There are 31 nickles, dimes, and quarters in the drawer with a value of $4.70. How many coins of each type are there if there are twice as many quarters as there are dimes?

Question

phi · Answer

can you write down any equations?

anonymous · Answer

Well I thought it might be:
Nn + Nd + Nq = 31
5Nn + 10Nd + 25Nq = 470
Nd = 3Nq

But when I plug the Nd = 3Nq into the first two and then try to match them up, all I get is fractions.

radar · Answer

Let x = number of dimes then
2x = number of quarters  well then we could further say that since there is a total of 31 coins
31-x-2x = number of nickels

phi · Answer

radar will show an approach.  But your original equations look ok except for the last one.  there are twice as many quarters as dimes.  that means more quarters than dimes:
Nq= 2*Nd  (and not Nd= 3Nq)

radar · Answer

Now for the money.
10x +25(2x) + 5(31-x-2x)=470
10x+50X+155-5x-10x=470
45x=315
x=7  (7 dimes)
2x=14 (quarters)
31-7-14=10 (10  nickels)  That sums up to 31 coins.

radar · Answer

Follow the money
dimes 70 cents
quarters 3.50
nickels 50 cents
Totals to $4.70      The money check, the answers are correct.

radar · Answer

Your approach was appropriate.  However, personally I prefer to keep the number of variables (or unknowns) to a minimum.  In this case "one"  (x)

radar · Answer

as phi pointed out you Nd=3Nq was incorrect here is why.  The problem there are twice as many quarters as dimes.  This would be expressed using your variables as
Nq=2Nd

radar · Answer

Inretrospect, you actually kept the unknown limted to one (N)