document.write( "Question 898956: Suppose you invest money in two accounts. One of the accounts pay 8% annual interest, whereas the other pays 9% annual interest. If you have $8,000 more invested at 9% than you invested at 8%, how much do you have invested in each account if the total amount of interest you earn in a year is $1,060?
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Algebra.Com's Answer #545108 by mananth(16946) $\"\"$ $\"About$

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One of the accounts pay 8% annual interest,-----------x
\n" ); document.write( "other pays 9% -----x+8000
\n" ); document.write( " the total amount of interest you earn in a year is $1,060\r
\n" ); document.write( "\n" ); document.write( "8%x+9%(x+8000) = 1060\r
\n" ); document.write( "\n" ); document.write( "multiply by 100
\n" ); document.write( "8x+9(x+8000( = 106000\r
\n" ); document.write( "\n" ); document.write( "8x+9x+72000=106000\r
\n" ); document.write( "\n" ); document.write( "17x=106000-72000\r
\n" ); document.write( "\n" ); document.write( "17x=34000\r
\n" ); document.write( "\n" ); document.write( "x=2000\r
\n" ); document.write( "\n" ); document.write( "2000+8000= 10,000\r
\n" ); document.write( "\n" ); document.write( "2000 at 8 %
\n" ); document.write( "10,000 at 9%\r
\n" ); document.write( "\n" ); document.write( "check\r
\n" ); document.write( "\n" ); document.write( "8%*2000 + 9%*10000=1060\r
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