ![]() You can put this solution on YOUR website! .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 \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |