document.write( "Question 1023076: A collection of nickels dimes and quarters amount to $10.00 there are 140 coins and there is twice as many dimes as quarters, find the number of nickels \n" ); document.write( "

Algebra.Com's Answer #638639 by ikleyn(52791) $\"\"$ $\"About$

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\n" ); document.write( "A collection of nickels dimes and quarters amount to $10.00 there are 140 coins and there is twice as many dimes as quarters, find the number of nickels
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document.write( "Your equations are\r\n" );
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document.write( " n +   d +   q =  140,    (1)\r\n" );
document.write( "5n + 10d + 25q = 1000,    (2)\r\n" );
document.write( "d = 2q.                   (3)\r\n" );
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document.write( "(three equations in 3 unknowns). Substitute (3) into (1) and (2). \r\n" );
document.write( "In this way you eliminate d and reduce the system to 2 equations in 2 unknowns:\r\n" );
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document.write( "n + 2q + q = 140,         (4)   (instead of (1))\r\n" );
document.write( "5n + 10*2q + 25q = 1000.  (5)   (instead of (2))\r\n" );
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document.write( "Now simplify equations (4) and (5)\r\n" );
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document.write( " n +  3q =  140,          (6)\r\n" );
document.write( "5n + 45q = 1000.          (7)\r\n" );
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document.write( "Next express n = 140 - 3q from (6) and substitute it into (7). You will get a single equation for q\r\n" );
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document.write( "5*(140-3q) + 45q = 1000.  (8)\r\n" );
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document.write( "Simplify (8)\r\n" );
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document.write( "700 - 15q + 45q = 1000,\r\n" );
document.write( "30q = 1000-700 = 300,\r\n" );
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document.write( "q =  = 10.\r\n" );
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document.write( "Now n = 140 - 3q = 140 - 3*10 = 140 - 30 = 110,  and\r\n" );
document.write( "    d = 2q = 2*10 = 20.\r\n" );
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document.write( "Answer. n = 110,  d = 20  and  q = 10.\r\n" );
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