SOLUTION: Steve has only nickels and quarters. If he has 5 more quarters than nickels and the total value of all the coins is $5.15, find the number of coins of each type. Define the variabl

 Q = N + 5             (1)     (counting coins: "5 more quarters than nickels")
5N + 25Q = 515         (2)     (counting cents)


Next substitute (1) into eq(2). You will get

5N + 25*(N+5) = 515    (3)  ====>

5N + 25N + 125 = 515  ====>  30N = 515-125 = 390  ====>  N =  = 13.


Answer.  13 nickels and 13+5 = 18 quarters.


Check.   13*5 + 18*25 = 515 cents  ! Correct !

Let N be the number of nickels. Then the number of quarters is (N+5), according to the condition.

The "money" equation is

5N + 25*(N+5) = 515   cents    - exactly as the equation (3)  above.


The solution and the answer are the same as in the part 1.

For a minute, take away (mentally) 5 quarters from the collection.

Then you have equal number of nickels and quarters in the updated collection which is worth 515-5*25 = 390 cents now.


You can group the coins in the updated collection in the sets consisting of 1 nickel and one quarter each.


Each set is worth 25+5 = 30 cents and the number of sets is  = 13.


Then it is clear that the original collection counts 13 nickels and 13+5 = 18 quarters.