SOLUTION: There is a total of 25 coins in Nickles, Dimes, and Quarters. There are four times as many dimes as quarters. The total vaule of the change is $2.05, how many of each coin is the

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Question 176676: There is a total of 25 coins in Nickles, Dimes, and Quarters. There are four times as many dimes as quarters. The total vaule of the change is $2.05, how many of each coin is there?
I have tried a few equations:
1) 5(x-5)+10(4x)+25(x+15)=205
2) 5(x-20)+10(4x-15)+25x=205
3) 5x+10(4x)+25x+=205
I know this can be solved with multiple variables, but I am suppose to solve it with one. Please help.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let n= number of nickels
Let d= number of dimes
Let q= number of quarters
given:
(1) n+%2B+d+%2B+q+=+25
(2) d+=+4q
In pennies:
(3) 5n+%2B+10d+%2B+25q+=+205
There are 3 equations and 3 unknowns, so it's solvable
Multiply (1) by 5 and subtract from (3)
5n+%2B+10d+%2B+25q+=+205
-5n+-+5d+-+5q+=+-125
--------------------------
5d+%2B+20q+=+80
And using (2),
5%2A4q+%2B+20q+=+80
40q+=+80
q+=+2
d+=+4q
d+=+8
(1) n+%2B+d+%2B+q+=+25
n+%2B+8+%2B+2+=+25
n+=+15
There are 15 nickels, 8 dimes and 2 quarters
check:
(3) 5n+%2B+10d+%2B+25q+=+205
5%2A15+%2B+10%2A8+%2B+25%2A2+=+205
75+%2B+80+%2B+50+=+205
205+=+205
OK