document.write( "Question 1193974: How can $69,000 be invested, part at 12% annual simple interest and the remainder at 11% annual simple interest, so that the interest earned by the two accounts will be equal? \n" ); document.write( "

Algebra.Com's Answer #826055 by ikleyn(52781) $\"\"$ $\"About$

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\n" ); document.write( "How can $69,000 be invested, part at 12% annual simple interest and the remainder at 11% annual simple interest,
\n" ); document.write( "so that the interest earned by the two accounts will be equal?
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document.write( "x dollars invested at 12%;  the rest (69000-x) dollars invested at 11%.\r\n" );
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document.write( "Write equation as you read the problem\r\n" );
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document.write( "    0.12x = 0.11*(69000-x).\r\n" );
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document.write( "Simplify and find x\r\n" );
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document.write( "    0.12x = 0.11*69000 - 0.11x\r\n" );
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document.write( "    0.12x + 0.11x = 0.11*69000\r\n" );
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document.write( "         0.23x    = 7590\r\n" );
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document.write( "             x    = 7590/0.23 = 33000.\r\n" );
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document.write( "ANSWER.  $33000 invested at 12%;  the rest  69000-33000 = 36000 invested at 11%.\r\n" );
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document.write( "CHECK.  0.12*33000 = 3960 dollars, the annual interest from the 12% investment.\r\n" );
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document.write( "        0.11*36000 = 3960 dollars, the annual interest from the 11% investment.\r\n" );
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document.write( "        the interest is the same, so the answer is correct.\r\n" );
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