Question 521739: Please help me solve this equation: Natalie has some nickels, Dirk has some dimes, and Quincy has some quarters. Dirk has five more dimes than Quincy has quarters. If Natalie gives Dirk a nickel,Dirk gives Quincy a dime, and Quincy gives Natalie a quarter, they will all have the same amount of money. How many coins did each have originally?
Answer by mamiya(56)   (Show Source):
You can put this solution on YOUR website! 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.
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