document.write( "Question 1101999: Michelle earned a total of $549 in simple interest from two separate accounts. In an account earning 9% interest, Michelle invested $2,300 more than twice the amount she invested in an account earning 1%. How much did she invest in each account? \n" ); document.write( "

Algebra.Com's Answer #716694 by ikleyn(52915) $\"\"$ $\"About$

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document.write( "From the condition, you have these two equations\r\n" );
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document.write( "x = 2300 + 2y,           (1)   (\"In an account earning 9% interest, Michelle invested $2,300 more than twice the amount \r\n" );
document.write( "                                 she invested in an account earning 1%.\")\r\n" );
document.write( "0.09x + 0.01y = 549      (2)   (\"earned a total of $549 in simple interest from two separate accounts\")\r\n" );
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document.write( "Modify the equations (the system) to the standard form.  On the way, multiply by 100 both sides in eq(2). You will get\r\n" );
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document.write( " x - 2y = 2300,          (3)\r\n" );
document.write( "9x +  y = 54900.         (4)\r\n" );
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document.write( "Multiply eq(4) by 2 (both sides), by keeping eq(3) without change. You will get\r\n" );
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document.write( "  x - 2y =   2300,       (5)\r\n" );
document.write( "18x + 2y = 109800.       (6)\r\n" );
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document.write( "Now add eq(5) and eq(6). The terms \"-2y\" and \"2y\" will cancel each other, and you will get a single equation for x:\r\n" );
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document.write( "19x      = 2300 + 109800 = 112100  ====>  x =  = 5900.\r\n" );
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document.write( "Thus 5900 dollars were invested at 9%.\r\n" );
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document.write( "Then from eq(1),  2y = 5900 - 2300 = 3600.   Hence,  y =  = 1800.\r\n" );
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document.write( "Answer.  $5900 invested at 9%  and  $1800 invested at 1%.\r\n" );
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document.write( "Check.   0.09*5900 + 0.01*1800 = 549.   ! the solution is correct !\r\n" );
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\n" ); document.write( "\n" ); document.write( "For similar problems on investments see the lesson\r
\n" ); document.write( "\n" ); document.write( " - Using systems of equations to solve problems on investment\r
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