Data Sufficiency

Hallie has only nickels,dimes and quarters in her pocket.If she has atleast 1 of each kind of coin and has a total of $2.75 in change,how many nickels does she have?

(1)She has a total of 21 coins,with twice as many dimes as nickels.
(2)She has $1.50 in quarters.

From the question we have,

Let x,y,z be the number of Nickels,Dimes and Quarters respectively
N(x)+D(y)+Q(z)=2.75 => 0.05(x)+0.10(y)+0.25(z)=2.75-----------------------------Equation 1

From statement 1,we get 2 more equations
x+y+z=21--------------------------------------------------------------------Equation 2

x=2y-------------------------------------------------------------------------Equation 3

Thus we have 3 equations and 3 variables and hence it can be solved.

Now my concern is what if I use another method--

0.05(x)+0.10(y)+0.25(z)=2.75

Lets substitute x=2y we got from statement 2

0.05(2y)+0.10(y)+0.25(z)=2.75 => 0.20(y)+0.25(z)=2.75---------------------Equation A

We also have from statement 1 ,x+y+z=21 again x=2(y) so we have 3(y)+z=21------Equation B

Solving Equation A and B gives us answers in decimals y=4.54 and z=7.36

The number of coins simply cannot be decimals.Please explain.I understand that one doesnt need to solve,just checking sufficiency is enough.But still,I wanted to solve this problem without using the N-Equations,N-Variables method since there are some problems out there that require solving.



Thanks people.

Dan