document.write( "Question 1147235: Suppose you invest a certain amount of money in an account that pays 6% interest annually and $1000 more in an account that pays 2% annually how much money was invested in each account if the total interest for a year is $660 \n" ); document.write( "

Algebra.Com's Answer #768558 by ikleyn(52914) $\"\"$ $\"About$

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document.write( "Let x be the amount invested at 6%.\r\n" );
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document.write( "Then the other amount is  (x+1000) dollars.\r\n" );
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document.write( "The total interest is the sum of partial interests\r\n" );
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document.write( "    0.06x + 0.02*(x+1000) = 660.\r\n" );
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document.write( "Express x and calculate\r\n" );
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document.write( "    x =  = 8000  dollars.\r\n" );
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document.write( "Answer.  $8000 invested at 6%  and  8000+1000 = 9000 dollars invested at 2%.\r\n" );
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document.write( "CHECK.  0.06*8000 + 0.02*9000 = 660  dollas.    ! Precisely correct !\r\n" );
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