document.write( "Question 1007644: Warren has 40 coins (all nickels, dimes, and quarters) worth $4.05. He has 7 nore nickels than dimes. How many quarters does Warren have? \n" ); document.write( "

Algebra.Com's Answer #623565 by ikleyn(52795) $\"\"$ $\"About$

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\n" ); document.write( "Warren has 40 coins (all nickels, dimes, and quarters) worth $4.05. He has 7 more nickels than dimes. How many quarters does Warren have?
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document.write( "The condition gives you these three equations for three unknowns n, d and q\r\n" );
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document.write( " n +   d +   q = 40,       (1)\r\n" );
document.write( "5n + 10d + 25q = 405,      (2)\r\n" );
document.write( " n             = d + 7     (3)\r\n" );
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document.write( "where n = number of nickels, d = number of dimes, q = number of quarters.\r\n" );
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document.write( "To solve it, first reduce (2) by factor (5). You will get (2') instead of (2)\r\n" );
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document.write( " n +  2d +  5q = 81        (2')\r\n" );
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document.write( "Next, substitute the expression (3) for n into (1) and (2'). You will get\r\n" );
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document.write( " (d + 7) +  d +  q = 40,   (3)\r\n" );
document.write( " (d + 7) + 2d + 5q = 81.   (4)\r\n" );
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document.write( "Simplify (3) and (4):\r\n" );
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document.write( " 2d +  q = 33,             (5)\r\n" );
document.write( " 3d + 5q = 74.             (6)\r\n" );
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document.write( "Multiply (5) by (-3) (both sides); multiply (6) by 2 and then add. In this way, you will exclude d and will get the single equation for q:\r\n" );
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document.write( " 10q - 3q = 2*74 - 3*33 = 148 - 99 = 49.\r\n" );
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document.write( "It gives\r\n" );
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document.write( "  7q = 49,\r\n" );
document.write( "   q =  = 7.\r\n" );
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document.write( "Now from (5)\r\n" );
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document.write( "   d = (33 - 7)/2 = 13,\r\n" );
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document.write( "and from (3)\r\n" );
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document.write( "   n = 20.\r\n" );
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document.write( "Answer. n = 20, d = 13, q = 7.\r\n" );
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