<PRE><B>Question 193326</B>
&lt;font face=&quot;Garamond&quot; size=&quot;+2&quot;&gt;


Let &lt;b&gt;&lt;i&gt;d&lt;/i&gt;&lt;/b&gt; represent the number of dimes.  Let &lt;b&gt;&lt;i&gt;n&lt;/i&gt;&lt;/b&gt; represent the number of nickels.


The &lt;b&gt;&lt;i&gt;value&lt;/i&gt;&lt;/b&gt; of each dime is 10 cents, so the value of the dimes in the collection is &lt;b&gt;10&lt;i&gt;d&lt;/i&gt;&lt;/b&gt;.  Likewise, the value of the nickels is &lt;b&gt;5&lt;i&gt;n&lt;/i&gt;&lt;/b&gt;.  Since we have established the value of the coins in cents, convert the total value of the collection from dollars and cents to just cents -- i.e. move the decimal point.  $6.05 equals 605 cents.


&lt;b&gt;Value equation:&lt;/b&gt;  The value of the dimes added to the value of the nickels must equal the value of the collection.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10d + 5n = 605]


Double the number of dimes is &lt;b&gt;2&lt;i&gt;d&lt;/i&gt;&lt;/b&gt;, so the value of double the number of dimes must be &lt;b&gt;20&lt;i&gt;d&lt;/i&gt;&lt;/b&gt;.  Decreasing the number of nickels by 10 is &lt;b&gt;&lt;i&gt;n&lt;/i&gt; - 10&lt;/b&gt;, so the value of the new amount of nickels must be *[tex \LARGE 5(n - 10)].


&lt;b&gt;&quot;What if&quot; value equation:&lt;/b&gt; The value of the new amount of dimes plus the value of the new amount of nickels must equal value of the &quot;what if&quot; collection (again, expressed in cents):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 20d + 5(n-10) = 985]


Put this second equation into standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 20d + 5n = 1035]


Multiply this equation by -1:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -20d - 5n = -1035]


Add this equation to the first value equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10d - 20d + 5n - 5n = 605 - 1035]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ - 10d + 0n = -430]


Notice that the &lt;b&gt;&lt;i&gt;n&lt;/i&gt;&lt;/b&gt; variable has been eliminated by choosing a multiplier for one of the equations such that the coefficients on &lt;b&gt;&lt;i&gt;n&lt;/i&gt;&lt;/b&gt; became additive inverses.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d = 43]


Substitute this value for &lt;b&gt;&lt;i&gt;d&lt;/i&gt;&lt;/b&gt; into either original value equation; let&#39;s use the first one:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10(43) + 5n = 605]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5n = 175]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n = 35]


So the collection contains 43 dimes worth $4.30, and 35 nickels worth $1.75.  $4.30 plus $1.75 is $6.05.  Double the number of dimes is 86, worth $8.60 and ten less nickels is 25, worth $1.25, $8.60 plus $1.25 is $9.85 so the answer checks.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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