SOLUTION: A vending​ machine's coin box contains​ nickels, dimes, and quarters. The total number of coins in the box is 296. The number of dimes is three times the number of nick

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Question 1061610: A vending​ machine's coin box contains​ nickels, dimes, and quarters. The total number of coins in the box is 296. The number of dimes is three times the number of nickels and quarters together. If the box contains 29 dollars and 30​cents, find the number of​ nickels, dimes and quarters that it contains.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
VARIABLES FOR HOW MANY OF EACH COIN
n    nickels
d    dimes
q    quarters


ACCOUNT FOR TOTAL COINS


ACCOUNT FOR TOTAL MONEY


FORM ANY OTHER DESCRIBED RELATIONSHIPS OF THE VARIABLES
The problem description describes one more equation.
The number of dimes is three times the number of nickels and quarters together.
What is the equation for that?

WRITE AND SOLVE THE SYSTEM
Can you do this?
Can you solve for the three variables?


--
Answer should be this solution:
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
A vending machine's coin box contains nickels, dimes, and quarters. The total number of coins in the box is 296.
The number of dimes is three times the number of nickels and quarters together. If the box contains 29 dollars and 30 cents,
find the number of nickels, dimes and quarters that it contains.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

Let "N" be the number of nickels and "Q" be the number of quarters.

Then the number of dimes is 3(N+Q).


The nickels  contribute   5N       cents to the total.
The quarters contribute  25Q       cents to the total.
The dimes    contribute  10*3(N+Q) cents to the total.    

Thus your equations are

N + Q + 3(N+Q) = 296,        (1)    ( for the total number of coins )
5N + 25Q + 30(N+Q) = 2930.   (2)    ( the "value" equation )

Simplify by canceling by 4 both sides in (1) and canceling by 5 both sides in (2). You will get

 N +   Q =  74,              (1')
7N + 11Q = 586.              (2')

Express N = 74-Q from (1') and then substitute it into (2') replacing N:

7(75-Q) + 11Q = 586.

Now you have a single equation for one unknown Q.

At this point I leave it to you to complete the solution on your own.



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