Question 521739
let x be the number of nickels Natalie has, y the number of dimes Dirk has and z the number of quarters Quincy has.

from the data, we know the number of dimes Dirk has is 5 more than the number of quarters Quincy has, so y= z+5  ...........................(1)
 
 if Nathalie gives a nickel to Dirk and receives a quarter from Quincy, Nathalie's money will be 5(x-1) + 25

 if Dirk gives a dime to Quincy and receives a nickel from Nathalie, Dirk's money will be 10(y-1) + 5

 if Quincy gives a quarter to Nathalie and receives a dime from Dirk, Quincy's money will be 25(z-1)+10

the data said if that happens , they will all have the same amount of money, 
 so 5(x-1)+25 = 10(y-1)+5 = 25(z-1)+10  .....................(2)


from both equations , we got the following  system of equations
            y=z+5
            10(y-1)+5= 25(z-1)+10

        
         y-z = 5
         10y-5 = 25z - 15

        y-z = 5
        25z-10y = 10

       y-z = 5
      5z-2y = 2

by adding up (2) times the first equation and the second equation, we got
 2y - 2z + 5z - 2y = 2*5 + 2
  3z= 12,  so z=4

  y = z+5= 4+5 = 9
 
from the equation (2), we have 5(x-1)+25= 10(y-1)+5 = 10y - 5
so, 5x+20= 10y - 5
     x+4= 2y - 1
    x= 2y - 1 -4 
     = 2*9- 5
     =13
  
     so, originally, Nathalie had 13 coins, Dirk 9 coins and Quincy 4 coins.


to check: let n be the among of Nathalie's money, d Dirk's and q Quincy's
n= 13 *5= 65 cents         d= 9*10=90 cents      q=4*25=100 cents
if Nathalie gives a 5 cents to dirk and receives 25 cents from Quincy, she will have 85 cents.
if Dirk gives 10 cents to Quincy and receives 5 cents from Nathalie, he will have 85 cents.
if Quincy gives a quarter to Nathalie and receives a dime from Dirk, He will have 85 cents.