document.write( "Question 416299: I need help on this problem. Lisa's piggy bank contained 21 dimes, nickels, and pennies. The total in the piggy bank was $1.21. Lisa removed all the dimes from the piggy bank. This left only 41 cents in the bank. How many pennies were in the bank? I did three equations:
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\n" ); document.write( "10d+5n+1p=121
\n" ); document.write( "5n+1p=41
\n" ); document.write( "I tried to subtract the frist two equations first, then work with the answer I got and tried to subtract the third one. Can you please show me how it's done correctly.
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Algebra.Com's Answer #291685 by Edwin McCravy(20059)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "The other tutor's answer is wrong. \r\n" );
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document.write( "  d +  n + p =  21\r\n" );
document.write( "10d + 5n + p = 121\r\n" );
document.write( "      5n + p =  41\r\n" );
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\n" ); document.write( "I tried to subtract the frist two equations first,
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document.write( "That was OK.  You eliminated p by subtracting the 1st from the second: \r\n" );
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document.write( "   10d + 5n + p = 121\r\n" );
document.write( " -(  d +  n + p =  21)\r\n" );
document.write( " ---------------------\r\n" );
document.write( "    9d + 4n     = 100 \r\n" );
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\n" ); document.write( "then work with the answer I got and tried to subtract the third one.
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document.write( "No that's no good because you eliminated p from the first two,\r\n" );
document.write( "so you must also eliminate p from a different pair of equations \r\n" );
document.write( "both of which have p in them.  What you got does not contain p.  So\r\n" );
document.write( "you must first eliminate p from the 1st and 3rd, by subtracting the \r\n" );
document.write( "3rd from the 1st:   \r\n" );
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document.write( "     d +  n + p =  21\r\n" );
document.write( "   -(    5n + p =  41)\r\n" );
document.write( "   -------------------\r\n" );
document.write( "     d - 4n     = -20\r\n" );
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document.write( "Now take what you got from subtracting the first two: \r\n" );
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document.write( "    9d + 4n     = 100\r\n" );
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document.write( "and put that with it and add the two equations:\r\n" );
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document.write( "    9d + 4n     = 100\r\n" );
document.write( "     d - 4n     = -20\r\n" );
document.write( "    -----------------\r\n" );
document.write( "   10d          =  80\r\n" );
document.write( "              d =   8\r\n" );
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document.write( "Now substitute 8 for d in  \r\n" );
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document.write( "         d - 4n = -20\r\n" );
document.write( "         8 - 4n = -20 \r\n" );
document.write( "            -4n = -28\r\n" );
document.write( "              n =   7\r\n" );
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document.write( "Now substitute both d = 8 and n = 7 in one of the original equations, \r\n" );
document.write( "say, the easiest one:\r\n" );
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document.write( "      d + n + p = 21\r\n" );
document.write( "      8 + 7 + p = 21\r\n" );
document.write( "         15 + p = 21 \r\n" );
document.write( "              p = 6\r\n" );
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document.write( "So the solution is (d,n,p) = (8,7,6)\r\n" );
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document.write( "Now we check:\r\n" );
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document.write( "8   dimes = 8 coins\r\n" );
document.write( "7 nickels = 7 coins\r\n" );
document.write( "6 pennies = 6 coins\r\n" );
document.write( "-------------------\r\n" );
document.write( "           21 coins, that checks!  \r\n" );
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document.write( "8 dimes   = 80 cents\r\n" );
document.write( "7 nickels = 35 cents\r\n" );
document.write( "6 pennies =  6 cents\r\n" );
document.write( "--------------------\r\n" );
document.write( "           121 cents or $1.21, that checks!\r\n" );
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document.write( "If you take the dimes away, you have only the 7 nickels\r\n" );
document.write( "and the 6 pennies\r\n" );
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document.write( "7 nickels = 35 cents\r\n" );
document.write( "6 pennies =  6 cents\r\n" );
document.write( "--------------------\r\n" );
document.write( "            41 cents, that checks!\r\n" );
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document.write( "So the answer is correct.\r\n" );
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document.write( "Edwin
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