Question 234633
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Let *[tex \Large d] represent the number of dimes.  Let *[tex \Large q] represent the number of quarters.  Each dime is worth 10 cents, so the value of the dimes, in cents is *[tex \Large 10d].  Likewise, the value of the quarters is *[tex \Large 25q].  And the total value of all the coins, in cents, is 515.


Now we can write an equation that describes the number of coins:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ +\ q\ =\ 35]


And another equation that describes the value of the coins:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10d\ +\ 25q\ =\ 515]


Giving you a system of two equations in two variables.  Solve for *[tex \Large d] and *[tex \Large q].  Since you have unity coefficients on the first equation, I would use the substitution method to solve this system.


Which leads me to the description of a shortcut method for solving coin and different value ticket problems.  Instead of defining a second variable as above, i.e. "Let *[tex \Large q] represent the number of quarters" you can realize up front that there are 35 coins, so if *[tex \Large d] represents the number of dimes, *[tex \Large 35\ -\ d] must represent the number of quarters.  Then you can just write the single equation for the value thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10d\ +\ 25(35\ -\ d)\ =\ 515]


Which is exactly the same place you would be if you solved the first equation for *[tex \Large q] and then substituted into the second equation.


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