Question 244705
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Let *[tex \Large q] represent the number of quarters.


Let *[tex \Large n] represent the number of nickels.


Let *[tex \Large p] represent the number of pennies.


Then *[tex \Large 25q] is the value of the quarters in cents.


And *[tex \Large 5n] is the value of the nickels in cents.


And *[tex \Large p] is the value of the pennies in cents.


The total value of the coins, in dollars, is $37.47.  Therefore the total value of the coins in cents is 3747.


There are 275 coins, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ q\ +\ n\ +\ p\ =\ 275]


The number of quarters is 46 less than twice the number of nickels, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ q\ =\ 2n\ -\ 46]


The face value of the coins, in cents, is 3747, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25q\ +\ 5n\ +\ p\ =\ 3747]


Now you have three equations in three variables.  Solve the system.  The coordinates of the ordered triple solution set, *[tex \LARGE \left(q,\,n,\,p\right)] are the answer to the question.


Hint:  I would proceed by eliminating the *[tex \LARGE q] variable in the first and third equations by substitution of the expression equal to *[tex \LARGE q] in the second equation.  After simplifying both the new equations for the first and third and putting them into standard form, I would multiply either one of them by -1 and then solve for *[tex \LARGE n] by elimination of *[tex \LARGE p]. 


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