Question 989611
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Gary has 42 coins in nickels, dimes, and quarters. 
If he has 8 more nickels than dimes and has $7.15 in all, how many of each does he have?
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            It is a typical problem to be solved using one single unknown and one single equation.


            See below on how to do it . . . 



<pre>
Let x be the number of nickels;

then the number od dimes is (x-8) and the number of quarters is  (42 - x - (x-8)) = (42+8 - 2x) = (50-2x).


At this point, you can write the total money equation


    5x + 10*(x-8) + 25*(50-2x) = 715  cents.


Simplify and solve


    5x + 10x - 80 + 1250 - 50x = 715

           -35x                = 715 + 80 - 1250

           -35x                = -455

              x                = {{{(-455)/(-35)}}} = 13.


<U>ANSWER</U>.  13 nickels;  13-8 = 5 dimes  and  the rest 42-13-5 = 24 quarters.


<U>CHECK</U>.   13*5 + 5*10 + 25*24 = 715 cents, in total.    ! Correct !
</pre>

Solved.