Question 247355
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Let *[tex \Large n] represent the number of nickels.


Let *[tex \Large d] represent the number of dimes.


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


So altogether you have


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n\ +\ d\ +\ q\ =\ 20]


total coins.


Then *[tex \Large 5n] is the value of the nickels in cents.  *[tex \Large 10d] is the value of the dimes in cents, and *[tex \Large 25q] is the value of the quarters in cents.


All together, you have


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5n\ +\ 10d\ +\ 25q\ =\ 335]


expressed in cents.


But if you had *[tex \Large n] quarters, you would have *[tex \Large 25n] cents, and if you had *[tex \Large d] nickels, you would have *[tex \Large 5d] cents, and if you had *[tex \Large q] dimes, you would have *[tex \Large 10q] cents, and that would all add up to 275 cents, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25n\ +\ 5d\ +\ 10q\ =\ 275]


Now you have a system of three equations in three variables to solve.  Since the problem only asks for the number of quarters, you really only need solve for *[tex \Large q], but I would solve for all three variables and plug the values back in to the original equations in order to check my work.



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