SOLUTION: Zachary had nickels, dimes, and quarters. He has 58 coin in total, value is 6.50. There are twice as many dimes as nickels. How many of each does he have.

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Question 1080827: Zachary had nickels, dimes, and quarters. He has 58 coin in total, value is 6.50. There are twice as many dimes as nickels. How many of each does he have.
Answer by ikleyn(52797)   (Show Source): You can put this solution on YOUR website!
.
Let N be the number of nickels and Q be the number of quarters.
Then the number of dimes is 2N.

Nickels contribute  5N             cents to the total.
Dimes contribute   10*(2N) = 20N   cents to the total.
Quarter contribute  25Q            cents.

Hence, the "value" equation is 

5N + 20N + 25Q = 650   cents,   or

25N + 25Q = 650,    or

N + Q = 26,

The "coins" equation is 

N + 2N + 5Q = 58,   or

3N + 5Q = 58.           


So, you have this system

 N +  Q = 26,     (1)
3N + 5Q = 58.     (2)


The setup is done.
Solve the system by any method you know (substitution, elimination).



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