Question 409090
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What you have done so far is absolutely spot on correct, but I'm going to show you a little trick that will get rid of all of the ugly decimal coefficients so that you are much less likely to make any arithmetic errors.


While it is true that a dime has a value of 0.10 dollars, it also can be said to have a value of 10 cents.  Likewise a nickel is worth 5 cents.  Furthermore, the total value of the coins in either situation can be said to be 395 cents or 840 cents.


Now we can write the equations as:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10d\ +\ 5n\ =\ 395]


And


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 20d\ +\ 5(n\ +\ 21)\ = 840]


Much tidier, no?


A little manipulation gets us:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10d\ +\ 5n\ =\ 395]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 20d\ +\ 5n\ = 735]


Multiply the first equation by -1:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -10d\ -\ 5n\ =\ -395]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 20d\ +\ 5n\ = 735]


Then add the two equations, term by term:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10d\ +\ 0n\ =\ 340]


And that leads us to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 34]


Meaning we started with 34 dimes, and it should follow directly from that fact that we started with 11 nickels accounting for the 55 cents remaining in the $3.95.


Checking our work:  If we have twice the dimes, or 68 dimes, we have $6.80 in dimes and an additional 21 nickels would be 55 cents plus $1.05 or $1.60.  Finally, $6.80 plus $1.60 is indeed $8.40.  Ta Da!


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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