Question 279801
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Let *[tex \Large n] represent the number of nickels. Let *[tex \Large d] represent the number of dimes.  Each nickel is worth 5 cents, so the value of the nickels in cents is *[tex \Large 5n].  Likewise, the value of the dimes is *[tex \Large 10d] cents.  Finally, the total amount of money is $3.20 which is the same as 320 cents.  Putting it all together you have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n\ +\ d\ =\ 38]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5n\ +\ 10d\ =\ 320]


If you solve the first equation for *[tex \Large n] in terms of *[tex \Large d] you get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n\ =\ 38\ -\ d]


giving you an expression involving *[tex \Large d] that is equal to *[tex \Large  n] and that can be substituted in place of *[tex \Large n] in the other equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\left(38\ -\ d\right)\ +\ 10d\ =\ 320]


Now solve for *[tex \Large d].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 190\ -\ 5d\ +\ 10d\ =\ 320]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 190\ +\ 5d\ =\ 320]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5d\ =\ 320\ -\ 190\ =\ 130]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 26]


So there are 26 dimes and therefore there must be *[tex \Large 38\ -\ 26\ =\ 12] nickels.


Check:  26 dimes are worth $2.60, and 12 nickels are worth $0.60.  $2.60 plus $0.60 is $3.20.  Checks.


<font size="+4"><b><i>Super Double Plus Extra Credit:</i></b></font>


Solve the system using either the Elimination Method or Cramer's Rule.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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