document.write( "Question 1207434: An investment adviser invested $14,000 in two accounts. One investment earned 6% annual simple interest, and the other investment earned 2.5% annual simple interest. The amount of interest earned for 1 year was $609. How much was invested in each account?
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Algebra.Com's Answer #845284 by ikleyn(52803) $\"\"$ $\"About$

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\n" ); document.write( "An investment adviser invested $14,000 in two accounts.
\n" ); document.write( "One investment earned 6% annual simple interest, and
\n" ); document.write( "the other investment earned 2.5% annual simple interest.
\n" ); document.write( "The amount of interest earned for 1 year was $609.
\n" ); document.write( "How much was invested in each account?
\n" ); document.write( "amount at 6% $
\n" ); document.write( "amount at 2.5%
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document.write( "Let x dollars are invested at 6%.\r\n" );
document.write( "Then the amount invested at 2.5% is (14000-x) dollars.\r\n" );
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document.write( "First amount earns  0.06x           dollars as a year interest.\r\n" );
document.write( "Second amount earns 0.025(14000-x)  dollars as a year interest.\r\n" );
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document.write( "Together, these amounts earn 0.06x + 0.025(14000-x) dollars as a year interest.\r\n" );
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document.write( "So, we write this equation\r\n" );
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document.write( "    0.06x + 0.025(14000-x) = 609  dollars.\r\n" );
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document.write( "The setup is complete.\r\n" );
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document.write( "To solve this equation, simplify it step by step\r\n" );
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document.write( "    0.06x + 0.025*14000 - 0.025x = 609\r\n" );
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document.write( "    0.06x - 0.025 = 609 - 0.025*14000\r\n" );
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document.write( "        0.035x    =     259\r\n" );
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document.write( "             x    =     259/0.035 = 7400.\r\n" );
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document.write( "So, $7400 was invested at 6% and the rest, $14000 - $7400 = $6600  was invested at 0.025%.    ANSWER\r\n" );
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document.write( "CHECK.  0.06*7400 + 0.025*6600 = 609 dollars as a total annual interest.  ! precisely correct !\r\n" );
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