document.write( "Question 217380: Mary Auburn makes three investm,ents. The investment that pays 12% is twice the amount of the account that pays 9%. The amount inveted at 10% is $500 more than the amount invested at 9%. If the annual interest income is $1555, how much money is invested at each rate?\r
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Algebra.Com's Answer #164030 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! three investments. \n" ); document.write( "The investment that pays 12% is twice the amount of the account that pays 9%. \n" ); document.write( "The amount inveted at 10% is $500 more than the amount invested at 9%. \n" ); document.write( "If the annual interest income is $1555, how much money is invested at each rate? \n" ); document.write( ": \n" ); document.write( "Let x = amt invested at 9% \n" ); document.write( ": \n" ); document.write( "\"The investment that pays 12% is twice the amount of the account that pays 9%.\" \n" ); document.write( "therefore: \n" ); document.write( "2x = amt invested at 12% \n" ); document.write( ": \n" ); document.write( "\"The amount invested at 10% is $500 more than the amount invested at 9%.\" \n" ); document.write( "therefore: \n" ); document.write( "(x+500) = amt invested at 10% \n" ); document.write( ": \n" ); document.write( "If the annual interest income is $1555, how much money is invested at each rate? \n" ); document.write( ".09x + .12(2x) + .10(x+500) = 1555 \n" ); document.write( ".09x + .24x + .10x + 50 = 1555 \n" ); document.write( ".43x = 1555 - 50 \n" ); document.write( ".43x = 1505 \n" ); document.write( "x = \n" ); document.write( "x = $3500 invested at 9% \n" ); document.write( "then \n" ); document.write( "2(2500) = $7000 invested at 12% \n" ); document.write( "and \n" ); document.write( "3500 + 500 = $4000 invested at 10% \n" ); document.write( "; \n" ); document.write( ": \n" ); document.write( "Is that true? \n" ); document.write( ".09(3500) + .12(7000) + .10(4000) \n" ); document.write( "315 + 840 + 400 = 1555 \n" ); document.write( " |