Question 1187432
.
My piggy bank has 28 coins in it that are only nickels, dimes, and quarters. I have $3.25 total.
The number of quarters is one less than the number of nickels.
How many of each coin do I have?
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            My loving/preferable method is different.

            It uses only one unknown and only one single equation.



<pre>
Let  x  be the number of nickels.

Then the number of quarters is  (x-1) and the number of dimes is (28-x - (x-1)) = 29-2x.


Now write the total money equation 

    5x + 10*(29-2x) + 25*(x-1) = 325   cents.


Simplify and find x

    5x - 20x + 25x = 325 - 10*29 + 25

         10x       =       60

           x       =       60/10 = 6.


<U>ANSWER</U>.  6 nickels, 6-1 = 5 quarters and the rest coins, 28 - 6 - 5 = 17, are dimes.
</pre>

Solved &nbsp;(without using any systems of algebraic equations), &nbsp;at the level,

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