document.write( "Question 1167264: Larry mitchel invested part of his $13,000 advance at 7% annual simple interest and the rest at 3% annual simple interest. If his total yearly interest from both accounts was $870, find the amount invested at each rate \n" ); document.write( "

Algebra.Com's Answer #791873 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "Let x = invested at 7%\r\n" );
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document.write( "Then the amount invested at 3% is  (13000-x)  dollars.\r\n" );
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document.write( "The total interest equation is\r\n" );
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document.write( "    0.07x + 0.03*(13000-x) = 870  dollars.\r\n" );
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document.write( "From the equation\r\n" );
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document.write( "    x =  = 12000.\r\n" );
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document.write( "ANSWER.  $12000 invested at 7% and the rest  $1000 invested at 3%.\r\n" );
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document.write( "CHECK.  0.07*12000 + 0.03*1000 = 870  dollars of total interest.  ! Correct !\r\n" );
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