The equations you need are: 

Substitute 3Q for D in eqs (i) & (ii) to get a system in 2 variables (N & Q). Solve that system for N & Q, the number

of nickels and quarters, respectively. Knowing Q, you will then be able to find D, the number of dimes [eq (iii)]. 

Answer by ikleyn(52867)   (Show Source): You can put this solution on YOUR website!
 .

How do you set up this problem for solving? Steve is cashing in his jar of spare nickels dimes and quarters. 

When he gets to the bank he receives a total of $14.70. 

He learned he had 133 coins in all, and that there were 3 times as many dimes as quarters. 

How many of each type of each type of coin did he save?

~~~~~~~~~~~~~~~~~~~~~





Let n = # of nickels, d = # of dimes, q = # of quarters.

Then immediately from the condition you have these system of three equations for three unknowns:

 n +   d +   q =  133,       (1)      ("he had 133 coins in all")
5n + 10d + 25q = 1470,       (2)      ("When he gets to the bank he receives a total of $14.70")
d = 3q.                      (3)

Next, we reduce this 3x3 system to the more simple 2x2 system of two equations for two unknowns:

For it, I substitute (replace) "d" in equations (1) and (2) by 3q based on equation (3). I will get

 n +     3q  +   q =  133,   (4)
5n + 10*(3q) + 25q = 1470.   (5)

or

 n +  4q =  133,             (6)
5n + 55q = 1470.             (7)

OK. So, you have now much simpler system (6), (7). To solve it, express n = 133-4q  from (6) and substitute it into (7). You will get

5*(133-4q) + 55q = 1470,  or

665 - 20q + 55q = 1470  --->  35q = 1470 - 665  --->  35q = 805  --->  q =  = 23.

So we just found the number of quarters. It is 23. There were 23 quarters.

Then the number of dimes is trice of it: there were 3*23 = 69 dimes.

Now the number of nickels is simply 133 - 69 - 23 = 41.

Check.  5*41 + 10*69 + 23*25 = 1470.  Correct!

Answer.  41 nickels, 69 dimes and 23 quarters.






Please ignore what the other tutor, "josgarithmetic" wrote in his post.

He is not able to solve word problems correctly, as well to explain the solutions.

I don't know what he is doing in this site.





Regarding the coin problems, see the lessons



    - Coin problems



    - More Coin problems



    - Solving coin problems without using equations



    - Kevin and Randy Muise have a jar containing coins



    - Typical coin problems from the archive

in this site.





Also, you have this free of charge online textbook in ALGEBRA-I in this site



    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.





 

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