document.write( "Question 989611: Gary has 42 coins in nickels, dimes, and quarters. If he has 8 more nickels than dimes and has $7.15 in all, how many of each does he have? \n" ); document.write( "
Algebra.Com's Answer #807877 by ikleyn(52831)\"\" \"About 
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\n" ); document.write( "Gary has 42 coins in nickels, dimes, and quarters.
\n" ); document.write( "If he has 8 more nickels than dimes and has $7.15 in all, how many of each does he have?
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\n" ); document.write( "\n" ); document.write( "            It is a typical problem to be solved using one single unknown and one single equation.\r
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document.write( "Let x be the number of nickels;\r\n" );
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document.write( "then the number od dimes is (x-8) and the number of quarters is  (42 - x - (x-8)) = (42+8 - 2x) = (50-2x).\r\n" );
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document.write( "At this point, you can write the total money equation\r\n" );
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document.write( "    5x + 10*(x-8) + 25*(50-2x) = 715  cents.\r\n" );
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document.write( "Simplify and solve\r\n" );
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document.write( "    5x + 10x - 80 + 1250 - 50x = 715\r\n" );
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document.write( "           -35x                = 715 + 80 - 1250\r\n" );
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document.write( "           -35x                = -455\r\n" );
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document.write( "              x                = \"%28-455%29%2F%28-35%29\" = 13.\r\n" );
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document.write( "ANSWER.  13 nickels;  13-8 = 5 dimes  and  the rest 42-13-5 = 24 quarters.\r\n" );
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document.write( "CHECK.   13*5 + 5*10 + 25*24 = 715 cents, in total.    ! Correct !\r\n" );
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